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ACCUPLACER MATH 0310

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MATH 0310

Page

Linear Equations 2

Linear Equations – Exercises 5

Linear Equations – Answer to Exercises 6

Linear Inequalities 8

Equations Involving Absolute Value 10 Inequalities Involving Absolute Value 11

Linear Inequalities and Absolute Value – Exercises 13

Linear Inequalities and Absolute Value – Answer to Exercises 14 Linear Functions 15

Linear Functions – Exercises 21

Linear Functions – Answer to Exercises 22 Exponents 23

Exponents – Exercises 26

Exponents – Answer to Exercises 27 Polynomials 28 Polynomials – Exercises 35

Polynomials – Answer to Exercises 36

Factoring 37

Factoring – Exercises 42

Factoring – Answer to Exercises 43

2

LINEAR EQUATIONS A linear equation can be defined as an equation in which the highest exponent of the equation variable is one. When graphed, the equation is shown as a single line. Example: 42 =+x ← Linear equation, highest exponent of the variable is 1. A linear equation has only one solution. The solution of a linear equation is equal to the value of the unknown variable that makes the linear equation true. Example: 42 =+x ← The value of the unknown variable that makes the equation 24 −=x true is 2: 422 =+ 2=x There are different forms of writing a linear equation. Each form has a different way of solving that makes the process easier. There is only one rule that applies to any form of linear equations, whatever you do on one side you need to do on the other side. That is, if you add on the left side, you need to add on the right side, if you multiply on the left side you need to multiply on the right side, and so on. 1. Equations of the form bax =+ or bax =− .

Linear equations in this form are solved by adding or subtracting the same quantity to both sides with the idea of leaving the variable by itself. Examples: 105 =−x 51055 +=+−x ← Add 5 to both sides to eliminate the number on 15=x the left side of the equation and leave x alone. 1.42.3 =+y 2.31.42.32.3 −=−+y ← Subtract 3.2 to both sides to eliminate the 9.0=y number on the left side and leave y alone.

2. Equations of the form bax = .

Linear equations in the form of multiplication are solved by dividing both sides of the equation by the number multiplying the variable. When a fraction is multiplying the variable, multiply both sides of the equation by the reciprocal of the fraction attached to the variable.

3

Examples: 183 =x

3

1833

=x ←Divide both sides by 3 to leave x by itself.

6=x

52

43

=x

34

52

43

34

⋅=⋅ x ←Multiply both sides by the reciprocal of 43 , which is

34 .

158

=x

3. Equations of the form cbax =+ .

This type of linear equation presents a combination of the first two forms of linear equations. First step in solving this form of linear equations is to eliminate the stand alone number on the variable side by adding or subtracting the same quantity to both sides of the equation. Next, eliminate the number attached to the variable by dividing or multiplying, depending on the case, to both sides of the equation. The idea is to leave variables on one side and constants on the other side. Examples: 2573 =+y 725773 −=−+y ←Subtract 7 to both sides to remove it from the left side. 183 =y

3

1833

=y ←Divide both sides by 3 to leave y alone and solve for y.

6=y

1258

−=+x ←Follow the steps described above to solve for x.

815

85

13

58

85

358

212258

−=

⋅−=⋅

−=

−−=−+

x

x

x

x

4

4. Equations of the form dcxbax +=+ .

In this form of linear equations the main goal is to leave the variables on one side and the constants on the other side. To do this, perform all the steps required to solve the previous forms of linear equations step by step. Do not try to do all the steps at once, this can only result in confusion and mistakes. Example: 138156 +−=− xx ←Start by moving the variable from the right

side to the left side. 13881586 ++−=−+ xxxx ←Combine the two variables together. 131514 =−x ←Add 15 to both sides to eliminate the 1513151514 +=+−x constant on the left side of the equation. 2814 =x ←Divide both sides of the equation by 14.

1428

1414

=x ←Solve for x.

2=x Distributive Property: ( ) acabcba +=+ The distributive property is used to remove grouping symbols in linear equations. Grouping symbols are terms enclosed in parentheses. To ungroup, distribute the term outside the parentheses to each of the terms inside the parentheses by multiplication. Example: ( ) 7132 −=+−x ←Multiply the term outside the parentheses to each of 7162 −=+−x the terms inside the parentheses. 752 −=−x ←Eliminate the constant on the left side of the equation 57552 +−=+−x by adding 5 to both sides of the equation. 22 −=x

22

22

−=x ←Divide both sides by 2 and solve for x.

1−=x

5

LINEAR EQUATIONS – EXERCISES

Solve for x: 1. 8 6 12x − = 2. 4 5 2 7x x− = + 3. x x x− − − = −1 3 5 2 2 2 5( ) ( )

4. 2 44

4 33

x x−=

+

5. 2 4x ab

x ba

−=

−

6. 31

45

11

6x x+

− =+

−

7. 31

41

81x x x−

++

=+

8. 1 1 1ax bx c

+ =

9. 3 43

10x x− =

10. 2 15

46

23

x x++

−=

11. 3 2 1 5 3( )x x x− + = + 12. 3 5 2 5 2− − = −( )x 13. − + − = −3 5 7 4 10( )x x 14. 8 3 2 5 1+ + = −( ) ( )x x 15. 5 2 4 1 7( ) ( )x x+ − + =

16. 12 1

34

72 1x x+

− =+

17. 72 3

42 3

6x x−

+−

=

6

LINEAR EQUATIONS – ANSWERS TO EXERCISES

1. 8 6 12x − = 2. 4 5 2 7x x− = + 3. ( ) ( )1 3 5 2 2 2 5x x x− − − = −

8 6 6 1 268 18

94

xx

x

− + = +=

=

4 2 7 52 12

6

x xx

x

− = +==

x x xx xx

x

− − + = −− = − + +==

1 15 6 4 107 4 10 1 153 6

2

4. 2 44

4 33

x x−=

+ 5. 2 4x ab

x ba

−=

−

( ) ( )3 2 4 4 4 36 12 16 126 16 12 12

10 24125

x xx xx x

x

x

− = +

− = +− = +

− =

= −

( ) ( )

( )

2 2

2 2

2 2

2 2

2 4

2 42 4

2 4

2 4

a x a b x b

ax a bx bax bx a b

x a b a b

a bxa b

− = −

− = −

− = −

− = −

−=

−

6. 31

45

11

6x x+

− =+

− 7. 31

41

81x x x−

++

=+

( ) ( )

( ) ( )

3 4 15 1 5 1 61 5 1

15 4 1 5 30 115 4 4 5 30 3026 36

1813

x xx x

x xx x

x

x

+ − = + − + + − + = − +

− − = − −= −

= −

( )( ) ( )

( ) ( ) ( )

3 4 81 1 11 1 1

3 1 4 1 8 13 3 4 4 8 87 8 1 8

77

x x xx x x

x x xx x xx xx

x

− + + = + − + + + + − = −

+ + − = −− = −

− = −=

8. 1 1 1ax bx c

+ = 9. 3 43

10x x− =

7

( )

1 1 1abcx abcxax bx c

bc ac abxbc ac x

abc a b

xab

+ =

+ =+

=

+=

( )3 43 3 103

9 4 3016

x xx x

x

x

− = − =

=

10. 2 15

46

23

x x++

−= 11. ( )3 2 1 5 3x x x− + = +

2 1 4 230 305 6 3

x x+ − + =

6 3 5 37 5 3 3

x x xx x− + = +− = +

( ) ( )6 2 1 5 4 2012 6 5 20 20

x xx x

+ + − =

+ + − =

2 63

xx

==

17 34

2x

x=

=

12. 3 5 2 5 2− − = −( )x 13. − + − = −3 5 7 4 10( )x x

3 10 25 210 2 3 2510 30

3

xxx

x

− + = −− = − − −− = −=

15 21 4 1015 10 21 45 25

5

x xx x

xx

− − − = −− + = +− == −

14. 8 3 2 5 1+ + = −( ) ( )x x 15. 5 2 4 1 7( ) ( )x x+ − + =

8 3 6 5 53 5 5 8 6

2 19192

x xx x

x

x

+ + = −− = − − −

− = −

=

5 10 4 4 7

7 10 41

x xxx

+ − − == − +=

16. 12 1

34

72 1x x+

− =+

17. 72 3

42 3

6x x−

+−

=

( ) ( )

( )

1 3 74 2 1 4 2 12 1 4 2 1

4 3 2 1 284 6 3 28

6 3 24276

x xx x

xx

x

x

+ − = + + + − + =

− − =− − =

= −

( ) ( )

( )

7 42 3 2 3 62 3 2 3

7 4 6 2 311 12 1811 18 12

2912

x xx xx

xx

x

− + = − ⋅ − − + = −

= −+ =

=

8

LINEAR INEQUALITIES When we use the equal sign in an equation we are stating that both sides of the equation are equal to each other. In an inequality, we are stating that both sides of the equation are not equal to each other. It can also be seen as an order relation; that is, it tells us which one of the two expressions is smaller, or larger, than the other one. A linear inequality is an equation in which the highest variable exponent is one. Examples: 3<x → less than, < 352 −≤+ xx → less than or equal to, ≤ 2−>x → greater than, > 623 −≥− xx → greater than or equal to, ≥ The solution to an inequality is the value of the variable which makes the statement, or the inequality, true. Examples: 3<x → This inequality is telling us that “x is less than 3”. Therefore, any

number less than three is a possible solution. Remember that on the number line, any number to the left is less than a given number, and any number to the right of that given number is greater.

–3 – 2 –1 0 1 2 3

352 −≤+ xx → This inequality is telling us that the equation 52 +x is

less than or equal to the equation 3−x . Solve for x to find which number, or numbers, make this equation true.

352 −≤+ xx → Solve for x by leaving the variables on one side 53552 −−≤−+ xx and the numbers on the other side, just as you

82 −≤ xx would solve a linear equation. 82 −−≤− xxxx

8−≤x –9 –8 –7 –6 –5 –4 –3 –2 –1 0

2−>x → This inequality is telling us that “x is greater than –2”. Therefore, any number greater than negative two is a possible solution.

Notice how 3 is marked with an empty dot, this means that the answer does not include the number 3.

Notice how –8 is marked with a black dot, this means that the answer does include the number –8.

9

–3 –2 –1 0 1 2 3

623 −≥− xx → Solve for x by leaving the variables on one side 26223 +−≥+− xx and the numbers on the other side.

43 −≥ xx 43 −−≥− xxxx

42 −≥x

24

22 −

≥x

2−≥x

–3 –2 –1 0 1 2 3 As you can see, solving linear inequalities is very similar to solving linear equations. There is only one thing you need to keep in mind when multiplying or dividing a negative number: the direction or sense of the inequality is reversed. For example, if the sign is >, after multiplying or dividing by a negative number the sign changes to <.

Example: 2421

−≥+ xx → Solve for x as in the previous examples.

424421

−−≥−+ xx

621

−≥ xx

621

−−≥− xxxx

621

−≥− x →

126

21

12

−⋅−≥−⋅− x

12≤x

Multiply times the negative reciprocal to leave the x alone and get an answer. However, since we are multiplying by a negative the inequality sign must change from ≥ to ≤ .

10

EQUATIONS INVOLVING ABSOLUTE VALUE The absolute value of a number is always the positive value of that number. Since the absolute value is the distance of a number from the origin, and since distances are always positive, the absolute value is always a positive value. The same definition applies to equations involving absolute value; any equation inside the absolute value bars must always be equal to a positive number. Example: 84 −=−x → This absolute value equation does not have a real answer.

Any equation inside the absolute value bars must always be equal to a positive number. In this case, for 4−x to be equal to 8− , x must be 4− : 844 −=−− However, since the equation 4−x is inside the absolute value bars it will always yield a positive answer 88 =− and never a negative answer.

When the equation involving absolute value is equal to a positive number, the value of the variable inside the absolute value bars can be negative or positive; therefore, the equation must be made equal to a positive and a negative answer. The answer will have two different possible values of x, and both will make the statement valid. Examples: 1263 =+− x → Make the equation equal to 12 and 12− and solve

for x in each case. 1263 =+− x and 1263 −=+− x

23

63

363

612663

−=−

=−−

=−−=−+−

x

xxx

63

183

3183

612663

=−−

=−−

−=−−−=−+−

x

xxx

The answer to 1263 =+− x is 2−=x and 6=x . 10656 =−+ x →

465 =−x

465 =−x and 465 −=−x

105

64665=

+=+−xx

25

64665=

+−=+−xx

2=x 52=x

The answer is 2=x and 52=x .

Leave the absolute value expression by itself on the left side of the equation. Then solve for x as in the previous example.

11

INEQUALITIES INVOLVING ABSOLUTE VALUE

Inequalities involving absolute value are solved in a similar form as equations involving absolute value. Since the variable inside the absolute value sign can be either positive or negative, both positive and negative values are used to solve for the variable. However, in the case of linear inequalities involving absolute value, the inequality is placed between the two values rather than having two different equations. Examples: 2<x → To solve this linear inequality use both positive and

negative values of 2 and place the inequality in the middle without the absolute value sign. The inequality sign will be the same on both sides.

22 <<− x → This notation is read: x is greater than 2− but less than 2

– 3 –2 –1 0 1 2 3 2104 ≥+− x → Write both positive and negative values of 2 on each

side of the inequality and leave it in the middle. 21042 ≥+−≥− x → Solve for x by moving each term to both sides of

the inequality.

10210104102 −≥−+−≥−− x 8412 −≥−≥− x

48

44

412

−−

≥−−

≥−− x → Remember that when dividing by a negative number

the direction of the inequality sign changes.

23 ≤≤ x → This notation is read: x is greater than or eqal to 3 but less than or equal to 2.

– 3 –2 –1 0 1 2 3 4 5

12

We explained previously that equations that contain an absolute value cannot be equal to a negative number since an absolute value will always give you a positive number. In the case of linear inequalities, if an inequality is greater than or equal to a negative number, then there are infinite solutions. If the inequality is less than or equal to a negative number, then there is no real solution to the inequality. Examples: 22 −≥+x → The absolute value will give you a positive number which

will always be greater than any negative number. The inequality is true and has infinite solutions.

22 −≤+x → The absolute value will give you a positive number and

no positive number will ever be less than or equal to a negative number. The inequality is false and has no real solution.

13

LINEAR INEQUALITIES AND ABSOLUTE VALUE EQUATIONS – EXERCISES

Solve for x: 1. 3 27x ≤ 2. − ≤5 40x 3. − + ≥6 3 21x

4. − ≥23

10x

5. 3 2 8 3− ≥ − +x x 6. 3 2 1 8( )x + < − 7. − + − > −( )x x x1 4 4 3 8. − ≤ + ≤3 4 2 10x

9. − ≤ + + ≤3 2 13

4 6x x

10. 3 7 5 12x − − =

11. 23

4− =x

12. x ≥ 4 13. 2 3 6x − ≤

14

LINEAR INEQUALITIES AND ABSOLUTE VALUE EQUATIONS – ANSWER TO

EXERCISES 1. 3 27x ≤ 2. − ≤5 40x 3. − + ≥6 3 21x x ≤ 9 x ≥ −8 6 21 3x− ≥ −

6 18

3x

x− ≥≤ −

4. − ≥23

10x 5. 3 2 8 3− ≥ − +x x 6. 3 2 1 8( )x + < −

− ≥≤ −2 30

15x

x

− − ≥ − −− ≥ −

≤

2 3 8 35 11

115

x xx

x

6 3 86 11

116

xx

x

+ < −< −

< −

7. − + − > −( )x x x1 4 4 3 8. − ≤ + ≤3 4 2 10x 9. − ≤ + + ≤3 2 13

4 6x x

1 4 4 3x x x− − − > − 3 2 4 10 2x− − ≤ ≤ − 9 6 12 18x x− ≤ + + ≤

9 2

29

x

x

− > −

<

5 4 85 24

x

x

− ≤ ≤

− ≤ ≤ 9 12 7 18 12

21 7 6x

x− − ≤ ≤ −− ≤ ≤

637

x− ≤ ≤

10. 3 7 5 12x − − = 11. 23

4− =x

3 7 12 5x − = + 2 43x

− = 2 43x

− = −

3 7 17x − = 23x

− = 63x

− = −

3 7 17x − = 3 7 17x − = − 6x = − or 18x =

3 248

xx

==

or 3 10

103

x

x

= −

= −

12. x ≥ 4 13. 2 3 6x − ≤ 44 ≥≥− x 6 2 3 6x− ≤ − ≤

6 3 2 6 3

3 2 93 9 2 2

xx

x

− + ≤ ≤ +− ≤ ≤

− ≤ ≤

15

LINEAR FUNCTIONS

As previously described, a linear equation can be defined as an equation in which the highest exponent of the equation variable is one. A linear function is a function of the form ( )f x ax b= + . The graph of a linear equation is a graphical view of the set of all points that make the equation true. The graph of any linear function is a straight line. A linear function can be represented in two ways, standard form and slope-intercept form. Standard form is a formal way of writing a linear equation, while slope-intercept form makes the equation easier to graph.

Form Equation Note Standard Ax By C+ = A and B are not 0. 0A >

Slope-intercept y mx b= + m is the slope of the line and b is the y-intercept.

To graph a linear function we must first identify the x and y intercepts. The x-intercept is the point where the graph crosses the x-axis and the y-intercept is the point where the graph crosses the y-axis.

To find the x-intercept:

1. Set y = 0 in the equation. 2. Solve for x. The value obtained is the x-coordinate of the x-intercept. 3. The x-intercept is the point (x, 0), with x the value found in step 2.

16

To find the y-intercept:

1. Set x = 0 in the equation. 2. Solve for y. The value obtained is the y-coordinate of the y-intercept. 3. The y-intercept is the point (0, y), with y the value found in step 2.

Vertical Lines Equations of the form x = a are vertical lines. The x-coordinate of every point on the vertical line x = a has the value "a," always, for any given value.. Horizontal Lines

Equations of the form y = a are horizontal lines. The y-coordinate of every point on the horizontal line y = b has the value "b," always, for any given x value.

17

−3 −2 −1 1 2 3 4 5 6

4

−3

−2

−1

1

2

3

4

5

x

y

Slope The slope of a line refers to the slant or inclination of the line. The slope is the ratio of the vertical change to the horizontal change between two points on the line. The slope can also be called the rise over run ratio because it tells you how many spaces to move up or down and how many spaces to move to the right. A positive sign will move the line up and a negative sign will move the line down. One important thing to remember is that the run will always be to the right, regardless of the sign.

change in change in

rise y xmrun x y

∆= = =

∆

The formula to find the slope of a line passing through the points ( ) ( )1 1 2 2 and x , y x , y is:

2 1

2 1

y ymx x−

=−

Note: A horizontal line has slope of 0, while a vertical line has an undefined slope. Example:

Find the slope of the line:

You can use any two points on a line to calculate its slope, your answer will be the same no matter which points you choose.

Choosing the points (1, 2) and (2, 0):

212

1220

12

12 −=−

=−−

=−−

=xxyym

18

Parallel Lines In the y-intercept form equation, bmxy += , m is the slope and b is the y-intercept. Two lines are parallel if their slopes are the same ( 21 mm = ) and their y-intercepts are different ( 1 2b b≠ ). Example: 1) 92 += xy 1 2m = 1 9b =

2) 12 −= xy 2 2m = 2 1b = −

In the previous example the slope of both equations is the same and their y-intercepts are different; therefore, the lines are parallel.

−8 −6 −4 −2 2 4 6 8 10

10

−8

−6

−4

−2

2

4

6

8

x

y

Perpendicular Lines

Two lines are perpendicular if the slopes are the negative reciprocal of each other: 2

11

mm −= .

Example: 1) 74 += xy 1 4m = 1 7b =

2) 241

+−= xy 214

m = − 2 2b =

The slope of equation 2 is the negative reciprocal of the slope of equation 1; therefore, the lines are perpendicular.

−8 −6 −4 −2 2 4 6 8 10

10

−8

−6

−4

−2

2

4

6

8

x

y

19

Finding the equation of a line To find the equation of a line when only points or a slope is given, use the point-slope form of a linear equation formula:

( )11 xxmyy −=−

where m is the slope of the line and ( )11, yx is a point on the line. Examples:

Find the equation of the line which passes through the point (–4, 2) and whose slope is 5. Using the point-slope form equation we obtain:

( )11 xxmyy −=− ( )( )452 −−=− xy 2052 +=− xy 225 += xy ← Resultant Equation

Find the equation of the line through the points (–2, 5) and (–6, 4). First find the slope between these two points using the slope equation:

( ) 41

41

2654

12

12 =−−

=−−−

−=

−−

=xxyym

With the slope obtained and one of the two points given, use the point-slope form equation to find the equation of the line.

( )

( )( )

21

415

2415

11

+=−

−−=−

−=−

xy

xy

xxmyy

2

1141

+= xy ← Resultant Equation

Eventually, you may be asked to find the equation of a line that is parallel or perpendicular to a given line. Just remember that the slopes of two parallel lines are exactly the same and that the slopes of two perpendicular lines are the negative reciprocals of each other.

20

Examples:

Find the equation of a line that is parallel to 4 2y x= + and that passes through the point (2, 7).

4 2y x= + ← Since the equation is in slope-intercept form, the coefficient of x is

1 4m = the slope of the line. Remember y=mx+b (m is the slope!).

Since the lines must be parallel, use the same slope and the given point to find the equation of the parallel line: 1 2 4m m= =

( )

( )1 1

7 4 27 4 8

y y m x x

y xy x

− = −

− = −

− = −

4 1y x= − ← Parallel line to 4 2y x= + .

Find the equation of a line perpendicular to 4 2 9x y+ = and that passes through the point (–1, –1).

924 =+ yx ← This equation is in standard form, we need to convert it to slope-intercept form: y=mx+b.

2

94942

+−=

+−=xy

xy

922

y x= − + ← The slope is 1 2m = − .

Since the second equation must be perpendicular to the first equation, find the negative reciprocal.

21

1mm

= − → 21 12 2

m = − =−

Using the slope 212

m = and the point (–1, –1):

( )

( ) ( )( )1 1

11 12

1 112 2

y y m x x

y x

y x

− = −

− − = − −

+ = +

1 12 2

y x= − ← Perpendicular line to 4 2 9x x+ = .

21

LINEAR FUNCTIONS – EXERCISES 1. Identify the false statement below

a) The standard form of a linear equation is y ax b= + b) The point (4, 0) is on the graph of 3 4 12x y− = c) The graph of 3 4 12x y− = passes the vertical line test for functions d) The y-intercept of 3 4 12x y− = is (0, –3)

2. Identify the false statement below

a) To find the x-intercept of a graph, set y equal to zero and solve the equation for x b) The graph of x = 3 is a horizontal line three units above the x-axis c) Generally speaking, it is a good idea to plot three points when constructing the graph of a linear

function d) To find the y-intercept of a graph, set x equal to zero and solve the equation for y

3. The standard form of the equation 5 6 30y x− = is

a) 6 5 30x y− + = b) 6 65

y x= +

c) 5 6 30y x− = d) 6 5 30x y− = − 4. Identify the true statement below

a) All linear graphs are functions b) The graph of 4y = is a function c) To find the y-intercept of a function, let y equal to zero and solve for x d) The graph of 4x = is a function

5. Determine the slope and y-intercept of each equation: a) 6 7 14x y+ = b) 2 5 0x + = c) 3 9 0y + = 6. Use the point slope form to find the equation of the line which passes through (2, –1) and whose

slope is 37

.

7. Write in standard form the equation of the line passing through the point (–3, –4) with slope equal to

32

m = .

8. Which of the following is the equation of a line in standard form passing through the point (3, 0) and perpendicular to the line 2 1y x− = − ?

a) 2 6x y+ = b) 2 3x y+ = c) 2 6x y− = d) 2 3x y− =

9. Find the equation of a line in standard form that passes through the points (–1, 4) and (3, 2). 10. Determine whether the two lines are parallel, perpendicular or either: 3 5 6 and 6 10 7x y x y− = − =

22

LINEAR FUNCTIONS – ANSWERS TO EXERCISES 1. A is the false statement. The standard form of a linear equation is Ax By C+ = . 2. B is the false statement. The graph of x = 3 is a vertical line 3 units to the right of the y-axis. 3. D: 6 5 30 0x y , A− = − > 4. B: 4y = is an example of a linear function.

5. a) 6 and 27

m b= − =

b) and does not existm undefinded b= c) 0 and 3m b= = −

6. 3 137 7

y x= −

7. 3 2 1x y− = − 8. B: 2 3x y+ = is a perpendicular line to 2 1y x− = − 9. 2 7x y+ = 10. The lines are parallel.

23

EXPONENTS Exponents are used to write long multiplications in a short way. The exponent will tell you how many times the number or variable needs to be multiplied. In this case the variable is ‘a’. Examples: ( )( ) 2666 = 4aaaaa =⋅⋅⋅ In exponential notation, the number or variable being multiplied several times is called the base. The exponent, or number that tells you how many times you need to multiply, is called the power.

32 → 2 is the base, and 3 is the power Exponents are mostly used when dealing with variables, or letters, since it is easier and simpler to write

4x than xxxx ⋅⋅⋅ . Also, letters are used for they represent an unknown number, and so, the term: variable. There are a few rules used for simplifying exponents: Zero Exponent Rule – Any number or letter raised to the zero power is always equal to 1. Example: 130 = 10 =a Product Rule – When multiplying the same base, the exponents are added together. Example: 43 xx ⋅ ← Same base, x, add the exponents, 3 + 4 = 7 743 xxx =⋅ Quotient Rule – When dividing the same base, subtract the exponents.

Example: 2

5

xx ← Same base, x, subtract the exponents, 5 – 2 = 3

32

5

xxx

=

24

Power Rule – When the operation contains parentheses, multiply the exponent in the inside with the exponent on the outside. Example: ( )26y ← Multiply the exponents, 1226 =× ( ) 1226 yy = When there is a fraction inside the parentheses, the exponent multiplies on the current power of the numerator and the denominator. However, this rule does not apply if you have a sum or difference within the parentheses; in that case a different rule will apply.

Examples: 169

43

43

2

22

==

( )

( ) 8

12

42

434

2

3

yx

yx

yx

==

!Be careful: 2

++

dcba is not the same as 22

22

dcba

++ !

In fact: 2 2 2

2 2

22

a b a ab bc d c cd d

+ + + = + + +

Negative Signs – If the negative sign is outside the parentheses, perform the operations inside the parentheses and carry out the negative sign to the final answer. Example: ( )33 (3)(3)(3) (27) 27− = − = − = − However, if the negative sign is inside the parentheses, the negative sign will be affected by the exponent. Example: 3( 3) ( 3)( 3)( 3) 27− = − − − = − If the negative sign is inside the parentheses and the exponent is an even number, the answer will be positive. If the exponent is an odd number, then the answer will be negative. Examples: ( ) 82 3 −=− ← Since the negative sign is outside the parentheses, carry it

out to the final answer. ( ) ( )( ) 25555 2 =−−=− ←Since the negative sign is inside the parentheses, it

needs to be carried out through the operation. ( ) ( )( ) 16444 2 =−−=− ← Even number of exponents, positive answer. ( ) ( )( )( ) 644444 3 −=−−−=− ← Odd number of exponents, negative answer.

25

Negative Exponents – Whenever the problem or the answer to the problem contains negative exponents, they need to be changed to positive. An answer with negative exponents will most likely be counted wrong. To change negative exponents into positive exponents, get the reciprocal fraction. In simpler words, if the negative exponent is on the top, move it down; if the negative exponent is on the bottom, move it up. Examples: 2−x ← Get the reciprocal, or move the negative exponent down.

2

22 1

1 xxx ==−

−

35 −y ← Get the reciprocal of only the base with the negative exponent, the

number stays in its place.

3

33 5

155

yyy ==−

−

4

2

−ba ← Get the reciprocal of the base with the negative exponent, the base

with the positive exponent stays in its place.

4242

4

2

1baba

ba

⋅=⋅

=−

26

EXPONENTS – EXERCISES

Simplify: 1. x x5 7⋅ 2. x0 (for x ≠ 0) 3. ( )0ba + (for a b+ ≠ 0)

4. 5 3− 5. 22

2

3 6. ( )23x

7. 2

25 −

8.

2

25

9. ( )23x

10. x yxy

2

11. ( ) 223 − 12. 9 92⋅

13. ( )42a− 14. ( )42a− 15. c c cc c c

7 2

5 2 3

⋅ ⋅⋅ ⋅

16. ( ) ( )43222 abcba − 17. ( ) 11 −−a 18. ( ) ( )33222 34 baba

19. 3

2

3

xx 20.

21 −

k 21. ( )232 −xa

22. ( )( )314

232

−

−

baxa 23. ( )51− 24. ( )91−

25. 32

278 −

27

EXPONENTS – ANSWERS TO EXERCISES

1. x x x5 7 12⋅ = 2. x0 1=

3. ( ) 10 =+ ba 4. 5 15

1125

33

− = =

5. 2

2 3 13

2 12 22 2

− −= = = 6. ( )2 2 2 23 3 9x x x= ⋅ =

7. 2 2 2

2

5 2 2 42 5 5 25

− = = =

8. 2 2

2

5 5 252 2 4

= =

9. ( ) ( )( )2 3 23 6x x x= = 10. ( ) ( )2

2 1 1 1 1 0x y x y x y xxy

− −= ⋅ = ⋅ =

11. ( )811

3133 4

422 === −− 12. 9 9 92 3⋅ =

13. ( ) ( )( )( )( )42 2 2 2 2 8a a a a a a− = − = − 14. ( ) ( )( )( )( )42 2 2 2 2 8a a a a a a− = − − − − =

15. c c cc c c

cc

7 2

5 2 3

10

10 1⋅ ⋅⋅ ⋅

= = 16. ( ) ( ) 6101043222 cbaabcba =−

17. ( ) aa =

−− 11 18. ( ) ( ) 111033222 43234 bababa =

19. 33 9

32 6

x x xx x

= =

20.

2 22

2

1 1 kk k

− −

− = =

21. ( ) 6

464232

xaxaxa == −− 22. ( )

( ) 68

3

312

64

314

232

xab

baxa

baxa

== −

−

−

−

23. ( )51 1− = − 24. ( )91 1− = −

25. 2 2 2 233

33

8 8 8 2 927 27 3 427

− −− − = = = =

28

POLYNOMIALS

Polynomials can be defined as the sum or difference of terms or expressions. Each term can be either a constant or variable, have one or more terms, and be composed of like terms or different terms. The expressions that can exist in a polynomial are defined as: Integer- a negative or positive whole number including zero. Example: -5, -4….-1,0,1,…5,6,7…etc… Constants – A single number in the equation that does not contain any variable. Example: 4, 6 Coefficient – The numerical part of a monomial. Example: 7 is the coefficient of 37x y Degree – The highest power to which a variable is raised. Examples: Identify each term in the following polynomial: 42 3 −x Constant: 4− → the single number in the equation Coefficient: 2 → the number in front of the variable x Degree: 3 → the highest power of the variable Give the degree of the following polynomial: 81135 246 +−− xxx Degree: 6 → It is a sixth degree polynomial because the highest exponent of x is 6. There are also different types of polynomials: Monomial – A constant, or the product of a constant, and one or more variables raised to an integer. Example: 2 33x y z− Polynomial – Any finite sum (or difference) of terms. Example: 3 2 2 34 3 9 2x y z x y xz− + − Binomial – A polynomial consisting of exactly two terms. Example: 2 7x − Trinomial – A polynomial consisting of exactly three terms. Example: 3 4x x− + There are special binomial rules that can be followed that can make them easier to solve.

29

( )( ) 222

222

2

2

bababa

bababa

+−=−

++=+

( )( ) 32233

32233

33

33

babbaaba

babbaaba

−+−=−

+++=+

( )( ) 22 bababa −=−+ Examples: ( )22+x → Solve using the binomial properties ( ) 222 2 bababa ++=+

( ) ( )( ) 222 2222 ++=+ xxx ( ) 442 22 ++=+ xxx ( )33−y → Solve using the binomial properties ( ) 32233 33 babbaaba −+−=− ( ) ( ) ( ) ( )( ) 32233 333333 −+−=− yyyy ( ) 272793 233 −+−=− yyyy

Combining like Terms

Polynomials can be short expression or really long ones. Probably the most common thing you will be doing with polynomials is combining like terms in order to simplify long polynomial expressions. Combining “like terms” is the process by which we combine exact same terms containing the same variable with the same exponent together to shorten the expression. Examples:

3 4x x+ → Both terms contain the same variable. Therefore, they can be combined.

3 4 7x x x+ =

2 22 3 4 9x x x x+ − − + + → Combine like terms together and then simplify.

( ) ( ) ( )

2 2

2 2

2

2 3 4 9

2 3 4 9

4 5

x x x x

x x x x

x x

+ − − + +

− + + + − +

+ +

30

Evaluating Polynomials To evaluate polynomials substitute the variables for the given value and solve the equation. Examples: Evaluate 3 22 4 2x x x− − + at 3−=x .

( ) ( ) ( )3 22 3 3 4 3 2− − − − − + → Plug in –3 for x; be careful with parentheses and negative signs.

( ) ( )2 27 9 12 254 9 1463 1449

− − + +

− − +− +−

If ( ) 25 4 7P x x x= − + , find ( )2P .→ This is the same thing as plugging a 2 for ‘x’

( ) ( ) ( )( )

( )

22 5 2 4 2 7

5 4 8 7

2 19

P

P

= − +

= − +

=

Addition and Subtraction of Polynomials

The addition and subtraction of polynomials consist of combining like terms by grouping together the same variable terms with the same degrees. In the case of subtraction, if the subtraction sign (or negative sign) is outside parentheses, the first thing to do is to distribute the negative sign to each of the terms inside the parentheses. Examples:

Simplify: ( ) ( )3 2 3 23 3 4 5 2 4x x x x x x+ − + + − + −

( ) ( )3 2 3 23 3 4 5 2 4x x x x x x+ − + + − + − → Combine same degree terms together.

( ) ( ) ( ) ( )

3 2 3 2

3 3 2 2

3 2

3 3 4 5 2 4

3 3 2 4 5 4

4 3 1

x x x x x x

x x x x x x

x x x

+ − + + − + −

+ + − + − + + −

= + − +

Simplify: ( ) ( )2 5 3 2x y x y+ + −

31

2 5 3 2x y x y+ + − → Combine x’s with x’s and y’s with y’s. ( ) ( )2 3 5 2x x y y+ + − 5x + 3y Simplify: ( ) ( )3 2 3 23 5 4 3 8 5 6x x x x x x+ + − − − − +

( ) ( )3 2 3 23 5 4 3 8 5 6x x x x x x+ + − − − − +

( ) ( ) ( ) ( )

3 2 3 2

3 3 2 2

3 2

3 5 4 3 8 5 6

3 3 8 5 5 4 6

2 11 10 10

x x x x x x

x x x x x x

x x x

+ + − − + + −

− + + + + + − −

= − + + −

Multiplication of Polynomials To multiply polynomials the same rules apply as with exponents since we are dealing with variables with different exponents. Remember that when multiplying, the exponents add together. Example: ( )( )23 103 xx → Multiply the constant terms together first and then the variables. ( )( )23103 xx ⋅⋅= → Add exponents together. 530x= Next is the one-term polynomial times a multi-term polynomial. Distribute the one-term polynomial outside the parentheses to each term inside the polynomial. Example: ( )23 4 10x x x− − + → Distribute the –3x to all the terms inside the parentheses.

( ) ( ) ( )103343 2 xxxxx −−−−= xxx 30312 23 −+−=

Multiplication of two two-term polynomials is a little more complex. To make it easier we use the FOIL method: a method that simplifies the process by following a pattern. FOIL stands for First, Outer, Inner and Last. Start with the first terms in each parenthesis, followed by the outer terms, then the inner terms, and finally the last terms. This method makes it easier to remember which terms to multiply and reduces the chance of forgetting to multiply some terms. Example:

Use the FOIL method to simplify ( )( )2 5 3x x+ + "first": ( )( ) 22 2x x x= ( )( )2 5 3x x+ +

32

"outer": ( )( )2 3 6x x= 22 6 5 15x x x+ + +

"inner": ( )( )5 5x x= 22 11 15x x= + +

"last": ( )( )5 3 15= In order to multiply one multi-term polynomial by another multi-term polynomial, break the smaller polynomial apart and multiply each individual term by the longer polynomial. Example:

( )( )3 22 3 4 17x x x x+ + + − → Take the smaller polynomial and multiply each of its individual terms by the largest polynomial.

( )( ) ( )( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

3 2 3 2

3 2 3 2

4 3 2 3 2

4 3 3 2 2

4 3 2

3 4 17 3 4 17 2

3 4 17 2 3 2 4 2 17 2

3 4 17 2 6 8 34

3 2 4 6 17 8 34

5 10 9 34

x x x x x x x

x x x x x x x x x x

x x x x x x x

x x x x x x x

x x x x

+ + − + + + −

+ + − + + + −

+ + − + + + −

+ + + + + − + −

= + + − −

Division of Polynomials When dealing with polynomials divided by a single term the division can be treated as a simplification problem and the action is just to reduce its terms to the lowest possible.

Examples: Simplify: 2 42

x + → In this case, there is a common factor in the numerator (top)

and denominator (bottom), so it is easy to reduce this fraction.

There are two ways of proceeding. Split the division into two fractions, each with only one term on top, and then reduce:

2 4 2 4 2

2 2 2x x x+

= + = +

Or factor out the common factor from the top and bottom and then cancel:

( )2 22 4 22 2

xx x++

= = +

Either way, the answer is the same: 2x +

Simplify: x

xx9

3618 34 +

33

Method 1: 233434

429

369

189

3618 xxxx

xx

xxx

+=+=+

Method 2: If you recognize a common factor that can be taken out of the

parentheses and used to cancel the term in the denominator use it.

( ) 2323

429

429 xxx

xxx+=

+

The answer is the same by any method: 23 42 xx +

Long Division

If you come about a more complicated polynomial division you can use the long division method to do the operation. Long division of polynomial works just as a regular long division, with the exception that in this case variables are included. Example: Divide 1262 −− xx by 2+x . Step 1 – Set up the division.

1262 2 −−+ xxx

Step 2 – Let’s look at the first term inside, 2x . On the outside we have an x, so to get to

2x we need to multiply xx ⋅ . Write this term on top.

1262 2x

xxx −−+

Step 3 – Multiply the term on the top by the term outside and write it at the bottom,

( ) xxxx 22 2 +=+ . However, to be able to eliminate the terms we need to change the sign, xx 22 −− .

xx

xxxx

2

1262

2

2

−−

−−+

Step 4 – Perform the required operations.

34

xxx

xxxx

8 2

1262

2

2

−−−

−−+

Step 5 – Bring down the next term inside the division, in this case 12− .

128

2

1262

2

2

−−−−

−−+

xxx

xxxx

Step 6 – Repeat step 2 through 4 to eliminate the remaining terms.

4 168 128

2

8 1262

2

2

+−−

−−

−−−+

xxxx

xxxx

The solution to the long division is 8−x . Divide 3 23 5 10 3x x x− + − by3 1x + :

2

3 22 4

3 1 3 5 10 3x x

x x x x− +

+ − + −

3 23x x− − ( )2 3 23 1 3x x x x+ = + , to subtract, ( )3 2 3 23 3x x x x− + = − −

26 10x x− + Subtract polynomials and carry down the next term, which is 10x 2 6 2 x x+ ( ) 22 3 1 6 2x x x x− + = − − , to subtract, ( )2 26 2 6 2x x x x− − − = + 12 3x − Subtract polynomials and carry down the next term, 3−

12 4x− − ( )4 3 1 12 4x x+ = + , to subtract, ( )12 4 12 4x x− + = − − 7−

Since the division did not come out even, the answer is the polynomial on top of the division plus the

remainder as a fraction with the divisor as the denominator: 2 72 43 1

x xx−

− + ++

.

35

POLYNOMIALS – EXERCISES Perform the indicated operations: 1. ( ) ( )7 4 8 3x x+ + −

2. ( ) ( )2 26 3 4 2 7x x x x+ + + − −

3. ( ) ( )4 3 21 3 3 10x x x+ + − + +

4. ( ) ( ) ( )22 2 3 2x x x− + + + +

5. ( ) ( )3 2 22 6 9 2 7x x x x x+ − + + − +

6. ( ) ( )2 21 08 3 6x x x x− + − − + +

7. ( ) ( )2 28 13 7 4 7 1x x x x+ − − + −

8. ( ) ( )2 22 5 4 4 7 1x x x x− + − − −

9. ( ) ( ) ( )2 2 26 2 2 5 3 3 5 5x x x x x x+ + + − − − − + +

10. ( ) ( ) ( )2 22 1 3 2 2 7x x x x− + − + − +

11. ( )( )1 4 3x x+ +

12. ( )( )22 3 3 1x x x+ + +

13. ( )( )24 1 3 7 2x x x− + −

14. ( )( )2 22 3 2x y x xy y− + +

15. ( )( )2 26 2 2x y x xy y+ − − Solve by long division: 16. ( ) ( )24 8 2 2x x x− + ÷

17. ( ) ( )3 28 16 20 4 4x x x x+ + + ÷

18. ( ) ( )27 2 8 1x x x− + ÷ +

19. ( ) ( )2 25 6x xy y x y− − ÷ +

20. ( ) ( )28 2 15 4 5x x x− − ÷ +

36

POLYNOMIALS – ANSWER TO EXERCISES

1. ( ) ( )7 4 8 3 15 1x x x+ + − = +

2. ( ) ( )2 2 26 3 4 2 7 5 4 4x x x x x x+ + + − − = + −

3. ( ) ( )4 3 2 4 3 21 3 3 10 3 3 11x x x x x x+ + − + + = − + +

4. ( ) ( ) ( )2 22 2 3 2 2 6x x x x x− + + + + = + +

5. ( ) ( )3 2 2 3 22 6 9 2 7 3 8 16x x x x x x x x+ − + + − + = + − +

6. ( ) ( )2 2 210 8 3 6 2 13 2x x x x x x− + − − + + = − +

7. ( ) ( )2 2 28 1 37 4 7 1 4 6 6x x x x x x+ − − + − = + −

8. ( ) ( )2 2 22 5 4 4 7 1 2 2 5x x x x x x− + − − − = − + +

9. ( ) ( ) ( )2 2 2 26 2 2 5 3 3 5 5 6 4 6x x x x x x x x+ + + − − − − + + = − −

10. ( ) ( ) ( )2 2 22 1 3 2 2 7 5 8x x x x x x− + − + − + = − − −

11. ( )( ) 21 4 3 4 7 3x x x x+ + = + +

12. ( )( )2 3 22 3 3 1 2 9 11 3x x x x x x+ + + = + + +

13. ( )( )2 3 24 1 3 7 2 12 25 15 2x x x x x x− + − = + − +

14. ( )( )2 2 3 2 2 32 3 2 2 6x y x xy y x x y xy y− + + = − + −

15. ( )( )2 2 3 2 2 36 2 2 2 11 8 12x y x xy y x x y xy y+ − − = + − −

16. ( ) ( )2 24 8 2 2 2 42

x x x xx

− + ÷ = − +

17. ( ) ( )3 2 2 48 16 20 4 4 2 4 54

x x x x x xx

+ + + ÷ = + + +

18. ( ) ( )2 177 2 8 1 7 91

x x x xx

− + ÷ + = − ++

19. ( ) ( )2 25 6 6x xy y x y x y− − ÷ + = −

20. ( ) ( )28 2 15 4 5 2 3x x x x− − ÷ + = −

37

FACTORING Factoring is similar to breaking up a number into its multiples. For example, 10=5*2. The multiples are ‘5’ and ‘2’. In a polynomial it is the same way, however, the procedure is somewhat more complicated since variables, not just numbers, are involved. There are different ways of factoring an equation depending on the complexity of the polynomial. Factoring out the greatest common factor The first thing to do when factoring is to look at all terms and break up each term into its multiples: Examples: Factor: xxx 1048 23 ++

( )1048 2 ++ xxx → In this polynomial the only variable in common to all is x.

2x(4x+2x+5) → The two is also common to all terms. Therefore, this

is as far as the polynomial can be factored.

Factor: xxx 1686 23 ++

xxx 1686 23 ++ → The common variable is x, and the smallest exponent is 1. The common multiple is 2.

( )8432 2 ++ xxx Therefore, the greatest common factor is 2x. You can also look at it this way: ( )( ) 32 632 xxx =

( )( )( )( ) xx

xxx1682

842 2

==

Factoring by grouping

38

There will be occasions when there are no common factors for all terms, but there will be terms that have a variable in common while other terms have a different variable in common. In this case, we can factor by grouping together these common terms. Examples: Factor: bybxayax +++ bybxayax +++ → Notice how there are two variables, x and y. Group

byaybxax +++ together the x’s and the y’s and then ( ) ( )baybax +++ factor out the common factor in each group.

Factor: 3322 224 yyxx −+−

3322 224 yyxx −+− → There are two variables present, x and y. Start by grouping two terms at a time.

( ) ( )222 232 −+− xyx → Notice how the two polynomials inside the parentheses are similar. To make them the same change the sign in the second polynomial.

( ) ( )232 222 xyx −−− → Now that they are the same, factor out the common term.

( )( )32 22 yx −− → Now it is completely factored out. Factoring a difference of squares For a difference of squares, 22 ba − , the factors will be ( )( )baba +− . Examples: ( )( )yxyxyx +−=− 22

( ) ( ) ( )( ) ( )( )( )224 2 2 2 21 1 1 1 1 1 1x x x x x x x− = − = − + = − + + Factoring the sum and difference of two cubes For the addition or subtraction of two cubes, the following formulas apply:

( )( )( )( )

3 3 2 2

3 3 2 2

a b a b a ab b

a b a b a a bb

+ = + − +

− = − + +

Examples: 3 8x + ( ) ( )3 32x= + → Manipulate to be in “ 3 3a b+ ” form.

( ) ( )( )( )2 22 2 2x x x= + − + → follow formula.

39

( )( )22 2 4x x x= + − + 38 27y − → Manipulate to be in “ 3 3a b− ” form.

( ) ( )3 32 3y= − → follow formula.

( ) ( ) ( )( )( )2 22 3 2 2 3 3y y y= − + +

( )( )22 3 4 6 9y y y= − + + Factoring trinomials Factoring trinomials is based on finding the two integers whose sum and product meet the given requirements. Example: Find two integers whose sum is –11 and whose product is 30. Step 1 – The product is positive, the sum is negative, therefore, both integers must be

negative. ( )( ) ( ) ( ) and − − = + − + − = − Step 2 – Possible pairs of factors that will give a positive answer:

1. ( )( )1 30 30− − = 3. ( )( )3 10 30− − =

2. ( )( )2 15 30− − = 4. ( )( )5 6 30− − =

Step 3 – The pair whose sum is –11 is –5 and –6. Therefore, the critical integers are –5 and –6, ( )5 6 11− − = − .

Factoring trinomials of the form x bx c2 + + To factor trinomials of the form x bx c2 + + follow the same principle described above. Find two integers whose sum equals the middle term and whose product equals the last term. Examples: Factor: 62 −− xx

Step 1 – Find two integers whose sum is equal to –1 and whose product is equal to –6. The product is negative; therefore, we need integers that have different sign. The sum is negative, so the bigger integer is negative.

Step 2 – Possible pairs of factors that will give us –6:

1. ( )( ) 623 −=− 2. ( )( ) 623 −=− 3. ( )( ) 661 −=− 4. ( )( ) 661 −=−

Step 3 – The pair whose sum is –1 is –3 and 2, ( )123 −=+− .

The critical integers are –3 and 2.

40

Step 4 – The factors for 62 −− xx are )2)(3( +− xx .

Factor: x x2 16 60+ +

Step 1 – Find two integers whose sum is 16 and whose product is 60. Both integers must

be positive since their sum and product are positive. Step 2 – Possible pairs of factors:

1. ( )( )1 60 60= 3. ( )( )3 20 60= 5. ( )( )5 12 60=

2. ( )( )2 30 60= 4. ( )( )4 15 60= 6. ( )( )6 10 60= Step 3 – The pair whose sum is 16 is (6, 10). Therefore, the critical integers are 6, 10. Step 4 – The factors for 2 16 60x x+ + are ( )( )x x+ +6 10 . Factoring trinomials of the form ax bx c2 + + In this type of trinomials the coefficient of x2 is not 1, therefore we must consider the product of ca ⋅ . Examples: Factor: 2 7 42x x− − Step 1 – Find two integers whose sum is –7 and whose product is ( )( )2 4 8− = − . The

integers must be of different signs since their product is a negative number. The larger integer is negative since their sum is negative.

Step 2 – Possible pairs of factors: 1. ( )( )1 8 8− = − 3. ( )( )2 4 8− = −

2. ( )( )1 8 8− = − 4. ( )( )2 4 8− = − Step 3 – The pair whose sum is –7 is 1 and –8. The critical integers are 1 and –8. Step 4 – Use the critical integers to break the first degree term into two parts:

2 7 4 2 8 42 2x x x x x− − = − + −

Step 5 – Factor by grouping the first two and last two terms:

2 8 4 2 4 1 42x x x x x x− + − = − + −( ) ( )

41

There are now two terms in the expression. The binomial factor in each of these two terms should be the same; otherwise, there is an error. In this case, the common binomial factor is ( )4x − .

Step 6 – Factor out the common binomial factor: 2 4 1 4 4 2 1x x x x x( ) ( ) ( )( )− + − = − +

Factor: 6 23 202x x− + Step 1 – Find two integers whose sum is –23 and whose product is 6(20) = 120. Since

the product is positive and the sum is negative, the two integers are negative. Step 2 – Possible pairs of factors:

1. ( )( )1 120 120− − = 4. ( )( )4 30 120− − = 7. ( )( )8 15 120− − =

2. ( )( )2 60 120− − = 5. ( )( )5 24 120− − = 8. ( )( )10 12 120− − =

3. ( )( )3 40 120− − = 6. ( )( )6 20 60− − = Step 3 – The sum of –8 and –15 is –23. Therefore, the critical integers are –8 and –15. Step 4 – Use the critical integers to break the first degree term into two parts:

6 23 20 6 15 8 202 2x x x x x− + = − − + Step 5 – Factor, separately, the first two and last two terms:

6 15 8 20 3 2 5 4 2 52x x x x x x− − + = − − −( ) ( ) Step 6 – Factor out the common binomial factor.

3 2 5 4 2 5 2 5 3 4x x x x x( ) ( ) ( )( )− − − = − −

42

FACTORING – EXERCISES

1. 8103 2 −− xx 2. ( )3(3)3(5)32 2 −−−−− xxxxx 3. 1892 23 −−+ xxx 4. 22 2510 yxx −+− 5. 66 yx −

6. 3 18

a −

7. 4)5(4)5( 2 ++++ xx 8. 6321 tt +− 9. 234 27183 xxx +− 10. 600100100 2 −− xx 11. 1528 2 −− xx 12. 22 23 yxyx ++ 13. )2(15)2(13)2(2 2 +++++ xxxxx 14. 122 −+ xx 15. 623 22 +++ xyyx 16. 5018 2 −a 17. 1582 ++ xx 18. xxx 502016 23 −− 19. 30339 2 ++ xx 20. 2212 2 −+ xx

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FACTORING – ANSWER TO EXERCISES 1. 8103 2 −− xx 2. ( ) ( ) ( )22 3 5 3 3 3x x x x x− − − − −

82123 2 −+− xxx )352)(3( 2 −−− xxx ( ) ( )3 4 2 4x x x− + − )362)(3( 2 −+−− xxxx

( )( )3 2 4x x+ − ))3(1)3(2)(3( −+−− xxxx

( )( )( ) ( ) ( )23 2 1 3 3 2 1x x x x x− + − = − + 3. 1892 23 −−+ xxx 4. 22 2510 yxx −+− )2(9)2(2 +−+ xxx 22 )2510( yxx −+− )2)(9( 2 +− xx 22)5( yx −− )3)(3(92 −+=− xxx Difference of squares: ))(()( 22 bababa +−=− )2)(3)(3( ++− xxx where ( )5 and a x b y= − =

)5)(5( yxyx +−−−

5. 66 yx − 6. 3 18

a −

( ) ( )2 26 6 3 3x y x y− = − Difference of cubes:

Difference of squares: ))(()( 22 bababa +−=− ( )( )3 3 2 2a b a b a a bb− = − + +

where 3 3 and a x b y= = aa = and 21

=b

( )( )6 6 3 3 3 3x y x y x y− = − + ( )2

21 1 12 2 2

a a a = − + +

Sum & Difference of Cubes: 21 12 2 4

aa a = − + +

( )( )3 3 2 2a b a b a ab b+ = + − +

( )( )3 3 2 2a b a b a a bb− = − + +

= ))()()(( 2222 yxyxyxyxyxyx ++−+−+ = ))()()(( 2222 yxyxyxyxyxyx +−++−+

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7. ( ) ( ) ( )( ) ( )2 25 4 5 4 7 7 7x x x x x+ + + + = + + = +

8. ( )( ) ( ) ( )223 6 3 3 21 2 1 1 1 1t t t t t t t− + = − − = − + + 9. ( ) ( )24 3 2 2 2 23 18 27 3 6 9 3 3x x x x x x x x− + = − + = − 10. ( )( )2100 100 600 100 3 2x x x x− − = − + 11. ( )( )28 2 15 4 5 2 3x x x x− − = + − 12. ( )( )2 23 2 2x xy y x y x y+ + = + + 13. ( ) ( ) ( ) ( )( )( )22 2 13 2 15 2 2 2 3 5x x x x x x x x+ + + + + = + + + 14. ( )( )2 12 4 3x x x x+ − = + − 15. ( )( )2 2 23 2 6 3 2x y y x x y+ + + = + + 16. ( )( )218 50 2 3 5 3 5a a a− = + − 17. ( )( )2 8 15 5 3x x x x+ + = + + 18. ( )( )3 216 20 50 2 2 5 4 5x x x x x x− − = − + 19. ( )( )29 33 30 3 2 3 5x x x x+ + = + + 20. ( )( )212 2 2 2 3 1 2 1x x x x+ − = − +