D.E.V. Project

Doug Bradfield

Factoring Polynomials The first step when factoring polynomials is to identify a greatest

common factor. For example, if you are given the function, 9x3-18x2+27x-54, then a 9 would be factored out first to create the

function 9(x3-2x2+3x-6).• Next, the number of terms must be determined, in this case there

are four terms. With four terms, factor by grouping if possible.• 9(x3-2x2+3x-6) now, break the problem into two parenthesis. We

can set the 9 aside for now if we remember to bring it back.• (x3-2x2)+(3x-6) Then factor out anything in the two new parenthesis.• x2(x-2)+3(x-2) x-2 must remain one factor and the other is

developed by the coefficients. This becomes 9(x2+3)(x-2)

Factoring Cont. All together this problem looks like this:

9x3-18x2+27x-54 9(x3-2x2+3x-6) x2(x-2)+3(x-2) 9(x2+3)(x-2)

Unfortunately. Some problems are more difficult and require us to find a rational root and use that to set up a long division problem.

For example, x3+21x2+99x+135 is a polynomial that can’t be grouped. One rational root must be found first. In this case x=-3 so we use x+3 to divide out the polynomial.

Factoring Cont. x+3 goes outside the division bracket and x3+21x2+99x+135 goes

inside the division bracket. Now the problem looks like this

x+3 x3+21x2+99x+135

Now, we find what times x equals x3 and put it on top of the division bracket. In this case it is x2. put that on top.

x+3 x3+21x2+99x+135

Then, multiply x2 by x and x2 by 3. those numbers are then subtracted from the overall equation.

x2

Factoring Cont. x+3 x3+21x2+99x+135 -(x3+3x2) 18x2

Now, 99x must be dropped and step one must be repeated. 18x times x equals 18x2. now 18x goes on top of the division bracket, and is then multiplied by x and by 3. The quantity is subtracted from the new value of 18x2+99x.

x+3 x3+21x2+99x+135 -(x3+3x2) 18x2+99x -(18x2+54x) 45x

x2

x2+18x

Factoring Cont. Continue these steps until all numbers have been used and there is

a remainder of zero. Here is the entire division problem.

x+3 x3+21x2+99x+135 -(x3+3x2) 18x2+99x -(18x2+54x) 45x+135 -(45x+135) 0 Now, factor the function on top (x2+18x+45) directly into

parenthesis. 15x3=45 and 15+3=18. therefore factors are 3 and 15. Don’t forget

about the x=-3 that was used to divide out.

x2+18x+45

Factoring Cont. We factored x2+18x+45 into (x+3)(x+15) but after adding the x=-3

we get the final factored form of x3+21x2+99x+135 to be (x+3)2(x+15)

Finding Inverse Functions First, the equation must be written in y= form. For

example, f(x)=16x-7 must be written as y=16x-7.

Next, you must solve for x. Add 7 to both sides and divide by 16. Now we have (y+7)/16=x

Lastly, we switch the x and y which leaves us with the inverse. f(x)-1=(x+7)/16

f(x)=(x+7)/16 and f(x)=16x-7 are inverses.

Inverses Cont. If I am given the function f(x)= then I must write it in y= form.

This becomes y=

Now you must solve for x. To do so, multiply both sides by the quantity (2x+4). This leaves you with y(2x+4)=13x+9. After distributing, you are left with 2xy+4y=13x+9.

In order to get the y by itself, subtract 2xy from both sides and also subtract 9 from both sides. Now 4y-9=13x-2xy. Now, all the x’s are on the same side of the equation.

From the quantity 13x-2xy an x can be factored out, leaving x(13-2y).

13x+92x+4

13x+92x+4

Inverses Cont. The equation now looks like this: 4y-9=x(13-2y)

Divide both sides by the quantity (13-2y). Now you have x=

After switching the x and y, you are left with the inverse function

f-1(x)= f(x)=

4y-913-2y

4x-913-2x

13x+92x+4

Inverses Cont. f(x)= 2x+4 ∙ y= ∙ 2x+4

y(2x+4)=13x+9 2xy+4y=13x+9 4y-9=x(13+2y)

= x f-1(x)=

13x+92x+4

13x+92x+4

-2xy -2xy-9 -9 (13+2y)(13+2y)

4y-913-2y

13x+92x+4

Simplifying Rational Expressions The first step is always to factor out each expression whether you

are dealing with multiplication, addition, division, or subtraction. If I am given the expression: ∙

That can be factored into ∙

In multiplication, Any similar expressions can then be divided out, which would leave us with (2x-4)∙(4x+4). In this particular case, both denominators were able to divide out. After factoring out a 2, the final expression in factored form is: 2(x-2)(2x+2)

In the case of addition or subtraction, we would first have to find common denominators before dividing out and in the case of division, we multiply the first part of the expression by the reciprocal of the second part.

4x2+12x+8 x+4

2x2+4x-16 x+2

4(x+2)(x+1) x+4

2(x+4)(x-2) x+2

Simplifying Rationals Cont. If we are given the expression: ÷ The first step, as

always, is to factor. This factors nicely into the expression:

÷ We now must flip the second part and multiply, called multiplying by the reciprocal. This leaves us with:

• The next step is to divide out. In this case, you

This is the final simplified rational expression.

x2+2x-15x2+3x-18

x+5x2-1x-6

x+5(x-3)(x+2)

(x+5)(x-3)(x-3)(x+6)

x+5(x-3)(x+2)

(x-3)(x+6)(x+5)(x-3)

can divide out an (x-3) and an (x+5). Now we are left with: x+6(x-3)(x+2)

Simplifying Rationals Cont. Here is all the work put together.

÷

÷

•

•

x+5x2-1x-6

x2+2x-15x2+3x-18

x+5(x-3)(x+2)

(x+5)(x-3)(x-3)(x+6)

x+5(x-3)(x+2)

(x-3)(x+6)(x+5)(x-3)

x+6(x+2)(x-3)

x+5(x-3)(x+2)

(x-3)(x+6)(x+5)(x-3)

Graphing Rational Functions The first step to graphing a rational function is always to factor. If

we are given the function: we can factor the numerator and denominator.

Now we have:

Next, we need to look to see if any factors will divide out. In this case (x+1) divides out leaving us with:

There will now be a hole where we divided out. We then have to set (x+1) equal to zero. This leaves us with x=-1

Now we need to find any asymptotes and intercepts.

x2+10x+9x2+3x+2

(x+9)(x+1)(x+2)(x+1)

x+9x+2

Graphing Cont. In order to find Vertical asymptotes, we need to look for any

solutions that cause the denominator to be equal to zero. Our equation only has one vertical asymptote at x=-2.

Next, we would look for any horizontal asymptotes. In order to find the horizontal asymptote, if the x’s are to the same power in the numerator and denominator, we divide their leading coefficients. In this case our horizontal asymptote is at y=1.

To find any y intercepts, simply set x=0 in the equation and solve. In this equation when x=0, it gives us also known as 4.5. our y-intercept can be found at (0,4.5)

Lastly we must find the x-intercept. To do so we must find what number, if plugged in for x would make the numerator=0. Our x-intercept is -9 and can be found at (-9,0).

92

Graphing Cont.

The graph would look something like this:

1

-2

4.5

-9

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _

-1

hole

Graphing Cont. Now lets try one a little more difficult.

following the steps, we must factor before we do anything else.

This time we have nothing to divide out and are left with more factors.

Now in order to make the denominator 0, we need x to equal -12 and -2. Therefore we have 2 vertical asymptotes. One at x=-12 and one at x=-2.

In this situation there is a higher power in the denominator than the numerator. This means our horizontal asymptote is y=0.

x+7x2+14x+24

x+7(x+12)(x+2)

Graphing Cont. Now we must find intercepts. To find the y-intercept we plug zero in

for x. After that, we are left with: y-intercept is y=

Lastly we must find the x-intercept by setting the numerator equal to zero. The x- intercept is at x=-7. So far we have everything except the direction of our arrows. To find this we must set up a chart to find the direction as our asymptotes are approached.

724

724

-12.1

-11.9

-2.1

-1.9

x+7(x+12)(x+2)

(-)(-)(-)

-∞ (-)(+)(-)∞

(+)(+)(-)

-∞ (+)(+)(+)∞

x

y

Graphing Cont.

The graph would look something like this:

.29-2-

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Reflection This project was something that I decided I would not stop until I believed I

have put all I could into it. I tried to use problems that were somewhat challenging and show a small variety of possibilities that could occur in each of the topics. As it turns out, I am very glad that I chose my due date during break because of the amount of time and effort that it took to complete and read back through checking my work. As a whole, I have gotten much better at the topics that I covered after going back through notes, and researching, and in some cases just figuring it out. A couple of these topics were not my strongest during the year, but I now feel that they are definitely my best four topics. Although this was a project that for the longest time was nothing but confusing, I strongly recommend that you continue to do it due to the amount of knowledge I have gained in the process. The most difficult part for me was to come up with numbers that would work to factor by dividing. Eventually I developed a strategy but other than making symbols in power point, it was the most time consuming. All in all, I would do this project again.