systems of non

7/27/2019 Systems of Non

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When you are solving a system of equations, you are looking for the points that are solutions for allof thesystem's equations. In other words, you are looking for the points that are solutions forallof the equationsat once. What does this mean...?

Suppose you have the following:

Solve the system by graphing:

y= x2

y= 8x2

I can graph each of these equationsseparately:

y =x2

y = 8x2

..and each point on each graph is a solution to that graph's equation .


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Now look at the graph of the system:

y =x

2

y = 8x2

A solution to the system is any point that is a solution forboth equations. In other words, asolution point for this system is any point that is on both graphs. In other words:

"SOLUTIONS" FOR SYSTEMS ARE INTERSECTIONS OF THE LINES

Then, graphically, the solutions for thissystem are the red-highlighted pointsat right:

That is, the solutions to this system are the points (2, 4) and (2, 4).

So when you're trying to solve a system of equations, you're trying to find the coordinates of theintersection points. Copyright 2002-2011 Elizabeth Stapel All Rights Reserved

The system shown abovehas two solutions, becausethe graph shows twointersection points. A system


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can have one solution:

...lots of solutions:

...or no solutions at all:

(In this last situation, where there was no solution, the system of equations is said to be "inconsistent".)

When you look at a graph, you can onlyguess at an approximation to the solution.Unless the solutions points are nice neatnumbers (and unless you happen to knowthis in advance), you can't get the solutionfrom the picture. For instance, you can't tellwhat the solution to the system graphed atright might be:
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...because you're having to guess from a picture. As it happens, the solution is (x, y) = (13/7,

9/14), but

you would have no possible way of knowing that from this picture.

Advisory: Your text will almost certainly have you do some "solve by graphing" exercises. You may safelyassume for these exercises that answers are nice and neat, because the solutions mustbe if you are to

be able to have a chance at guessing the solutions from a picture.

This "solving by graphing" can be useful, in that it helps you get an idea in picture form of what is going onwhen solving systems. But it can be misleading, too, in that it implies that all solutions will be "neat" ones,when most solutions are actually rather messy.

To find the exactsolution to a system of equations, you must use algebra. Let's look at thatfirst systemagain:

Solve the following system algebraically:

y= x2

y= 8

x

2

Since I am looking for the intersection points, I am therefore looking for the points where theequations overlap, where they share the same values. That is, I am trying to find any spots where

y =x2

equalsy = 8x2:

y =x2

=y = 8x2

The algebra comes in when I manipulate useful bits of this last equation. I can pick out whicheverparts I like. (They're all equal, after all -- at least at the intersection points, but the intersectionpoints are the only points that I care about anyway!) So I can pick out any of the following:

y =x2

y = 8x2

y =y

x2

= 8x2

Each of these sub-equations is true, but only the last one is usefully new and different:

x2

= 8x2
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I cansolve thisfor thex-values that make the equation true:

x2

= 8x2

2x2

= 8

x2

= 4

x =

2, +2

Then the solutions to the original system will occur when x =2 and whenx = +2.

What are the correspondingy-values? To find them, I plug thex-values back in to eitherof thetwo original equations. (It doesn't matter which one I pick because I only care about the pointswhere the equations spit out the same values. So I can pick whichever equation I like better.) I'll

plug thex-values into the first equation, because it's the simpler of the two:

x =2:

y =x2

y = (

2)2 = 4

x = +2:

y =x2

y = (+2)2

= 4

Then the solutions (as we already knew) are (x, y) = (2, 4) and (2, 4).

In this case, the solutions were "neat" values; no fractions or decimals. But solutions will not always beneat, so, while the pictures can be very useful for giving you a "feel" for what is going on, graphing is not

as accurate as doing the algebra. Warning: Students are often taught nowadays to "round" absolutelyeverything, and are thus implicitly taught that all answers will be "neat" answers. But this is wrong; don'tfall for it. For instance:

Solve the following system:

y= x2

+ 3x+ 2

y= 2x+ 3

I can solve this in the same manner as we did on the previous problem. The "solution" to the

system will be any point(s) that the lines share; that is, any point(s) where the x-value and

correspondingy-value fory =x2

+ 3x + 2 is the same as thex-value and correspondingy-value

fory = 2x + 3; that is, where the lines overlap or intersect; that is, wherey =x2 + 3x + 2 equalsy = 2x + 3. Copyright 2002-2011 Elizabeth Stapel All Rights Reserved
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Looking at the graph of the system:

...I can see that there appear to be solutions at around (x, y) = (1.5,0.25) and (x, y) =

(0.5, 4.25). But I cannot assume that this is the answer! The picture can give me a goodidea, but only the algebra can give me the actual answer.

I'll set the equations equal, and solve:2

+ 3x + 2 = 2x + 32

+x1 = 0

Using theQuadratic Formulagives me:

Then I have one solution:

...which has a corresponding y-value of:

The other solution (from the "" in front of thesquare root) is:

....which gives me ay-value of:
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So the solutions are:

For purposes of graphing, the approximate solutions are:

(x, y) = (1.62,0.24) and (0.62, 4.24).

In other words, while our guess from the picture was close, it was not entirely correct. (However, if thealgebra had given me answers that are far afield of these picture-based guesses, I would have been ableto safely assume that I had messed up the math somewhere. In this way, the graph can be helpful forchecking your work.)


y= 2x2

+ 3x+ 4

y= x2

+ 2x+ 3

As before, I'll set these equations

equal, and solve for the values ofx:

2x + 3x + 4 =x + 2x + 32

+x + 1 = 0

Using theQuadratic Formula:

But I can't graph that negative insidethe square root! What's going on here?

Take a look at the graph:


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The lines do not intersect. Since there is no intersection, then there is no solution. That is, this isan inconsistent system. My final answer is: no solution: inconsistent system.

In general, themethod of solutionfor general systems of equations is to solve one of the equations (youchoose which) for one of the variables (again, you choose which). Then you plug the resulting expressioninto the otherequation for the chosen variable, and solve for the values of the othervariable. Then youplug those solutions back into the first equation, and solve for the values of the first variable. Here aresome additional examples: Copyright 2002-2011 Elizabeth Stapel All Rights Reserved


y=x3

x2

+ y2

= 17

Graphically, this system is a straight line crossing a circle centered at the origin:
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There appear to be two solutions. I'll proceed algebraically to confirm this impression, and to getthe exact values.

Since the first equation is already solved fory, I will plug "x3" in for "y" in the second

equation, and solve for the values ofx:

x2

+y2

= 17

x2

+ (x3)2

= 17

x2

+ (x3)(x3) = 17x

2+ (x

2+ 6x + 9) = 17

2x2

+ 6x + 9 = 17

2x2 + 6x

8 = 0x

2+ 3x4 = 0

(x + 4)(x1) = 0

x =4,x = 1

Whenx =4, y =x3 =(4)3 = 43 = 1

Whenx = 1, y =x3 =(1)3 =4

Then the solution consists of the points (4, 1) and (1,4).

Note the procedure: I solved one of the equations (the first equation looked easier) for one of thevariables (solving for"y=" looked easier), and then plugged the resulting expression back into the other

equation. This gave me one equation in one variable (the variable happened to be x), and a one-variableequation is something I know how to solve. Once I had the solution values forx, I back-solved for the

correspondingy-values. I emphasize "corresponding" because you have to keep track of which y-value

goes with whichx-value. In the example above, the points (4,4) and (1, 1) are notsolutions. Eventhough I came up withx =4 and 1 andy =4 and 1, thex =4 did not go with they =4, and thex =

1 did not go with the y = 1. Warning: You must match thex-values andy-values correctly!


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Solve the following system of equations:

y= (1/2)x5

y= x2

+ 2x15

Since both equations are already solved fory

, I'll set them equal and solve for the values ofx

:

(1/2)x5 =x

2+ 2x15

x10 = 2x2

+ 4x30

0 = 2x2

+ 3x200 = (2x5)(x + 4)

x =5/2,x =4

Whenx =5/2:

y = (

1

/2)x

5 = (

1

/2)(

5

/2)

5 =

5

/4

20

/4 =

15

/4 =

3.75

Whenx =4:

y = (1/2)x5 = (

1/2)(4)5 =25 =7

Then the solutions are the points (5/2,

15/4 ) and (4, 7).

Graphically, the above system looks like this:

Solve the following system of equations:

xy= 1

x+ y= 2

Taking a quick look at the graph, I see that there appears to be only one solution:


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I guess I'll solve the second equation fory, and plug the result into the first equation:

x +y = 2

y =x + 2

Then:

xy = 1

x(x + 2) = 1

x2

+ 2x = 1x

2+ 2x1 = 0

x22x + 1 = 0

(x1)(x1) = 0

x = 1

Then:

x +y = 2

(1) +y = 2

y = 1

Then the solution is the point(1, 1).

Solve the system of nonlinear equations:

y= x2

x

2

+ (y

2)

2

= 4

From the form of the equations, you should know that this system contains a parabola and acircle.


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According to the graph, there should be three solutionsto this system:

The solution at the origin is a "neat" one, but the othertwo intersection points may be messy.

From the first equation, I think I'll plug in "y" for "x2" in the second equation, and solve:

x2

+ (y2)2

= 4y + (y2)

2= 4

y + (y2

4y + 4) = 4

y23y = 0

y(y3) = 0

y = 0, y = 3

Now I need to find the correspondingx-values. Wheny = 0:

y =x2

0 =x2

0 =x

(This is the solution at the origin that we'd been expecting.) Wheny = 3:

y =x2

3 =x2

sqrt(3) =x

Then the solutions are the points , (0, 0), and .


3x2

+ 2y2

= 354x

23y

2= 24

I can rearrange the first equation to get:


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This tells me that the first equation is an ellipse. However, rather than graphing this using ellipseformulae, you could also solve to get a "plus-minus" expression that you can graph as twoequations: Copyright 2002-2011 Elizabeth Stapel All Rights Reserved

(You would plug this into your graphing calculator as two graphs, one graph for the top "plus" partof the ellipse, and another for the bottom "minus" part.) The second equation rearranges as:

...which is an hyperbola. The second equation also solves (for your graphing calculator) as:

Whatever format you use (the ellipse and the hyperbola center-vertex forms, or the "plus-minus"for-calculator forms), this system graphs as:

As you can see, there appear to be four solutions. To find them algebraically, I will choose to

solve the second equation forx2

(rather than justx), and plug the resulting expression into the

first equation, which I will then solve fory:. (It's okay that I "only" solve forx2, because neither

equation has anx-term. There is no need, in this particular case, to do any more solving.)

4x23y

2= 24

4x2

= 3y2

+ 24x

2= (

3/4 )y

2+ 6

Then, subsituting into the first equation for thex2, I get:


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3x2

+ 2y2

= 35

3((3/4 )y

2+ 6) + 2y

2= 35

(9/4 )y

2+ 18 + 2y

2= 35

9y2

+ 72 + 8y2

= 14017y

2= 68

y

2

= 4y = 2

When y =2:

x2

= (3/4 )y

2+ 6

= (3/4 )(2)

2+ 6

= (3/4 )(4) + 6

= 3 + 6 = 9

x = 3

When y = 2:

x2

= (3/4 )y

2+ 6

= (3/4 )(2)

2+ 6

= (3/4 )(4) + 6

= 3 + 6 = 9

x = 3

Then the solution is the points (3,2), (3, 2), (3,2), and (3, 2), which may alse be written

as ( 3, 2), since all the "plus-minus" combinations are included.

The example below demonstrates how theQuadratic Formulais sometimes used to help in solving, andshows how involved your computations might get.

Solve the system:

x2xy+ y

2= 21

x2

+ 2xy8y2

= 0

This system represents an ellipse and a set of straight lines. If you solve each equation above for

y, you can enter the "plus-minus" equations into your graphing calculator to verify this:

x2xy +y

2= 21


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y2xy + (x

221) = 0

x2

+ 2xy8y2

= 0

0 = 8y22xyx

2

As you can see, I used theQuadratic Formulain each case to solve fory in terms ofx. This gave

me equations that I could graph. This technique doesn't come up that often, but it can be a life-saver when you can't seem to solve things any other way.

Where did the absolute-value bars come from? Recall that,technically, the square root ofx2

is

the absolute value ofx. That's how I did that simplification in the next-to-last line above.

The absolute value ofx in the second equation above gives two cases for the values ofy:

Ifx < 0, then |x | =x, soy =(x 3x)

/8 =x/2,

x/4

Ifx > 0, then |x | =x, soy =(x 3x)

/8 =x/4,

x/2.

In either case, y =x/4 ory =

x/2.

Since I derived these "y=" solution-equations from the second of the original equations, I willplug them into the first equation to solve for some actual numerical values:

Ify =x/4: Copyright 2002-2011 Elizabeth Stapel All Rights Reserved
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Then, plugging into the "y=" solution-equations above, I get:

Ify =x

/2:

Then:

Then the four solutions are:

, , ,


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Warning: Do not try to write the solution points as " " or " ", because this is not

correct. Not all combinations of thesex-value andy-values are solution points. Don't be sloppy; write thesolution out correctly.

By the way, the graph of the system looks like this:

(To graph the ellipse using the traditional methods, you would have to do a "rotation of axes", a processyou probably won't see until calculus, if at all.)

There is another way of proceeding in the above exercise, because the second equation happens to befactorable. (This factorability is NOT generally true, but you should try to remember to check, just in case.)

If you factor the second equation and solve forx in terms ofy, you get:

x2

+ 2xy8y2

= 0(x + 4y)(x2y) = 0x + 4y = 0 or x2y = 0x =4y or x = 2y

Plug these into the first equation fory, and solve for thex-values.

This last example (the first way I worked it) is about as complicated as these things ever get. But if you

plug away and work neatly and completely, you should be able to arrive at the solution successfully. Andif you have a graphing calculator (and the time), try doing a quick graph to verify your answers visually.

Special Facto r ing :

Facto r ing Differences of Squares(page 1 of 3)

Sections: Differences of squares,Sums and differences of cubes,Recognizing patterns
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When you learn tofactor quadratics, there are three other formulas that they usually introduce at thesame time. The first is the "difference of squares" formula.

Remember from yourtranslationskills that "difference" means "subtraction". So a difference of squares is

something that looks likex2

4. That's because 4 = 22, so you really havex2

22, a difference ofsquares. To factor this, do your parentheses, same as usual:

x24 = (x )(x )

You need factors of4 that add up to zero, so use 2 and +2:

x24 = (x2)(x + 2)

(ReviewFactoring Quadratics, if this example didn't make sense to you.)

Note that we hadx22

2, and ended up with (x2)(x + 2). Differences of squares (something squared

minus something else squared) always work this way:

Fora2b

2, do the parentheses:

( )( )

...put the first squared thing in front:

(a )(a )

...put the second squared thing in back:

(a b)(a b)

...and alternate the signs in the middles:

(ab)(a + b)

Memorize this formula! It will come in handy later, especially when you get to rational expressions(polynomial fractions), and you'll probably be expected to know the formula for your next test.

Here are examples of some typical homework problems:

Factorx216

This isx24

2, so I get:
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x216 =x

24

2= (x4)(x+ 4)

Factor4x225

This is (2x)25

2, so I get:

4x225 = (2x)

25

2= (2x5)(2x+ 5)

Factor9x6y

8

This is (3x3)2(y

4)2, so I get:

9x6y

8= (3x

3)2(y

4)2

= (3x3y

4)(3x

3+ y

4)

Factorx41Copyright Elizabeth Stapel 2000-2011 All Rights Reserved

This is (x

2

)

2

1

2

, so I get:

x41 = (x

2)21

2= (x

21)(x

2+ 1)

Note that I'm not done yet, becausex21 is itself a difference of squares, so I need to apply the

formula again to get the fully-factored form. Sincex21 = (x1)(x + 1), then:

x41 = (x

2)21

2= (x

21)(x

2+ 1)

= ((x)2(1)

2)(x

2+ 1)

= (x

1)(x+ 1)(x

2

+ 1)

The answer to this last exercise depended on the fact that 1, to any power at all, is still just 1.

Warning: Never forget that this formula is for the difference of squares; the sum of squares is alwaysprime (that is, it can't be factored).

The other two special factoring formulas are two sides of the same coin: the sum and difference of cubes.These are the formulas:

a3

+ b3

= (a + b)(a2ab + b

2)

a3b

3= (ab)(a

2+ ab + b

2)

You'll learn in more advanced classes how they came up with these formulas. For now, just memorizethem. First, notice that the terms in each factorization are the same; then notice that each formula has

only one "minus" sign. For the difference of cubes, the "minus" sign goes with the linear factor, ab; for

the sum of cubes, the "minus" sign goes in the quadratic factor, a2ab + b

2. Some people use

the mnemonic "SOAP" for the signs; the letters stand for "same" as the sign in the middle of the originalexpression, "opposite" sign, and "always positive".

a3

b3

= (a[same sign]b)(a2[opposite sign]ab[always positive]b

2)


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Whatever method helps you best keep these formulas straight, do it, because you should not assume thatyou'll be given these formulas on the test. You really should know them. Note: The quadratic part of eachcube formula does not factor, so don't attempt it.

When you have a pair of cubes, carefully apply the appropriate rule. By "carefully", I mean "usingparentheses to keep track of everything, especially the negative signs". Here are some typical problems:Copyright Elizabeth Stapel 2000-2011 All Rights Reserved

Factorx38

This isx32

3, so I get:

x38 =x

32

3

= (x2)(x2

+ 2x + 22)

= (x 2)(x2 + 2x+ 4)

Factor27x3

+ 1

Remember that 1 can be regarded as having been raised to any power you like, so this is really

(3x)3

+ 13. Then I get:

27x3

+ 1 = (3x)3

+ 13

= (3x + 1)((3x)2(3x)(1) + 1

2)

= (3x+ 1)(9x2 3x+ 1)

Factorx3y

664

This is (xy2)34

3, so I get:

x3y

664 = (xy

2)34

3= (xy

24)((xy

2)2

+ (xy2)(4) + 4

2)

= (xy2 4)(x

2y

4+ 4xy

2+ 16)

You can use the Mathway widget below to practice "Factoring Polynomials", subtopic "Sum of Cubes".Try the entered exercise, type in your own exercise, or select subtopic "Difference of Cubes". Then click"Answer" to compare your answer to Mathway's. (Or skip the widget andcontinuewith the lesson.)

(Clicking on "View Steps" on the widget's answer screen will take you to the Mathway site, where you can

register for a free seven-day trial of the software.)

There is one "special" factoring that can actually be done using theusual methodsfor factoring, but, forwhatever reason, many texts and instructors make a big deal of treating this case separately. "Perfect

square trinomials" are quadratics that you got by squaring a binomial. For instance, (x + 3)2

= (x + 3)(x

+ 3) =x2

+ 6x + 9 is a perfect square trinomial.
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Recognizing the pattern to perfect squares isn't a make-or-break issue, but it can be a time-saveroccasionally. The trick is really quite simple: If the first and third terms are squares, figure out what they're

squares of. Multiply those things, multiply that product by 2, and compare your result with thequadratic's middle term. If you've got a match, then you've got a perfect square.

Is x2

+ 10x+ 25 a perfect square trinomial? If so, write the trinomial as the square of a

binomial.

Well, the first term,x2, is the square ofx. The third term, 25, is the square of5. Multiplying, I get

5x. Multiplying this by 2, I get 10x. And this matches the middle term. So:

this quadratic is a perfect square: x2

+ 10x+ 25 = (x+ 5)2

Write 16x248x+ 36 as a squared binomial.

The first term, 16x2, is the square of4x, and the last term, 36, is the square of6. Actually, since

the middle term has a "minus" sign, the 36 is the square of6. Just to be sure, I'll make sure that

the middle term matches the pattern: (4x)(

6)(2) =

48x. It's a match, so this is a perfectsquare:

16x248x + 36 = (4x6)

2

Is 4x225x+ 36 a perfect square trinomial?

The first term, 4x2, is the square of2x, and the last term, 36, is the square of6 (or, in this case,

6, if this is a perfect square). Checking the middle term, I get (2x)(6)(2) =24x, which does notmatch the middle term. So:

this is no ta perfect square trinomial.

That's all there is to perfect squares.

You've learned the difference-of-squares formula and the difference- and sum-of-cubes formulas. Buthow do you know which formula to use, and when to use it?

First off, to use any of these formulas, you have to have only two terms in your polynomial. If you'vefactored out everything you can and you're still left with two terms with a square or a cube in them, then

you should look at using one of these formulas. For instance, 6x2

+ 6x is two terms, but you can factorout a 6x, giving you 6x

2+ 6x = 6x(x + 1). Since the bit inside the parentheses does not have a squared

or a cubed variable in it, you can't apply any of these special factoring formulas (and you don't need to,

since it's already fully factored -- you can't go further than just plain old "x"!).

On the other hand, 2x2162 = 2(x

281), andx

281 is a quadratic. When you see that you have a

two-term non-linear polynomial, check to see if it fits any of the formulas. In this case, you've got a

difference of squares, so apply that formula: 2x2162 = 2(x

281) = 2(x9)(x + 9).

Warning: Always remember that, in cases like 2x2

+ 162, all you can do is factor out the 2; the sum of

squares doesn't factor! 2x2

+ 162 = 2(x2

+ 81). (Your book may callx2

+ 81 "prime", "unfactorable", or"irreducible". These all mean the same thing.)


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There is one special case for applying these formulas. Take a look at x664. Is this expression a

difference of squares, being ( (x3)28

2), or a difference of cubes, being ( (x

2)34

3)? Actually, it's both.

You can factor this difference in either of two ways:

difference of squares, followed by difference of cubes:

x664 = (x

3)28

2= (x

38)(x

3+ 8)

= (x32

3)(x

3+ 2

3)

= (x2)(x2

+ 2x + 4)(x + 2)(x22x + 4)

difference of cubes, followed by difference of squares:

x664 = (x

2)34

3= (x

24)((x

2)

2+ 4x

2+ 4

2)

= (x2

4)(x4 + 4x2 + 16)= (x2)(x + 2)(x

4+ 4x

2+ 16)

You shouldget full credit for either answer, since you shouldn't be expected to know (or to guess) that the

quartic polynomialx4

+ 4x2

+ 16 factors as (x2

+ 2x + 4)(x22x + 4). But if you happen to notice that a

problem could be worked either way (as a difference of squares or as a difference of cubes), then youcan see from the above example that it might be best to apply the difference-of-squares formula first,because doing the squares first means that you'll get all four factors, not just three.

Since the hardest part of factoring usually comes in figuring out how to proceed with a given problem,

below are some factoring examples, with an explanation of which way you need to go with it to arrive atthe answer. Note: Not all solutions are provided.

Factorx2

+ 11x+ 18

This polynomial has three terms. Try to factor the "usual" way. Find factors of 18 that add up to11, and then fill in the parentheses:Copyright Elizabeth Stapel 2000-2011 All Rights Reserved

x2

+ 11x + 18 = (x+ 2)(x+ 9)

Factor16x249

This has two terms, and nothing factors out of both terms, so you have to be thinking "difference

of squares, or sum or difference of cubes". Since there are no cubes (and especially since the x

is squared), you should look for a difference of squares. Sixteen is a square, and so is 49, soapply the difference of squares formula to (4x)

27

2.

Factor3x312x


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You can factor out a 3x, giving you 3x(x24). This leaves two terms inside the parentheses,

where the two terms have a subtraction in the middle, and the x is squared. Apply the difference

of squares formula tox22

2, and don't forget the factor of "3x" when you write your final answer.

Factorx2

+ 6x+ 9

This is a quadratic with three terms. Factor in the "usual" way:

x2

+ 6x + 9 = (x+ 3)(x+ 3)

You might also have noticed that this is a perfect square trinomial, from the fact thatx2

is the square ofx,

9 is the square of3, and (x)(3)(2) = 6x, which matches the middle term. So this answer could also havebeen written as (x + 3)

2.

Factor27x38

This has two terms, and nothing comes out of both. It's a difference. Twenty-seven is a cube, and

so is 8. Apply the difference of cubes formula to (3x)3

23.

Factor7x756x

This has two terms, and a 7x comes out of both, giving you 7x(x68). Inside the parentheses

you still have two terms, and it's a difference. The first term,x6, could be a cube, (x

2)3, or a

square, (x3)2, but 8 can only be a cube, 2

3. Apply the difference of cubes formula to (x

2)32

3.

Factorx9

+ 1

This polynomial has two terms, and nothing factors out of both. Remember that you can put any

power you feel like on 1, so you just have to figure out what to do with the x9. Since this is a sum,not a difference, you have to hope that there is some way you can turn x

9into a cube. There is:

apply the sum of cubes formula to (x3)3

+ 13

to get (x3

+ 1)(x6x

3+ 1). Then notice that you

can apply the sum of cubes formula again, to the factorx3

+ 1.

Factor(x+ y)3

+ (xy)3

Yes, this is needlessly complex, but you might see something like this in an extra-creditassignment. This is just a big lumpy sum of cubes. Be very careful with your parentheses whenapplying the formula. As you can imagine, there are many opportunities for mistakes!

(x +y)

3

+ (x

y)

3

= [ (x +y) + (xy) ] [ (x +y)2(x +y)(xy) + (xy)2 ]= [ 2x ] [ (x

2+ 2xy +y

2)(x

2y

2) + (x

22xy +y

2) ]

= [ 2x ] [x2

+ 2xy +y2x

2+y

2+x

22xy +y

2]

= [ 2x] [ x2

+ 3y2

]

To successfully complete these problems, just take your time, and don't be afraid to rely on your instinctsand common sense.


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Synthetic division is a shorthand, or shortcut, method ofpolynomial divisionin the special case of dividingby a linear factor -- and it onlyworks in this case. Synthetic division is generally used, however, not fordividing out factors but for finding zeroes (or roots) of polynomials. More about this later.

If you are given, say, the polynomial equationy =x2

+ 5x + 6, you can factor the polynomial asy = (x +3)(x + 2). Then you can find the zeroes ofy by setting each factor equal to zero and solving. You will find

thatx =

2 andx =

3 are the two zeroes ofy.

You can, however, also work backwards from the zeroes to find the originating polynomial. For instance,

if you are given thatx =2 andx =3 are the zeroes of a quadratic, then you know thatx + 2 = 0, sox

+ 2 is a factor, andx + 3 = 0, sox + 3 is a factor. Therefore, you know that the quadratic must be of theformy = a(x + 3)(x + 2).

(The extra number "a" in that last sentence is in there because, when you areworking backwardsfrom

the zeroes, you don't know toward which quadratic you're working. For any non-zero value of "a", yourquadratic will still have the same zeroes. But the issue of the value of " a" is just a technical consideration;as long as you see the relationship between the zeroes and the factors, that's all you really need to know

for this lesson.)

Anyway, the above is a long-winded way of saying that, ifxn is a factor, thenx = nis a zero, and if

x = n is a zero, thenxn is a factor. And this is the fact you use when you do synthetic division.

Let's look again at the quadratic from above:y =x2

+ 5x + 6. From theRational Roots Test, you know

that 1, 2, 3, and 6 are possible zeroes of the quadratic. (And, from the factoring above, you know that

the zeroes are, in fact,3 and2.) How would you use synthetic division to check the potential zeroes?Well, think about howlong polynomial divisonworks. If we guess thatx = 1 is a zero, then this means

thatx1 is a factor of the quadratic. And if it's a factor, then it will divide out evenly; that is, if we divide

x2

+ 5x + 6 byx1, we would get a zero remainder. Let's check:

As expected (since we know thatx1 is not a factor), we got a non-zero remainder. What does this looklike in synthetic division? Copyright Elizabeth Stapel 2002-2011 All Rights Reserved

First, write the coefficients ONLY inside anupside-down division symbol:

Make sure you leave room inside, underneath the row ofcoefficients, to write another row of numbers later.
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Put the test zero,x = 1, at the left:

Take the first number inside, representing the

leading coefficient, and carry it down,unchanged, to below the division symbol:

Multiply this carry-down value by the testzero, and carry the result up into the nextcolumn:

Add down the column:

Multiply the previous carry-down value by thetest zero, and carry the new result up into thelast column:

Add down the column:

This last carry-down value is the remainder.

Comparing, you can see that we got the same result from the synthetic division, the same quotient(namely, 1x + 6) and the same remainder at the end (namely, 12), as when we did the long division:

The results are formatted differently, but you should recognize that each format provided us with the

result, being a quotient ofx + 6, and a remainder of12.

You already know (from the factoring above) thatx

+ 3 is a factor of the polynomial, and therefore that=3 is a zero. Now compare the results of long

division and synthetic division when we use the

factorx + 3 (for the long division) and the zerox =


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3 (for the synthetic division):

As you can see above, while the results are formatted differently, the results are otherwise the same:

In the long division, I divided by the factorx + 3, and arrived at the result ofx + 2 with a remainder of

zero. This means thatx + 3 is a factor, and thatx + 2 is left after factoring out thex + 3. Setting the

factors equal to zero, I get thatx =3 andx =2 are the zeroes of the quadratic.

In the synthetic division, I divided byx =3, and arrived at the same result ofx + 2 with a remainder of

zero. Because the remainder is zero, this means that x + 3 is a factor andx =3 is a zero. Also,

because of the zero remainder,x + 2 is the remaining factor after division. Setting this equal to zero, I getthatx =2 is the other zero of the quadratic.

I will return to this relationship between factors and zeroes throughout what follows; the two topics areinextricably intertwined.

Polynom ial Graphs: End Behavior(page 1 of 5)

Sections: End behavior,Zeroes and their multiplicities,"Flexing",Turnings & "bumps",Graphing

When you're graphing (or looking at a graph of) polynomials, it can help to already have an idea of whatbasic polynomial shapes look like. One of the aspects of this is "end behavior", and it's pretty easy. Lookat these graphs:

with a positiveleading coefficient

with a negativeleading coefficient
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Copyright Elizabeth Stapel 2005-2011 All Rights Reserved

with a positiveleading coefficient

with a negativeleading coefficient

As you can see, even-degree polynomials are either "up" on both ends (entering and then leaving thegraphing "box" through the "top") or "down" on both ends (entering and then leaving through the"bottom"), depending on whether the polynomial has, respectively, a positive or negative leadingcoefficient. On the other hand, odd-degree polynomials have ends that head off in opposite directions. Ifthey start "down" (entering the graphing "box" through the "bottom") and go "up" (leaving the graphing"box" through the "top"), they're positive polynomials; if they start "up" and go "down", they're negativepolynomials.


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All even-degree polynomials behave, on their ends, like quadratics, and all odd-degree polynomialsbehave, on their ends, like cubics.

Which of the following could be the graph of a polynomial

whose leading term is "3x4"?

The important things to consider are the sign and the degree of the leading term. The exponent

says that this is a degree-4 polynomial, so the graph will behave roughly like a quadratic: up onboth ends or down on both ends. Since the sign on the leading coefficient is negative, the graph

will be down on both ends. (The actual value of the negative coefficient, 3 in this case, isactually irrelevant for this problem. All I need is the "minus" part of the leading coefficient.)

Clearly Graphs A and C represent odd-degree polynomials, since their two ends head off inopposite directions. Graph D shows both ends passing through the top of the graphing box, just

like apositive quadratic would. The only graph with both ends down is: Graph B

Describe the end behavior of f(x) = 3x7

+ 5x+ 1004

This polynomial is much too large for me to view in the standard screen on my graphingcalculator, so either I can waste a lot of time fiddling with WINDOW options, or I can quickly usemy knowledge of end behavior.

This function is an odd-degree polynomial, so the ends go off in opposite directions, just likeevery cubic I've ever graphed. A positive cubic enters the graph at the bottom, down on the left,and exits the graph at the top, up on the right. Since the leading coefficient of this odd-degree

polynomial is positive, then its end-behavior is going to mimic a positive cubic.

The real (that is, the non-complex) zeroes of a polynomial correspond to thex-interceptsof the graph ofthat polynomial. So you can find information about the number of real zeroes of a polynomial by looking at

the graph, and conversely you can tell how many times the graph is going to touch or cross the x-axis bylooking at the zeroes of the polynomial (or at the factored form of the polynomial).

A zero has a "multiplicity", which refers to the number of times that its associated factor appears in the

polynomial. For instance, the quadratic (x + 3)(x2) has the zeroesx =3 andx = 2, each occuring
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once. The eleventh-degree polynomial (x + 3)4(x2)

7has the same zeroes as did the quadratic, but in

this case, thex =3 solution has multiplicity 4 because the factor(x + 3) occurs four times (the factor is

raised to the fourth power) and thex = 2 solution has multiplicity 7 because the factor(x2) occursseven times.

The point of multiplicities with respect to graphing is that any factors that occur an even number of time

(twice, four times, six times, etc) are squares, so they don't change sign. Squares are always positive.

This means that thex-intercept corresponding to an even-multiplicity zero can't cross thex-axis, becausethe zero can't cause the graph to change sign from positive (above the x-axis) to negative (below thex-axis), or vice versa. The practical upshot is that an even-multiplicity zero makes the graph just barely

touch thex-axis, and then turns it back around the way it came. You can see this in the followinggraphs: Copyright Elizabeth Stapel 2005-2011 All Rights Reserved

y= (x+ 6)(x7) y= (x+ 6)(x7)

x=6 once

x= 7 once

x=6 once

x= 7 twice

y= (x+ 6)(x7) y= (x+ 6) (x7)

x=6 twice

x= 7 once

x=6 twice

x= 7 twice

All four graphs have the same zeroes, atx =6 and atx = 7, but the multiplicity of the zero determineswhether the graph crosses thex-axis at that zero or if it instead turns back the way it came.

The following graph shows an eighth-degree polynomial. List the polynomial's zeroes withtheir multiplicities.


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I can see from the graph that there are zeroes at x =15,x =10,x =5,x = 0,x = 10, andx

= 15, because the graph touches or crosses thex-axis at these points. (At least, I'm assumingthat the graph crosses at exactly these points, since the exercise doesn't tell me the exact values.When I'm guessing from a picture, I do have to make certain assumptions.)

Since the graph just touches atx =10 andx = 10, then it must be that these zeroes occur aneven number of times. The other zeroes must occur an odd number of times. The odd-multiplicityzeroes might occur only once, or might occur three, five, or more times each; there is no way to

tell from the graph. (At least, there's no way to tell yet

we'll learn more about that on the nextpage.) And the even-multiplicity zeroes might occur four, six, or more times each; I can't tell bylooking.

But if I add up the minimum multiplicity of each, I should end up with the degree, becauseotherwise this problem is asking for more information than is available for me to give. I've got the

four odd-multiplicity zeroes (atx =15,x =5,x = 0, andx = 15) and the two even-multiplicity

zeroes (atx =10 andx = 10). Adding up their minimum multiplicities, I get 1 + 2 + 1 + 1 + 2 +1 = 8, which is the degree of the polynomial. So the minimum multiplicities are the correctmultiplicities.

x=15 with multiplicity 1,

x=

10 with multiplicity 2,x=5 with multiplicity 1,

x= 0 with multiplicity 1,

x= 10 with multiplicity 2, and

x= 15 with multiplicity 1

I was able to compute the multiplicities of the zeroes in part from the fact that the multiplicities will add upto the degree of the polynomial, or two less, or four less, etc, depending on how manycomplexzeroesthere might be. But multiplicity problems don't usually get into complex numbers.

There's an extra detail I'd like to mention regarding the multiplicity of a zero and the graph of thepolynomial: You can tell from the graph whether an odd-multiplicity zero occurs only once or if it occurs

more than once.

What is the multiplicity ofx= 5, given that the graph shows a fifth-degree polynomial with

all real-number roots, and the root x=5 has a multiplicity of2?
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The intercept atx =5 is of multiplicity 2. The polynomial is of degree 5, so the zero atx = 5, theonly other zero, must use up the rest of the multiplicities. Since 52 = 3, then

x= 5 must be of multiplicity 3.

The zero atx = 5 had to be of odd multiplicity, since the graph went through thex-axis. But the graph

flexed a bit (the "flexing" being that bendy part of the graph) right in the area ofx = 5. This flexing is whattells you that the multiplicity ofx = 5 had to be more than just 1. In this particular case, the multiplicity

couldn't have been 5 or7 or more, because the degree of the whole polynomial was only 5,but the

multiplicity certainly had to be more than just 1. Keep this in mind: Any odd-multiplicity zero that flexes at

the crossing point, like this graph did atx = 5, is of multiplicity 3 or more.

Notw: If you get that odd flexing behavior at some location on the graph that is off thex-axis (above orbelow the axis), then you're probably looking at the effect ofcomplexzeroes; namely, the zeroes that

you'd find by using theQuadratic Formula, the zeroes that don't correspond to the graph crossing the x-axis. Copyright Elizabeth Stapel 2005-2011 All Rights Reserved

Which of the following graphs could represent the polynomial

f(x) = a(xb)2(xc)

3?

Whatever this polynomial is, it is of degree 5. (I know this by adding the degrees on on the tworepeated factors: if I multiplied everything out, the degree on the leading term would be 5.)
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Since the polynomial is of odd degree, Graph A can't be correct, because its endsboth go thesame direction, meaning it is an even-degree polynomial.

Since my polynomial has two real-number zeroes (namely, zeroes atx = b and atx = c), I knowthat Graph C can't be right: it only crosses the x-axis once. So Graph C may be of odd degree,but it doesn't have enough zeroes.

From the end behavior, I can see that Graph D is of odd degree. Also, I know that the negativezero has an even multiplicity because the graph just touches the axis; this zero could correspond

tox = b. But there is no flexing where the graph crosses the positive x-axis, so the odd zero here

must have a multiplicity of only 1, and I need the multiplicity of this zero to be more than just 1.

So Graph D might have the right overall degree (if the zero x = b is of multiplicity 1), but themultiplicities of the two zeroes don't match up with what I need.

On the other hand, the ends of the graph tell me that Graph B is of odd degree, and the way the

graph touches or crosses thex-axis at the two graphedx-intercepts tells me that the polynomial

being graphed has one even-multiplicity zero and one odd-and-more-than-1-multiplicity zero. Thismatches what I need.

The correct graph is Graph B.

Find the degree-7 polynomial corresponding to the following graph, given that one of the

zeroes has multiplicity 3.

From the graph, I can see that there are zeroes of even multiplicity at x =4 andx = 4. The zeroatx =1 must be the zero of multiplicity 3. (This matches the graph, since the line goes throughthe axis, but flexes as it does so, telling me that the multiplicity must be odd and must be more

than 1.)

Since the total degree of the polynomial is7

, and I already have multiplicities of2

,2

, and

3 (which adds up to 7), then the zeroes at 4 and 4 must be of multiplicity 2, rather than

multiplicity 4 or multiplicity 6 or something bigger.

Working backwards from the zeroes, I get the following expression for the polynomial:

y = a(x + 4)2(x + 1)

3(x4)

2
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They marked that one point on the graph so that I can figure out the exact polynomial; that is, so I

can figure out the value of the leading coefficient "a". Plugging in thesex- andy-values from the

point (1,2), I get:

a(1 + 4)2(1 + 1)

3(14)

2= a(25)(8)(9) = 1800a =2

Then a =1/900, and the polynomial, in factored form, is:

y= (1

/900 )(x+ 4)2(x+ 1)

3(x4)

2

The exercise didn't say that I had to multiply this out, so I'm not going to. The factored form (especially forsomething as huge as this) should be a completely acceptable form of the answer.

Degrees, Turning s, and "Bum ps"(page 4 of 5)

Sections:End behavior,Zeroes and their multiplicities,"Flexing", Turnings & "bumps",Graphing

Graphs don't always head in just one directly, like niceneatstraight lines; they can turn around and head backthe other way. It isn't standard terminology, and you'lllearn the proper terms when you get to calculus, but Irefer to the "turnings" of a polynomial graph as its"bumps".

For instance, the following graph has three bumps, asindicated by the arrows:

Compare the numbers of bumps in the graphs below to the degrees of their polynomials:

degree two degree 3 degree 3 degree 4 degree 4

one bumpno bumps, butone flex point

two bumpsone (flattened)

bumpthree bumps

Copyright Elizabeth Stapel 2005-2011 All Rights Reserveddegree 5 degree 5 degree 5 degree 6 degree 6 degree 6
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no bumps,but one

flex point

two bumps(one flattened)

four bumpsone (flat)

bumpthree bumps

(one flat)five bumps

You can see from these graphs that, for degree n, the graph will have, at most, n1 bumps. The bumpsrepresent the spots where the graph turns back on itself and heads back the way it came. This change of

direction often happens because of the polynomial's zeroes or factors. But extra pairs of factors don'tshow up in the graph as much more than just a little extra flexing or flattening in the graph.

Because pairs of factors have this habit of disappearing from the graph (or hiding as a little bit of extraflexture or flattening), the graph may have two fewer, or four fewer, or six fewer, etc, bumps than youmight otherwise expect, or it may have flex points instead of some of the bumps. That is, the degree ofthe polynomial gives you the upper limit (the ceiling) on the number of bumps possible for the graph (thisupper limit being one less than the degree of the polynomial), and the number of bumps gives you thelower limit (the floor) on degree of the polynomial.

What is the minimum possible degree of the polynomial graphed below?

Since there are four bumps on the graph, and since the end-behavior says that this is an odd-

degree polynomial, then the degree of the polynomial is 5, or7, or9, or... But:

The min imumpossible degree is 5.

Given that a polynomial is of degree six, which of the following could be its graph?


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To answer this question, I have to remember that the polynomial's degree gives me the ceiling on

the number of bumps. In this case, the degree is 6, so the highest number of bumps the graph

could have would be 61 = 5. But the graph, depending on the multiplicities of the zeroes,might have only 3 or1 bumps.

(I would add1 or3 or5, etc, if I were going from the number of displayed bumps on the graph tothe possible degree of the polynomial, but here I'm going from the known degree of thepolynomial to the possible graph, so I subtract.)

Also, I'll want to check the zeroes (and their multiplicities) to see if they give me any additionalinformation.

Graph A: This shows one bump (so not too many), but only two zeroes, each looking like a

multiplicity-1 zero. This is probably just a quadratic, but it might possibly be a sixth-degreepolynomial (with four of the zeroes being complex).

Graph B: This has seven bumps, so this is a polynomial of degree at least 8, which is too high.

Graph C: This has three bumps (so not too many), it's an even-degree polynomial (being "up" onboth ends), and the zero in the middle is an even-multiplicity zero. Also, the bump in the middle

looks flattened, so this is probably a zero of multiplicity 4 or more. With the two other zeroes

looking like multiplicity-1 zeroes, this is a likely graph for a sixth-degree polynomial.

Graph D: This has six bumps, which is too many. On top of that, this is an odd-degree graph,since the ends head off in opposite directions. This can't be a sixth-degree polynomial.

Graph E: From the end-behavior, I can tell that this graph is from an even-degree polynomial.The one bump is fairly flat, so this is probably more than just a quadratic. This might be a sixth-degree polynomial.

Graph F: This is an even-degree polynomial, and it has five bumps (and a flex point at that thirdzero). But looking at the zeroes, I've got an even-multiplicity zero, a zero that looks like

multiplicity-1, a zero that looks like at least a multiplicity-3, and another even-multiplicity zero.That gives me a minimum of2 + 1 + 3 + 2 = 8 zeroes, which is too many for a degree-sixpolynomial. The bumps were right, but the zeroes were wrong. This can't be a degree-six graph.

Graph G: This is another odd-degree graph.


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Graph H: From the ends, I can see that this is an even-degree graph, and there aren't too manybumps, seeing as there's only the one. Looking at the two zeroes, they both look like at least

multiplicity-3 zeroes. So this could very well be a degree-six polynomial.

Graphs B, D, F, and G can't possibly be graphs of degree-six polynomials.Graphs A and E migh tbe degree-six, and Graphs C and H probablyare.

To help you keep straight when to add and when to subtract, remember your graphs of quadratics andcubics. Quadratics are degree-two polynomials and have one bump (always); cubics are degree-threepolynomials and have two bumps or none (having a flex point instead). So going from your polynomial toyour graph, you subtract, and going from your graph to your polynomial, you add. If you know yourquadratics and cubics very well, and if you remember that you're dealing with families of polynomials andtheir family characteristics, you shouldn't have any trouble with this sort of exercise.

Quick ie Graphing of Polynom ials(page 5 of 5)

Sections:End behavior,Zeroes and their multiplicities,"Flexing",Turnings & "bumps", Graphing

Once you know the basic behavior of polynomial graphs, you can use this knowledge to quickly sketchrough graphs, if required. This can save you the trouble of trying to plot a zillion points for a degree-sevenpolynomial, for instance. Once the graph starts heading off to infinity, you know that the graph is going tokeep going, so you can just draw the line heading off the top or bottom of the graph; you don't need toplot a bunch of actual points.

Without plotting any points other than intercepts, draw a graph of the followingpolynomial: Copyright Elizabeth Stapel 2005-2011 All Rights Reserved

y=(1/5600 )(x+ 5)

2(x+ 1)(x4)

3(x7)

This polynomial has already been put into factored form, which saves me the trouble of doing thesolving for the zeroes. I'll just solve the factors, noting the multiplicities as I go. The zeroes will be:

x =5, with multiplicity 2 (so the graph will be just touching thex-axis here)

x =1, with multiplicity 1 (so the graph will be crossing the axis here)

x = 4, with multiplicity 3 (so the graph will be crossing the axis here, but also flexing)

x = 7, with multiplicity 1 (so the graph will be just crossing the axis here)
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Also, adding the degrees of thefactors, I see that this is apolynomial of degree seven (thatis, an odd degree), so the endswill head off in oppositedirections. Because the leadingcoefficient is negative, the left-hand end will be "up" (comingdown from the top of the graph)and the right-hand end will go"down" (heading off the bottom ofthe graph). So I can start mygraph by pencilling in the zeroes,the behavior near the zeroes,and the ends, like this:

If I multiplied this polynomial out(and I'm not going to, so don'thold your breath), the constantterms of the factors would give

me 5 5 1 (4) (4) (

4) (7) = 11 200, which israther large. This would explainthe large denominator of theleading coefficient: by dividingthe polynomial by a sufficiently-large number, they made thispolynomial graphable. Otherwise,the graph would likely go off thepicture between the zeroes. (Notall texts notice this, so don't worryabout this consideration if itdoesn't come up in class.) When

= 0, I get ay-value of(1/5600)(11 200) =2, so Ican pencil this in, too:

I'm not supposed to find other plot points, so I'll just sketch in a rough guess as to what the graphlooks like. I'll go further from the axis where there is more space between the zeroes, and I won't

be so primitive as to assume that they-intercept point is the minimum point, what with the

midpoint between the two nearest zeroes,x =1 andx = 4, being atx = 1.5. Granted, the flexy

zero atx = 4 will probably push the graph a little to the left, but the bump is still probably to theright of they-axis.


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So I'll "rough in" an approximatedrawing, and then draw my finalanswer as a heavier line, erasingmy preliminary sketch-marks

before I hand in my solution.

This compares favorably with the actualgraph of the polynomial:

Some of my details (like my max and minpoints) were a little off, but my overall sketch

was still pretty good.

Complex Fract ion s(page 1 of 2)

I sometimes refer to complex fractions as "stacked" fractions, because they tend to have fractions stackedon top of each other, like this:


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Simplify the following expression:

This fraction is formed of two fractional expressions, one on top of the other. There are twomethods for simplifying complex fractions. The first method is fairly obvious: findcommondenominatorsfor the complex numerator and complex denominator, convert the complexnumerator and complex denominator to their respective common denominators, combineeverything in the complex numerator and in the complex denominator into single fractions, andthen, once you've got one fraction (in the complex numerator) divided by another fraction (in thecomplex denominator), you flip-n-multiply. (Rememberthat, when you are dividing by a fraction,you flip the fraction and turn the division into multiplication.)

This method looks like this:

Nothing cancels at this point, so this is the final answer.

(The "forx not equal to zero" part is because, in the original expression, "x = 0" would havecaused division by zero in the complex fraction. Depending on your book and instructor, you maynot need toaccount for this technicality. If you're not sure, ask now, before the test.)

The other method is to find one common denominator for all the fractions in the expression, andthen multiply both the complex numerator and complex denominator by this expression. Thensimplify.
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This method looks like this:

Then the final answer is:

By multiplying through, top and bottom, by the same thing, I was really just multiplying by 1. This is similar

to multiplying the fraction1/2 by

2/2 to convert it to

2/4. In my experience, books and teachers often use

the first method, but students generally prefer the second method. When I was in school, I was taught thefirst method. As soon as I encountered the second method, I switched to it. In the remaining examples, Iwill demonstrate this second method, but you can use either method you prefer.

(If your text or instructor requires that you find the restrictions on the domains [the "x not equal to zero"part in the above example], you might find it helpful to use the "flip-n-multiply" method covered first, sincethis will give you the full fraction form of the denominator at some point in the computations.)


Copyright Elizabeth Stapel 2003-2011 All Rights Reserved

Can I start by hacking off thex's? Or lopping off the 3's? (Hint: No!) I can only cancel off factors,not terms, so I can't do any canceling yet. The first thing I'll do is find the LCM for this expression.


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TheLCM(Least CommonMultiple, or, for us older types,the LCD, Lowest CommonDenominator) of the givendenominators within this complex

fraction is (x

1)(x + 4), so I'llmultiply through, top and bottom,by this expression:

(If you're not sure how I multipliedthose factors to get the cubicresults, reviewthis lessononmultiplying polynomials.)

Can I now cancel off thex3's? Or cancel the 6's into the 12? Can I go inside the adding and rip

out parts of some of the terms? (Hint: No!) Nothing cancels, so this is the final answer:

Complex Fract ion s: More Examp les(page 2 of 2)


Can I start by hacking off thex

3's? Can I cancel the 4 with the 12? Or the 3 with the 9 or the12? (Hint: No!)
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The common denominator for this

complex fraction would bex3, so I'llmultiply through, top and bottom, by that.

Clearly, nothing cancels, so my final answer is:

(Why the restrictions?)

It is highly unusual for a complex fraction to simplify this much, but it can happen. In this case, the "except

forx equal to 3" part is rather important, since the original fraction is not always equal to3/4. Indeed, it is

not even defined forx equal to 3 (since this would cause division by zero).


Can I start off by canceling like this:

DON'T DO THIS!

I can only cancel factors, not terms, so the above cancellations are not proper.
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Simplify sqrt(9).

(Warning: The step that goes through the third "equals" sign is " ", not " ". The i is

outside the radical.)

Simplify sqrt(25).

Simplify sqrt(18).

Simplifysqrt(6).

In your computations, you will deal with i just as you would withx, except for the fact thatx2 is justx2, buti2 is1:

Simplify 2i+ 3i.

2i + 3i = (2 + 3)i = 5i

Simplify 16i5i.

16i5i = (165)i = 11i

Multiply and simplify (3i)(4i).

(3i)(4i) = (34)(ii) = (12)(i2) = (12)(1) =12

Multiply and simplify (i)(2i)(3i).

(i)(2i)(

3i) = (2

3)(i i i) = (

6)(i2 i)

=(6)(1 i) = (6)(i) = 6i

Note this last problem. Within it, you can see that , because i2 = 1. Continuing, we get:


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This pattern of powers, signs, 1's, and i's is a cycle:

In other words, to calculate any high power of i, you can convert it to a lower power by taking the closestmultiple of4 that's no bigger than the exponent and subtracting this multiple from the exponent. For

example, a common trick question on tests is something along the lines of "Simplify i99", the idea being

that you'll try to multiply i ninety-nine times and you'll run out of time, and the teachers will get a goodgiggle at your expense in the faculty lounge. Here's how the shortcut works:

i99

= i96+3

= i(424)+3

= i3

=i

That is, i99 = i3, because you can just lop off the i96. (Ninety-six is a multiple of four, so i96 is just 1, which

you can ignore.) In other words, you can divide the exponent by 4 (using long division), discard theanswer, and use only the remainder. This will give you the part of the exponent that you care above. Hereare a few more examples:

Simplify i17

.

i17

= i16 + 1

= i4 4 + 1

= i1

=i

Simplify i120

.

i120

= i4 30

= i4 30 + 0

= i0

= 1

Simplify i64,002

.

i64,002

= i64,000 + 2

= i4 16,000 + 2

= i2

=1

Now you've seen how imaginaries work; it's time to move on to complex numbers. "Complex" numbershave two parts, a "real" part (being any "real" number that you're used to dealing with) and an "imaginary"

part (being any number with an "i" in it). The "standard" format for complex numbers is "a + bi"; that is,

real-part first and i-part last.

Graphing Logar i thm ic Funct ions : Int ro(page 1 of 3)


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By nature of thelogarithm, most log graphs tend to have the same shape, looking similar to a square-rootgraph:

y= sqrt(x) y= log2(x)

The graph of the square root starts at the point (0, 0) and then goes off to the right. On the other hand,the graph of the log passes through (1, 0), going off to the right but also sliding down the positive side ofthey-axis. Remembering that logs are theinversesofexponentials, this shape for the log graph makesperfect sense: the graph of the log, being the inverse of the exponential, would just be the "flip" of thegraph of the exponential:

y= 2x y= log2(x)

comparison of the two graphs,showing the inversion line in red
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It is fairly simple to graph exponentials. For instance, to graphy = 2x

, you would just plug in some values

forx, compute the correspondingy-values, and plot the points. But how do you graph logs? There aretwo options. Here is the first: Copyright Elizabeth Stapel 2002-2011 All Rights Reserved

Graph y= log2(x).

In order to graph this "by hand", I need first to remember that logs are not defined for negative x

or forx = 0. Because of this restriction on thedomain(the input values) of the log, I won't evenbother trying to findy-values for, say,x =3 orx = 0. Instead, I'll start withx = 1, and work fromthere, using thedefinitionof the log.

Since 20

= 1, then log2(1) = 0, and (1, 0) is on the graph.

Since 21


Since 3 is not a power of2, then log2(3) will be some messy value. So I won't bother with

graphingx = 3.

Since 22 = 4, then log2(4) = 2, and (4, 2) is on the graph.

Since 5, 6, and 7 aren't powers of2 either, I'll skip them and move up tox = 8.

Since 23

= 8, then log2(8) = 3, so (8, 3) is on the graph.

The next power of2 is 16: since 24


The next power of2,x = 32, is too big for my taste; I don't feel like drawing my graph that wide,

so I'll quit atx = 16.

The above gives me the point (1, 0) and some points to the right, but what do I do forx-values

between 0 and 1? For this interval, I need to think in terms of negative powers and reciprocals.Just as the left-hand "half" of the exponential function had few graphable points (the rest of them

being too close to thex-axis), so also the bottom "half" of the log function has few graphable

points, the rest of them being too close to the y-axis. But I can find a few:

Since 21

=1/2 = 0.5, then log2(0.5) =1, and (0.5,1) is on the graph.

Since 22

=1/4 = 0.25, then log2(0.25) =2, and (0.25,2) is on the graph.

Since 23

=1/8 = 0.125, thenlog2(0.125) =3, and (0.125,3) is on the graph.

The next power of2 (asx moves in this direction) is1/16 = 2

4, but thex-value for the point

(0.0625,4) seems too small to bother with, so I'll quit with the points I've already found.
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Listing these points gives memy T-chart:

Drawing my dots and then

sketching in the line(remembering notto go to

the left of they-axis!), I getthis graph:

Graphing Logar i thm ic Funct ions: Examples(page 2 of 3)

In theprevious example, I said there were two options for how to graph logs. The previous pagedemonstrated how to work from the concept of logarithms to find nice neat points to plot. However, theother option is that you can use your calculator to find plot points.

Graph y= log2(x)

To find my plot-points using my calculator(since my calculator can only computecommon, or base-10, and natural, or base-e,

logs), I will need to use thechange-of-baseformula, which gives me an equivalent

equation. The original equation,y = log2(x)

becomesy =ln(x)

/ln(2):
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Once I've entered the calculator-friendly formof the equation, I can get some plot pointsfrom my calculator's TABLE feature:

Copyright Elizabeth Stapel 2002-2011 All RightsReserved

(If you have an old TI-85, so you have no in-built "TABLE" utility, you can install any of thevarious after-market programs which domuch the same thing. The graphics to theright are screen shots of whatthe programon my TI-85 produced.)

I know I have to have all positive -valuesinside the log, so I start the TABLE listing at

= 1, and go from there.

I'd still like to have some plot points betweenzero and one, so I adjust the initial ("start")value and the increment (the "count by"

amount) to get some additional plot pointsbetweenx = 0 andx = 1:

Using these points (plottedto one or two decimalplaces in "accuracy"), I willend up with the same graphas before.
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Depending upon your calculator's software, you will either get blank spaces in your TABLE for the y-

values whenx = 0 and whenxis negative; or the slot will display "ERROR", "UNDEFINED", or some

other error code; or else the program will crash. (Mine crashes for undefinedy-values, which is why I wascareful to start my TABLE display above at a positive x-value.) This behavior in the TABLE featurereinforces the fact that logarithms are not defined for non-positive arguments.

(Regarding finding plot points betweenx = 0 andx = 1, if you do not know how to change your initialvalue fromx = 0 or how to change your increment from 1, consult your owner's manual; the instructionswill be somewhere in the chapter on graphing.)

If you are graphing the common (base-10) log or the natural (base-e) log, just use your calculator to getthe plot points. When working with the common log, you will quickly reach awkwardly large numbers if you

try to plot only whole-number points; for instance, in order to get as high asy = 2, you'd have to usex =100, and your graph would be ridiculously wide. When working with the natural log, the base e is anirrational number anyway, so there's no point in even tryingto find nice neat plot points, because, other

than (1, 0), there aren't any.

Sometimes the log graph is shifted a bit from the "usual" location (shown in the graph above), either up,down, right, left, or upside-down, or else some combination of these. But the general shape of the graphtends to remain the same.

Graph y= log3(x) + 2.

This is the basic log graph, but it's beenshifted upwardby two units. To find plot points for this

graph, I will plug in useful values ofx (being powers of3, because of the base of the log) andthen I'll simplify for the corresponding values ofy.

30

= 1, solog3(1) = 0, and log3(1) + 2 = 23

1= 3, solog3(3) = 1, and log3(3) + 2 = 3

32

= 9, solog3(9) = 2, and log3(9) + 2 = 43

3= 27, solog3(27) = 3, and log3(27) + 2 = 5

Moving in the other direction (to get somey-values forx between 0 and 1):

31

=1/3, solog3(

1/3 ) =1, and log3(

1/3 ) + 2 = 1

32

=1/9, solog3(

1/9 ) =2, and log3(

1/9 ) + 2 = 0

33 = 1/27, solog3( 1/27 ) =

3, and log3( 1/27 ) + 2 =

1

These are the only "neat" points that I'm going to bother finding for my graph. If I feel a need foradditional plot points, especially between any two of the points I found above, I can evaluate the

function "ln(x) / ln(3)" in my calculator.
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The graph ofy = log3(x) + 2looks like this:

Graphing Logar i thm ic Funct ions: Examples(page 3 of 3)

Graph y= log2(x+ 3).

This graph will be similar to the graph of log2(x), but it will beshifted sideways.

Since the "+ 3" is inside the log's argument, the graph's shift cannot be up or down. This means that theshift has to be to the left or to the right. But which way? You can keep track of the direction of the shift by

looking at the basic point (1, 0) ("basic" because it's neat and easy to remember). The log will be 0 when

the argument,x + 3, is equal to 1. When isx + 3 equal to 1? Whenx =2. Then the basic log-graph

point of(1, 0) will be shifted over to (2, 0) on this graph; that is, the graph is shifted three units to theleft. If you are not comfortable with this concept or these manipulations, please review how to work withtranslations of functions.

Since a log cannot have an argument of zero or less, then I must have x + 3 > 0, this tells me

that, for this graph,x must always be greater than

3.

The graph of the basic log functiony = log2(x) crawled up the positive side of they-axis to reachthex-axis, with the line never going to the left of the limitation thatx must be greater than zero.

To remind myself of the similar limitation of this log (wherex must always be greater than3), I

will insert a dashed line atx =3:
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A line like this, which marks off territory where the graph shouldn't go, is called a "verticalasymptote", or simply an "asymptote". I don't have to add this to the graph, but it can be veryhelpful, and might convince the grader that I know what I'm doing.

After I dash in the asymptote, I plot some points:

20 = 1, so log2(1) = 0;x + 3 = 1 forx =

2: (

2, 0)21 = 2, so log2(2) = 1;x + 3 = 2 forx =1: (1, 1)2

2= 4, so log2(4) = 2;x + 3 = 4 forx = 1: (1, 2)

23

= 8, so log2(8) = 3;x + 3 = 8 forx = 5: (5, 3)

Then, working in the other direction: Copyright Elizabeth Stapel 2002-2011 All Rights Reserved

21

= 0.5, so log2(0.5) =1;x + 3 = 0.5 forx =2.5: (2.5,1)

22

= 0.25, so log2(0.25) =2;x + 3 = 0.25 forx =2.75: (2.75,2)

23

= 0.125, so log2(0.125) =

3;x + 3 = 0.125 forx =2.875: (2.875,3)

Note that, to find each of these points, I did not start with an x-value and then puzzle my way to ay-value; that would be too hard, and I'm too lazy. Instead, I started with a simple exponentialstatement, switched it around to the corresponding logarithmic statement, and then figured out,

for that exponent (which is also myy-value), what thex-value needed to be. This method is, inmy view, much the simpler way to work these problems.

Plotting the points I've calculated, I get:

...and connecting the dots gives me thefollowing graph:
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If you check this in your calculator, first, remember to put the "x + 3" inside parentheses, or your

calculator will think you mean "log2(x) + 3", and you'll get the wrong answer; second, remember that yourcalculator can only follow its programming it can't think so the graph it displays will likely beincorrect, even if you enter the function correctly.

If you plugy = log2(x + 3) into a graphingcalculator (in the change-of-base formulation of

"ln(x + 3) / ln(2)"), you will likely get a graphthat looks something like this:

Now, you know full well that the log doesn't just "end" there at the left, hanging uselessly in space. Why isthe calculator doing it wrong? Because it's just a machine, and it's doing the best it can. The calculator

graphs in a manner similar to how you do: it picksx-values, computesy-values, plots the points, andconnects the dots. But, whereas you know that the log graph continues downward forever, getting

infinitesimally close to they-axis (or whatever the vertical asymptote happens to be), the calculator onlyknows that it tried onex-value on its list, got "ERROR" for an answer, tried the nextx-value on its list, and

got a validy-value. Since it has no other dots before that first one, and because it can't think, it starts thegraph with that first dot. This is another instance of "student smart; calculator stupid". Don't assume, justbecause the calculator displays a graph a certain way, that this is what the graph actually looks like. Useyour head!

To review: below are some different variations on the same basic logarithmic function, with the associatedgraph below each equation. Note that, even if the graph is moved left or right, or up or down, or is flippedupside-down, it still displays the same curve:

y = ln(x) y = ln(x) y =ln(x)

y = ln(x + 1) y = ln(x + 1) y =ln(x)


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y = ln(x)1 y = ln(x)1 y = ln(x1)

y = ln(x) + 1 y = ln(x) + 1 y = ln(x1)

Logari thms: Int rodu ct ion to

" The Relationship "(page 1 of 3)

Sections: Introduction to logs,Simplifying log expressions,Common and natural logs
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Logarithms are the "opposite" ofexponentials, just as subtraction is the opposite of addition and divisionis the opposite of multiplication. Logs "undo" exponentials. Technically speaking, logs are theinversesofexponentials.

In practical terms, I have found it useful to think of logs in terms of The Relationship:

The Relationship

y= bx

..............is equivalent to...............(means the exact same thing as)

logb(y) = x

On the left-hand side above is the exponential statement "y = bx

". On the right-hand side above, "logb(y)

=x" is the equivalent logarithmic statement, which is pronounced "log-base-b ofyequalsx"; The value of

the subscripted "b" is "the base of the logarithm", just asb is the base in the exponential expression "bx

".

And, just as the baseb in an exponential is always positive and not equal to 1, so also the baseb for a

logarithm is always positive and not equal to 1. Whatever is inside the logarithm is called the "argument"of the log. Note that the base in both the exponential equation and the log equation

systems of non

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