The University of Texas at El Paso
Tutoring and Learning Center
ACCUPLACER
MATH 0310
http://www.academics.utep.edu/tlc
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
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:
x+2=4
← 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:
x+2=4
x = 4−2
x=2
← The value of the unknown variable that makes the equation
true is 2: 2 + 2 = 4
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 x + a = b or x − a = b .
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:
x − 5 = 10
x − 5 + 5 = 10 + 5
x = 15
← Add 5 to both sides to eliminate the number on
the left side of the equation and leave x alone.
y + 3.2 = 4.1
y + 3.2 − 3.2 = 4.1 − 3.2
y = 0.9
← Subtract 3.2 to both sides to eliminate the
number on the left side and leave y alone.
2. Equations of the form ax = b .
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.
2
Examples:
3 x = 18
3
18
x=
3
3
x=6
← Divide both sides by 3 to leave x by itself.
3
2
x=
4
5
4 3
2 4
⋅ x= ⋅
3 4
5 3
8
x=
15
← Multiply both sides by the reciprocal of
3
4
, which is .
4
3
3. Equations of the form ax + b = c .
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:
3 y + 7 = 25
3 y + 7 − 7 = 25 − 7
3 y = 18
3
18
y=
3
3
y=6
8
x + 2 = −1
5
8
x + 2 − 2 = −1 − 2
5
8
x = −3
5
5 8
3 5
⋅ x=− ⋅
8 5
1 8
15
x=−
8
3
← Subtract 7 to both sides to remove it from the left side.
← Divide both sides by 3 to leave y alone and solve for y.
← Follow the steps described above to solve for x.
4. Equations of the form ax + b = cx + d .
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:
← Start by moving the variable from the right
side to the left side.
← Combine the two variables together.
← Add 15 to both sides to eliminate the
constant on the left side of the equation.
← Divide both sides of the equation by 14.
6 x − 15 = −8 x + 13
6 x + 8 x − 15 = −8 x + 8 x + 13
14 x − 15 = 13
14 x − 15 + 15 = 13 + 15
14 x = 28
14
28
← Solve for x.
x=
14
14
x=2
Distributive Property:
a(b + c ) = ab + ac
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:
2( x − 3) + 1 = −7
2 x − 6 + 1 = −7
2 x − 5 = −7
2 x − 5 + 5 = −7 + 5
2 x = −2
2
2
x=−
2
2
x = −1
← Multiply the term outside the parentheses to each of
the terms inside the parentheses.
← Eliminate the constant on the left side of the equation
by adding 5 to both sides of the equation.
← Divide both sides by 2 and solve for x.
4
LINEAR EQUATIONS – EXERCISES
Solve for x:
1.
8 x − 6 = 12
2.
4x − 5 = 2x + 7
3.
x − 1 − 3(5 − 2 x ) = 2 ( 2 x − 5)
4.
2x − 4 4x + 3
=
4
3
5.
2x − a 4x − b
=
b
a
6.
3
4
1
− =
−6
x +1 5 x +1
7.
3
4
8
+
=
x −1 x +1 x +1
8.
1 1 1
+
=
ax bx c
9.
3 4
−
= 10
x 3x
10.
2x +1 x − 4 2
+
=
5
6
3
11.
3( 2 x − 1) + x = 5 x + 3
12.
3 − 5( 2 x − 5) = −2
13.
−3(5 x + 7 ) − 4 = −10 x
14.
8 + 3( x + 2 ) = 5( x − 1)
15.
5( x + 2 ) − 4 ( x + 1) = 7
16.
1
3
7
− =
2x +1 4 2x +1
17.
7
4
+
=6
2x − 3 2x − 3
5
LINEAR EQUATIONS – ANSWERS TO EXERCISES
1. 8 x − 6 = 12
8 x − 6 + 6 = 1 +26
8 x = 18
9
x=
4
4.
3. x − 1 − 3 ( 5 − 2=
x ) 2 ( 2 x − 5)
x − 1 − 15 + 6 x = 4 x − 10
2. 4 x − 5 = 2 x + 7
4 x − 2 x =7 + 5
2 x = 12
x=6
2x − 4 4x + 3
=
4
3
3 ( 2 x − 4=
) 4 ( 4 x + 3)
7 x − 4 x = −10 + 1 + 15
3x = 6
x=2
5.
2ax − a 2 = 4bx − b 2
6 x − 12 = 16 x + 12
6 x − 16 x =12 + 12
−10 x =
24
12
x= −
5
6.
8.
3
4
1
− =
−6
x +1 5 x +1
4
 3
 1

− =
− 6
5 ( x + 1) 
5 ( x + 1) 
 x +1 5 
 x +1 
2x − a 4x − b
=
b
a
a ( 2x − a=
) b ( 4x − b)
2ax − 4bx =a 2 − b 2
x ( 2a − 4b ) =a 2 − b 2
x=
7.
a 2 − b2
2a − 4b
3
4
8
+
=
x −1 x +1 x +1
3
4 
8 
+
( x − 1)( x + 1) 
( x + 1) 
=

 x −1 x +1 
 x +1 
15 − 4 ( x + 1) =5 − 30 ( x + 1)
3 ( x + 1) + 4 ( x − 1)= 8 ( x − 1)
15 − 4 x − 4 = 5 − 30 x − 30
26 x = −36
18
x= −
13
3x + 3 + 4 x − 4 = 8 x − 8
7 x − 8 x =−
1 8
− x =−7
x=7
1 1 1
+
=
ax bx c
9.
3 4
−
= 10
x 3x
6
1 
 1
1
abcx  +  =
abcx  
 ax bx 
c
bc + ac =
abx
bc + ac
=x
ab
c (a + b)
x=
ab
2x +1 x − 4 2
10.
+
=
5
6
3
2
x
+
1
x
−
4


2
30 
+
30  
=
6 
 5
3
6 ( 2 x + 1) + 5 ( x − 4 ) =
20
3 4 
3x  −  =
3 x (10 )
 x 3x 
9−4 =
30 x
1
x=
6
11. 3 ( 2 x − 1) + x = 5 x + 3
6 x − 3 + x = 5x + 3
7 x − 5 x =3 + 3
2x = 6
x=3
12 x + 6 + 5 x − 20 =
20
17 x = 34
x=2
12. 3 − 5( 2 x − 5) = −2
3 − 10 x + 25 =
−2
−10 x =−2 − 3 − 25
−10 x =
−30
x=3
13. −3(5 x + 7 ) − 4 = −10 x
−15 x − 21 − 4 =−10 x
−15 x + 10 x =21 + 4
−5 x =
25
x = −5
14. 8 + 3( x + 2 ) = 5( x − 1)
8 + 3x + 6 = 5 x − 5
3 x − 5 x =−5 − 8 − 6
−2 x =
−19
19
x=
2
15. 5( x + 2 ) − 4 ( x + 1) = 7
16.
1
3
7
− =
2x +1 4 2x +1
3
 1
 7 
− = 4 ( 2 x + 1) 
4 ( 2 x + 1) 

 2x +1 4 
 2x +1 
4 − 3 ( 2 x + 1) =
28
4 − 6x − 3 =
28
−6 x − 3 =
24
27
x= −
6
7
5 x + 10 − 4 x − 4 = 7
x = 7 − 10 + 4
x =1
17.
7
4
+
=6
2x − 3 2x − 3
7
4 
+
( 2 x − 3) 
=
 2x − 3 2x − 3 
7 +=
4 6 ( 2 x − 3)
11 12 x − 18
=
11 + 18 =
12 x
29
x=
12
( 2 x − 3) ⋅ 6
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:
→
→
→
→
x<3
2x + 5 ≤ x − 3
x > −2
3x − 2 ≥ x − 6
less than, <
less than or equal to, ≤
greater than, >
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:
x<3
→
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.
Notice how 3 is marked with an
empty dot, this means that the answer
does not include the number 3.
–3 – 2 –1 0
2x + 5 ≤ x − 3
→
2x + 5 ≤ x − 3
2x + 5 − 5 ≤ x − 3 − 5
2x ≤ x − 8
2x − x ≤ x − x − 8
x ≤ −8
1
2
3
This inequality is telling us that the equation 2 x + 5 is
less than or equal to the equation x − 3 . Solve for x to find
which number, or numbers, make this equation true.
→
Solve for x by leaving the variables on one side
and the numbers on the other side, just as you
would solve a linear equation.
Notice how –8 is marked with a black
dot, this means that the answer does
include the number –8.
–9 –8 –7 –6 –5 –4 –3 –2 –1
x > −2
→
0
This inequality is telling us that “x is greater than –2”. Therefore,
any number greater than negative two is a possible solution.
8
–3 –2 –1
0
→
3x − 2 ≥ x − 6
3x − 2 + 2 ≥ x − 6 + 2
3x ≥ x − 4
3x − x ≥ x − x − 4
2 x ≥ −4
2x − 4
≥
2
2
x ≥ −2
1
–3 –2 –1
2
3
Solve for x by leaving the variables on one side
and the numbers on the other side.
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:
1
x+4≥ x−2
2
1
x+4−4≥ x−2−4
2
1
x ≥ x−6
2
1
x− x ≥ x− x−6
2
1
− x ≥ −6
2
2 1
2
− ⋅ − x ≥ −6 ⋅ −
1 2
1
x ≤ 12
9
→
Solve for x as in the previous examples.
→
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 ≤ .
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:
x − 4 = −8
→
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 x − 4 to be equal to − 8 , x must be − 4 : − 4 − 4 = −8
However, since the equation x − 4 is inside the absolute value bars it will always
yield a positive answer − 8 = 8 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:
− 3 x + 6 = 12
− 3 x + 6 = 12
− 3 x + 6 − 6 = 12 − 6
→
Make the equation equal to 12 and − 12 and solve
for x in each case.
and
− 3x = 6
6
− 3x
=
−3 −3
x = −2
− 3 x + 6 = −12
− 3 x + 6 − 6 = −12 − 6
− 3 x = −18
− 3 x − 18
=
−3
−3
x=6
The answer to − 3 x + 6 = 12 is x = −2 and x = 6 .
6 + 5 x − 6 = 10
→
5x − 6 = 4
5x − 6 = 4
5x − 6 + 6 = 4 + 6
and
5 x = 10
x=2
Leave the absolute value expression by itself
on the left side of the equation. Then solve
for x as in the previous example.
5 x − 6 = −4
5 x − 6 + 6 = −4 + 6
5x = 2
x=2
5
The answer is x = 2 and x = 2 .
5
10
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:
x <2
→
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.
−2< x< 2
→
This notation is read: x is greater than − 2 but less than 2
– 3 –2 –1
0
1
2
3
− 4 x + 10 ≥ 2
→ Write both positive and negative values of 2 on each
side of the inequality and leave it in the middle.
− 2 ≥ −4 x + 10 ≥ 2
→ Solve for x by moving each term to both sides of
the inequality.
− 2 − 10 ≥ −4 x + 10 − 10 ≥ 2 − 10
− 12 ≥ −4 x ≥ −8
− 12 − 4 x − 8
→ Remember that when dividing by a negative number
≥
≥
−4
−4 −4
the direction of the inequality sign changes.
3≤ x≤2
than or equal to 2.
→
This notation is read: x is greater than or eqal to 3 but less
– 3 –2 –1
11
0
1
2
3
4
5
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:
x + 2 ≥ −2
→
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.
x + 2 ≤ −2
→
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.
12
LINEAR INEQUALITIES AND ABSOLUTE VALUE EQUATIONS – EXERCISES
Solve for x:
1.
3 x ≤ 27
2.
−5 x ≤ 40
3.
−6 x + 3 ≥ 21
4.
2
− x ≥ 10
3
5.
3 − 2 x ≥ −8 + 3 x
6.
3( 2 x + 1) < −8
7.
− ( x + 1) − 4 x > 4 x − 3
8.
−3 ≤ 4 x + 2 ≤ 10
9.
1
−3 ≤ 2 x + x + 4 ≤ 6
3
10.
3 x − 7 − 5 = 12
11.
2−
12.
x ≥4
13.
2x − 3 ≤ 6
13
x
=4
3
LINEAR INEQUALITIES AND ABSOLUTE VALUE EQUATIONS – ANSWER TO
EXERCISES
1. 3 x ≤ 27
x≤9
2. −5 x ≤ 40
x ≥ −8
2
4. − x ≥ 10
3
5.
3. −6 x + 3 ≥ 21
−6 x ≥ 21 − 3
−6 x ≥ 18
x ≤ −3
6. 3( 2 x + 1) < −8
3 − 2 x ≥ −8 + 3 x
−2 x − 3 x ≥ −8 − 3
−5x ≥ −11
11
x≤
5
−2 x ≥ 30
x ≤ −15
7. − ( x + 1) − 4 x > 4 x − 3
−x −1− 4x > 4x − 3
−9 x > −2
2
x<
9
8. −3 ≤ 4 x + 2 ≤ 10
−3 − 2 ≤ 4 x ≤ 10 − 2
−5 ≤ 4 x ≤ 8
6 x + 3 < −8
6 x < −11
11
x<−
6
1
9. −3 ≤ 2 x + x + 4 ≤ 6
3
−9 ≤ 6 x + x + 12 ≤ 18
−9 − 12 ≤ 7 x ≤ 18 − 12
−21 ≤ 7 x ≤ 6
5
− ≤x≤2
4
−3 ≤ x ≤
11. 2 −
10. 3x − 7 − 5 = 12
3 x − 7 = 12 + 5
3x − 7 =
17
3x − 7 =
17
3 x = 24
x =8
12.
x ≥4
−4≥ x≥ 4
or
x
=4
3
x
2− =
4
3
x
− =
2
3
or
x = −6
6
7
3 x − 7 =−17
3 x = −10
10
x= −
3
2−
x
=
−4
3
x
=
−6
3
x = 18
−
13. 2 x − 3 ≤ 6
−6 ≤ 2 x − 3 ≤ 6
−6 + 3 ≤ 2 x ≤ 6 + 3
− 3 ≤ 2x ≤ 9
3
9
− ≤x≤
2
2
14
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
Standard
Slope-intercept
Equation
Ax + By =
C
=
y mx + b
Note
A and B are not 0. A > 0
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.
15
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.
16
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.
m
=
rise change in y ∆x
=
=
run change in x ∆y
The formula to find the slope of a line passing through the points ( x1 , y1 ) and
m=
( x2 , y2 ) is:
y2 − y1
x2 − x1
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.
y
5
Choosing the points (1, 2) and (2, 0):
4
3
m=
y 2 − y1 0 − 2 − 2
=
=
= −2
x 2 − x1 2 − 1
1
2
1
x
−3
−2
1
−1
−1
−2
−3
4
17
2
3
4
5
6
Parallel Lines
In the y-intercept form equation, y = mx + b , m is the slope and b is the y-intercept. Two lines are parallel
if their slopes are the same ( m1 = m2 ) and their y-intercepts are different ( b1 ≠ b2 ).
Example:
1) y = 2 x + 9
m1 = 2
b1 = 9
2) y = 2 x − 1
m2 = 2
b2 = −1
In the previous example the slope of both equations is the same and their y-intercepts are different;
therefore, the lines are parallel.
y
8
6
4
2
x
−8
−6
−4
2
−2
4
6
10
8
−2
−4
−6
−8
10
Perpendicular Lines
Two lines are perpendicular if the slopes are the negative reciprocal of each other: m1 = −
Example:
m1 = 4
1) y = 4 x + 7
1
2) y = − x + 2
4
m2 = −
1
.
m2
b1 = 7
1
4
b2 = 2
The slope of equation 2 is the negative reciprocal of the slope of equation 1; therefore, the lines are
perpendicular.
y
8
6
4
2
x
−8
−6
−4
−2
2
4
6
8
10
−2
−4
−6
−8
10
18
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:
y − y1 = m(x − x1 )
where m is the slope of the line and ( x1 , y1 ) 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:
y − y1 = m(x − x1 )
y − 2 = 5( x − (− 4 ))
y − 2 = 5 x + 20
y = 5 x + 22
← 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:
m=
y 2 − y1
−1 1
4−5
=
=
=
x 2 − x1 − 6 − (− 2 ) − 4 4
With the slope obtained and one of the two points given, use the point-slope form equation to find
the equation of the line.
y − y1 = m( x − x1 )
1
(x − (− 2))
4
1
1
y −5 = x+
4
2
1
11
y = x+
4
2
y −5 =
← 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.
19
Examples:
Find the equation of a line that is parallel to =
y 4 x + 2 and that passes through the point (2, 7).
=
y 4x + 2
m1 = 4
← Since the equation is in slope-intercept form, the coefficient of x is
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: m
=
m=
4
1
2
y − y1= m ( x − x1 )
y − 7= 4 ( x − 2 )
y − 7 = 4x − 8
← Parallel line to =
=
y 4x −1
y 4x + 2 .
Find the equation of a line perpendicular to 4 x + 2 y =
9 and that passes through the point
(–1, –1).
4x + 2 y = 9
2 y = −4 x + 9
− 4x + 9
y=
2
9
y=
−2 x +
2
← This equation is in standard form, we need to convert it
to slope-intercept form: y=mx+b.
← The slope is m1 = −2 .
Since the second equation must be perpendicular to the first equation, find the negative reciprocal.
1 1
1
→
m2 =
−
=
m2 = −
−2 2
m1
1
and the point (–1, –1):
2
y − y1= m ( x − x1 )
Using the slope m2 =
1
( x − ( −1) )
2
1
1
y + 1=
x+
2
2
1
1
← Perpendicular line to 4 x + 2 x =
=
y
x−
9.
2
2
y − ( −1=
)
20
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 x − 4 y =
12
c) The graph of 3 x − 4 y =
12 passes the vertical line test for functions
d) The y-intercept of 3 x − 4 y =
12 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 y − 6 x =
30 is
6
a) −6 x + 5 y =
b) =
y
x+6
30
5
c) 5 y − 6 x =
d) 6 x − 5 y =
30
−30
4. Identify the true statement below
a) All linear graphs are functions
b) The graph of y = 4 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 x = 4 is a function
5. Determine the slope and y-intercept of each equation:
a) 6 x + 7 y =
b) 2 x + 5 =
0
14
c) 3 y + 9 =
0
6. Use the point slope form to find the equation of the line which passes through (2, –1) and whose
3
slope is .
7
7. Write in standard form the equation of the line passing through the point (–3, –4) with slope equal to
3
m= .
2
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 y − 2 x =
−1 ?
a) 2 x + y =
b) x + 2 y =
6
3
c) 2 x − y =
d) x − 2 y =
6
3
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 x=
− 5 y 6 and 6 x =
− 10 y 7
21
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 x − 5 y =
−30 , A > 0
4. B: y = 4 is an example of a linear function.
6
5. a) m =
− and b =
2
7
b) m = undefinded and b does not exist
c) m = 0 and b = −3
6. =
y
3
13
x−
7
7
7. 3 x − 2 y =
−1
8. B: x + 2 y =
3 is a perpendicular line to y − 2 x =
−1
9. x + 2 y =
7
10. The lines are parallel.
22
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’.
(6)(6) = 6 2
Examples:
a ⋅ a ⋅ a ⋅ a = a4
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.
→
23
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
x 4 than x ⋅ x ⋅ x ⋅ x . 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:
30 = 1
a0 = 1
Product Rule – When multiplying the same base, the exponents are added together.
Example:
← Same base, x, add the exponents, 3 + 4 = 7
x3 ⋅ x4
x3 ⋅ x4 = x7
Quotient Rule – When dividing the same base, subtract the exponents.
Example:
x5
x2
← Same base, x, subtract the exponents, 5 – 2 = 3
x5
= x3
x2
23
Power Rule – When the operation contains parentheses, multiply the exponent in the inside with the
exponent on the outside.
Example:
(y )
6 2
(y )
6 2
← Multiply the exponents, 6 × 2 = 12
= y 12
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.
( )
( )
=
a2 + b2
a+b
 is not the same as 2

c +d2
c+d 
!
 x3
 2
y
2
Examples:
32
9
3
=
=
 
2
16
4
4
!Be careful:
4

x3
 =
y2

2
4
4
x 12
y8
2
2
 a + b   a + 2ab + b 
=

  2
2 
 c + d   c + 2cd + d 
2
In fact:
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.
3
Example:
− ( 3) =
−(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:
− (2 ) = −8
3
← Since the negative sign is outside the parentheses, carry it
out to the final answer.
(− 5)2 = (− 5)(− 5) = 25
← Since the negative sign is inside the parentheses, it
needs to be carried out through the operation.
(− 4)2 = (− 4)(− 4) = 16
← Even number of exponents, positive answer.
(− 4)3 = (− 4)(− 4)(− 4) = −64
← Odd number of exponents, negative answer.
24
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:
x −2
← Get the reciprocal, or move the negative exponent down.
x −2
1
x −2 =
= 2
1
x
5 y −3
← Get the reciprocal of only the base with the negative exponent, the
number stays in its place.
5 y −3
5
−3
y
=
= 3
5
1
y
a2
b −4
← Get the reciprocal of the base with the negative exponent, the base
with the positive exponent stays in its place.
a2
a2 ⋅ b4
=
= a2 ⋅ b4
−4
1
b
25
EXPONENTS – EXERCISES
Simplify:
1. x 5 ⋅ x 7
2. x 0 (for x ≠ 0 )
4. 5−3
5.
5
7.  
2
−2
5
8.  
2
x2 y
10.
xy
( )
4
13. − a 2
16.
(a b c ) (− ab)
2 2 2 3
 x3 
19.  2 
x 
3
4
4 −1 3
 8 
25.  
 27 
−
6.
(3x )2
9.
(x )
2
(for a + b ≠ 0)
3 2
(3 )
12. 9 ⋅ 9 2
14.
(− a )
15.
c7 ⋅ c ⋅ c2
c5 ⋅ c 2 ⋅ c 3
17.
(a )
18.
(4a b) (3a b )
21.
(a x )
24.
( −1)
2 −2
2 4
−1 −1
1
20.  
k
(a x )
(a b )
(a + b )0
11.
2 −3 2
22.
22
23
3.
23.
−2
(− 1)5
2
2
2
3 3
2 −3 2
9
2
3
26
EXPONENTS – ANSWERS TO EXERCISES
1. x 5 ⋅ x 7 = x 12
2. x 0 = 1
3.
(a + b )0 = 1
4. 5−3 =
5.
22
1
−3
−1
22=
2=
=
3
2
2
6.
−2
2
x )
(=
11.
(3 )
3
2 −2
= 3− 4 =
( )
= 32 ⋅ x 2 = 9 x 2
2
( 3)( 2 )
x=
x6
2
2
52 25
5
8.  = =
22
4
2
22
4
5
2
7.  =  = =
2
5
25
2
5
9.
( 3x )
1
1
=
3
5 125
10.
1
1
=
4
3
81
x2 y
= x( 2−1) ⋅ y (1−1) = x1 ⋅ y 0 = x
xy
12. 9 ⋅ 9 2 = 9 3
15.
c 7 ⋅ c ⋅ c 2 c10
=
=1
c5 ⋅ c 2 ⋅ c 3 c10
16.
a
( −a ) =
( −a )( −a )( −a )( −a ) =
(a b c ) (− ab) = a b c
17.
(a )
18.
(4a b) (3a b )
13. − a 2
( )( )( )( )
4
−1 −1
=
− a2 a2 a2 a2 =
−a8
=a
21.
(a x )
23.
( −1)
5
=a x
−
2
4
2
2 2 2 3
4
2
2
2
2 3 3
−2
 8 
3
=

 27 
22.
(a x )
(a b )
24.
( −1)
2 −3 2
a4
= 6
x
−2
−2
 38 
9
2
=
 =
 
3
4
3
 27 
4 −1 3
9
=
=
−1
2
10 10 6
= 432a10b11
1−2
1
20.  = =
k2
−2
k
k
 
=
−1
 8  3
25. =
 
 27 
27
4 −6
2
−2
3
 x3 
x9
19.  2 =
=
x3

6
x
x
 
2 −3 2
14.
a 4 x−6
b3
=
a12b − 3 a8 x 6
2
8
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 7x3 y
Degree – The highest power to which a variable is raised.
Examples:
Identify each term in the following polynomial:
Constant: − 4
Coefficient: 2
Degree: 3
→ the single number in the equation
→ the number in front of the variable x
→ the highest power of the variable
Give the degree of the following polynomial:
Degree: 6
2x3 − 4
5 x 6 − 3 x 4 − 11x 2 + 8
→ 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: −3x 2 y 3 z
Polynomial – Any finite sum (or difference) of terms. Example: 4 x 3 y 2 − 3 z + 9 x 2 y − 2 xz 3
Binomial – A polynomial consisting of exactly two terms. Example: 2 x − 7
Trinomial – A polynomial consisting of exactly three terms. Example: x 3 − x + 4
There are special binomial rules that can be followed that can make them easier to solve.
28
(a + b )2 = a 2 + 2ab + b 2
(a − b )2 = a 2 − 2ab + b 2
(a + b )3 = a 3 + 3a 2 b + 3ab 2 + b 3
(a − b )3 = a 3 − 3a 2 b + 3ab 2 − b 3
(a + b )(a − b ) = a 2 − b 2
Examples:
( x + 2 )2
→ Solve using the binomial properties
(a + b )2 = a 2 + 2ab + b 2
(x + 2)2 = x 2 + 2(x )(2) + 2 2
( x + 2 )2 = x 2 + 4 x + 4
( y − 3)3
→ Solve using the binomial properties
(a − b )3 = a 3 − 3a 2 b + 3ab 2 − b 3
( y − 3)3 = y 3 − 3( y )2 (3) + 3( y )(3)2 − 33
( y − 3)3 = y 3 − 9 y 2 + 27 y − 27
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:
→
3x + 4 x
3x + 4 x =
7x
Both terms contain the same variable. Therefore, they can be
combined.
2 x 2 + 3x − 4 − x 2 + x + 9
→
2 x 2 + 3x − 4 − x 2 + x + 9
( 2x
2
)
− x 2 + ( 3 x + x ) + ( −4 + 9 )
x2 + 4x + 5
29
Combine like terms together and then simplify.
Evaluating Polynomials
To evaluate polynomials substitute the variables for the given value and solve the equation.
Evaluate 2 x 3 − x 2 − 4 x + 2 at x = −3 .
Examples:
2 ( −3) − ( −3) − 4 ( −3) + 2
3
2
→
2 ( −27 ) − ( 9 ) + 12 + 2
Plug in –3 for x; be careful with
parentheses and negative signs.
−54 − 9 + 14
−63 + 14
−49
If P ( x ) = 5 x 2 − 4 x + 7 , find P ( 2 ) . → This is the same thing as plugging a 2 for ‘x’
P ( 2) = 5 ( 2) − 4 ( 2) + 7
2
= 5 ( 4) − 8 + 7
P ( 2 ) = 19
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 x 3 + 3 x 2 − 4 x + 5 + x 3 − 2 x 2 + x − 4
( 3x
3
) (
+ 3x − 4 x + 5 + x − 2 x + x − 4
2
3
2
)
)
→ Combine same degree terms together.
3x3 + 3x 2 − 4 x + 5 + x3 − 2 x 2 + x − 4
( 3x
3
) (
)
+ x 3 + 3 x 2 − 2 x 2 + ( −4 x + x ) + ( 5 − 4 )
= 4 x3 + x 2 − 3x + 1
Simplify: ( 2 x + 5 y ) + ( 3x − 2 y )
30
→ Combine x’s with x’s and y’s with y’s.
2 x + 5 y + 3x − 2 y
( 2 x + 3x ) + ( 5 y − 2 y )
5x + 3y
Simplify:
(x
3
(x
3
) (
+ 3x 2 + 5 x − 4 − 3x3 − 8 x 2 − 5 x + 6
) (
+ 3x 2 + 5 x − 4 − 3x3 − 8 x 2 − 5 x + 6
)
)
x + 3x + 5 x − 4 − 3x + 8 x + 5 x − 6
3
(x
2
3
3
) (
2
)
− 3 x3 + 3 x 2 + 8 x 2 + ( 5 x + 5 x ) + ( −4 − 6 )
=
−2 x3 + 11x 2 + 10 x − 10
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:
(3x )(10 x )
3
2
(
= (3 ⋅ 10 ) x 3 ⋅ x 2
= 30x 5
→ Multiply the constant terms together first and then the variables.
)
→ Add exponents together.
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:
(
−3 x 4 x 2 − x + 10
)
→ Distribute the –3x to all the terms inside the parentheses.
( )
= −3 x 4 x 2 − 3 x(− x ) − 3 x(10 )
= −12 x 3 + 3 x 2 − 30 x
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 x + 5 )( x + 3)
"first":
31
( 2 x )( x ) = 2 x 2
( 2 x + 5)( x + 3)
"outer":
"inner":
( 2 x )( 3) = 6 x
( 5)( x ) = 5 x
( 5)( 3) = 15
2 x 2 + 6 x + 5 x + 15
= 2 x 2 + 11x + 15
"last":
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:
( x + 2 ) ( x3 + 3x 2 + 4 x − 17 )
→ Take the smaller polynomial and multiply each of its
individual terms by the largest polynomial.
(x
3
)
(
)
+ 3 x 2 + 4 x − 17 ( x ) + x 3 + 3 x 2 + 4 x − 17 ( 2 )
x 3 ( x ) + 3 x 2 ( x ) + 4 x ( x ) − 17 ( x ) + x 3 ( 2 ) + 3x 2 ( 2 ) + 4 x ( 2 ) − 17 ( 2 )
x 4 + 3 x 3 + 4 x 2 − 17 x + 2 x 3 + 6 x 2 + 8 x − 34
(
) (
)
x 4 + 3 x3 + 2 x 3 + 4 x 2 + 6 x 2 + ( −17 x + 8 x ) − 34
= x 4 + 5 x 3 + 10 x 2 − 9 x − 34
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:
2x + 4
2
→ 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:
2x + 4 2x 4
= + =x + 2
2
2 2
Or factor out the common factor from the top and bottom and then cancel:
2x + 4 2 ( x + 2)
=
= x+2
2
2
Either way, the answer is the same: x + 2
Simplify:
18 x 4 + 36 x 3
9x
32
Method 1:
Method 2:
18 x 4 + 36 x 3 18 x 4 36 x 3
=
+
= 2x3 + 4x 2
9x
9x
9x
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.
(
)
9x 2x3 + 4x 2
= 2x3 + 4x 2
9x
The answer is the same by any method: 2 x 3 + 4 x 2
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 x 2 − 6 x − 12 by x + 2 .
Step 1 – Set up the division.
x + 2 x 2 − 6 x − 12
Step 2 – Let’s look at the first term inside, x 2 . On the outside we have an x, so to get to
x 2 we need to multiply x ⋅ x . Write this term on top.
x
x + 2 x 2 − 6 x − 12
Step 3 – Multiply the term on the top by the term outside and write it at the bottom,
x(x + 2 ) = x 2 + 2 x . However, to be able to eliminate the terms we need to change
the sign, − x 2 − 2 x .
x
x + 2 x 2 − 6 x − 12
− x 2 − 2x
Step 4 – Perform the required operations.
33
x
x + 2 x 2 − 6 x − 12
− x 2 − 2x
− 8x
Step 5 – Bring down the next term inside the division, in this case − 12 .
x
x + 2 x 2 − 6 x − 12
− x 2 − 2x
− 8 x − 12
Step 6 – Repeat step 2 through 4 to eliminate the remaining terms.
x −8
x + 2 x 2 − 6 x − 12
− x 2 − 2x
− 8 x − 12
8 x + 16
4
The solution to the long division is x − 8 .
Divide 3 x 3 − 5 x 2 + 10 x − 3 by 3 x + 1 :
x2 − 2x + 4
3 x + 1 3 x 3 − 5 x 2 + 10 x − 3
 x 2 ( 3 x + 1) = 3 x 3 + x 2 , to subtract, − ( 3 x 3 + x 2 ) =
−3 x 3 − x 2
−3x 3 − x 2
−6 x 2 + 10 x
6x2 + 2x
 Subtract polynomials and carry down the next term, which is 10x
 −2 x ( 3 x + 1) =
−6 x 2 − 2 x , to subtract, − ( −6 x 2 − 2 x=
) 6x2 + 2x
12 x − 3
−12 x − 4
 Subtract polynomials and carry down the next term, −3
−12 x − 4
 4 ( 3 x + 1) = 12 x + 4 , to subtract, − (12 x + 4 ) =
−7
Since the division did not come out even, the answer is the polynomial on top of the division plus the
−7
remainder as a fraction with the divisor as the denominator: x 2 − 2 x + 4 +
.
3x + 1
34
POLYNOMIALS – EXERCISES
Perform the indicated operations:
1. ( 7 x + 4 ) + ( 8 x − 3)
2. ( x 2 + 6 x + 3) + ( 4 x 2 − 2 x − 7 )
3. ( x 4 + 1) + ( −3 x3 + 3 x 2 + 10 )
4. ( 2 − x ) + ( x 2 + 2 ) + ( 3 x + 2 )
5. ( x3 + 2 x 2 − 6 x + 9 ) + ( x 2 − 2 x + 7 )
6. ( x 2 − 1 x +08 ) − ( − x 2 + 3 x + 6 )
7. ( 8 x 2 + 13 x − 7 ) − ( 4 x 2 + 7 x − 1)
8. ( 2 x 2 − 5 x + 4 ) − ( 4 x 2 − 7 x − 1)
9. ( x 2 + 6 x + 2 ) + ( 2 x 2 − 5 x − 3) − ( −3 x 2 + 5 x + 5 )
10. ( x 2 − 2 x + 1) − ( 3 x + 2 ) − ( 2 x 2 + 7 )
11. ( x + 1)( 4 x + 3)
12. ( 2 x + 3) ( x 2 + 3x + 1)
13. ( 4 x − 1) ( 3 x 2 + 7 x − 2 )
14. ( 2 x − 3 y ) ( x 2 + xy + 2 y 2 )
15. ( x + 6 y ) ( 2 x 2 − xy − 2 y 2 )
Solve by long division:
16. ( 4 x 2 − 8 x + 2 ) ÷ ( 2 x )
17. ( 8 x3 + 16 x 2 + 20 x + 4 ) ÷ ( 4 x )
18. ( 7 x 2 − 2 x + 8 ) ÷ ( x + 1)
19. ( x 2 − 5 xy − 6 y 2 ) ÷ ( x + y )
20. ( 8 x 2 − 2 x − 15 ) ÷ ( 4 x + 5 )
35
POLYNOMIALS – ANSWER TO EXERCISES
1. ( 7 x + 4 ) + ( 8 x − 3) = 15 x + 1
2. ( x 2 + 6 x + 3) + ( 4 x 2 − 2 x − 7 ) = 5 x 2 + 4 x − 4
3. ( x 4 + 1) + ( −3 x3 + 3 x 2 + 10 ) =x 4 − 3 x3 + 3 x 2 + 11
4. ( 2 − x ) + ( x 2 + 2 ) + ( 3 x + 2 ) = x 2 + 2 x + 6
5. ( x3 + 2 x 2 − 6 x + 9 ) + ( x 2 − 2 x + 7 ) = x 3 + 3x 2 − 8 x + 16
6. ( x 2 − 10 x + 8 ) − ( − x 2 + 3 x + 6=
) 2 x 2 − 13x + 2
7. ( 8 x 2 + 1 x − 37 ) − ( 4 x 2 + 7 x − 1)= 4 x 2 + 6 x − 6
8. ( 2 x 2 − 5 x + 4 ) − ( 4 x 2 − 7 x − 1) =
−2 x 2 + 2 x + 5
9. ( x 2 + 6 x + 2 ) + ( 2 x 2 − 5 x − 3) − ( −3 x 2 + 5 x + 5 ) = 6 x 2 − 4 x − 6
10. ( x 2 − 2 x + 1) − ( 3 x + 2 ) − ( 2 x 2 + 7 ) =
− x2 − 5x − 8
11. ( x + 1)( 4 x + 3) = 4 x 2 + 7 x + 3
12. ( 2 x + 3) ( x 2 + 3x + 1) = 2 x 3 + 9 x 2 + 11x + 3
13. ( 4 x − 1) ( 3 x 2 + 7 x − 2 )= 12 x 3 + 25 x 2 − 15 x + 2
14. ( 2 x − 3 y ) ( x 2 + xy + 2 y 2 ) = 2 x3 − x 2 y + xy 2 − 6 y 3
15. ( x + 6 y ) ( 2 x 2 − xy − 2 y 2 ) = 2 x 3 + 11x 2 y − 8 xy 2 − 12 y 3
16. ( 4 x 2 − 8 x + 2 ) ÷ ( 2 x ) = 2 x − 4 +
2
2x
17. ( 8 x 3 + 16 x 2 + 20 x + 4 ) ÷ ( 4 x )= 2 x 2 + 4 x + 5 +
4
4x
18. ( 7 x 2 − 2 x + 8 ) ÷ ( x + 1) = 7 x − 9 +
19. ( x 2 − 5 xy − 6 y 2 )
17
x +1
÷ ( x + y ) =x − 6 y
20. ( 8 x 2 − 2 x − 15 ) ÷ ( 4 x + 5 ) = 2 x − 3
36
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: 8 x 3 + 4 x 2 + 10 x
(
)
x 8 x 2 + 4 x + 10 →
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: 6 x 3 + 8 x 2 + 16 x
6 x 3 + 8 x 2 + 16 x
(
2 x 3x 2 + 4 x + 8
)
→
The common variable is x, and the smallest
exponent is 1. The common multiple is 2.
Therefore, the greatest common factor is 2x.
You can also look at it this way:
Factoring by grouping
37
(2 x )(3x 2 ) = 6 x 3
(2 x )(4 x ) = 8 x 2
(2 x )(8) = 16 x
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: ax + ay + bx + by
→
ax + ay + bx + by
ax + bx + ay + by
x(a + b ) + y (a + b )
Notice how there are two variables, x and y. Group
together the x’s and the y’s and then
factor out the common factor in each group.
Factor: 4 − 2 x 2 + x 2 y 3 − 2 y 3
4 − 2x 2 + x 2 y 3 − 2 y 3 →
(
)
(
)
→
(
)
(
)
→
2 2 − x2 + y3 x2 − 2
2 2 − x2 − y3 2 − x2
(2 − x )(2 − y )
2
There are two variables present, x and y. Start
by grouping two terms at a time.
Notice how the two polynomials inside the
parentheses are similar. To make them the
same change the sign in the second
polynomial.
Now that they are the same, factor out the
common term.
Now it is completely factored out.
→
3
Factoring a difference of squares
For a difference of squares, a 2 − b 2 , the factors will be (a − b )(a + b ) .
x 2 − y 2 = (x − y )(x + y )
Examples:
( ) =(1 − x )(1 + x ) =(1 − x )(1 + x ) (1 + x )
1 − x 4 =(1) − x 2
2
2
2
2
2
Factoring the sum and difference of two cubes
For the addition or subtraction of two cubes, the following formulas apply:
(
= (a − b) (a
a 3 + b3 = ( a + b ) a 2 − ab + b 2
a 3 − b3
2
+ a +bb 2
)
)
x3 + 8
Examples:
( x ) + ( 2)
=+
( x 2 ) ( x 2 − ( x )( 2 ) + 22 )
=
3
3
→ Manipulate to be in “ a 3 + b3 ” form.
→ follow formula.
38
(
= ( x + 2) x2 − 2 x + 4
)
→ Manipulate to be in “ a 3 − b3 ” form.
8 y 3 − 27
=
→
( 2 y ) − ( 3)
2
=
( 2 y − 3) ( ( 2 y ) + ( 2 y )( 3) + 32 )
3
=
3
follow formula.
( 2 y − 3) ( 4 y 2 + 6 y + 9 )
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 2 + bx + c
To factor trinomials of the form x 2 + bx + c follow the same principle described above. Find two integers
whose sum equals the middle term and whose product equals the last term.
Examples:
Factor: x 2 − x − 6
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. (− 3)(2 ) = −6
3. (− 1)(6 ) = −6
2. (3)(− 2 ) = −6
4. (1)(− 6 ) = −6
Step 3 – The pair whose sum is –1 is –3 and 2, (− 3 + 2 = −1) .
The critical integers are –3 and 2.
39
Step 4 – The factors for x 2 − x − 6 are ( x − 3)( x + 2) .
Factor: x 2 + 16 x + 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 x 2 + 16 x + 60 are ( x + 6)( x + 10) .
Factoring trinomials of the form ax2 + bx + c
In this type of trinomials the coefficient of x 2 is not 1, therefore we must consider the product of a ⋅ c .
Examples:
Factor: 2 x 2 − 7 x − 4
−8 . The
Step 1 – Find two integers whose sum is –7 and whose product is ( 2 )( −4 ) =
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:
−8
1. ( −1)( 8 ) =
−8
3. ( −2 )( 4 ) =
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 x2 − 7 x − 4 = 2 x2 − 8x + x − 4
Step 5 – Factor by grouping the first two and last two terms:
2 x 2 − 8 x + x − 4 = 2 x ( x − 4 ) + 1( x − 4 )
40
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 ( x − 4 ) .
Step 6 – Factor out the common binomial factor: 2 x ( x − 4) + 1( x − 4) = ( x − 4)( 2 x + 1)
Factor: 6 x 2 − 23 x + 20
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
120
2. ( −2 )( −60 ) =
120
5. ( −5 )( −24 ) =
120
8. ( −10 )( −12 ) =
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 x 2 − 23 x + 20 = 6 x 2 − 15 x − 8 x + 20
Step 5 – Factor, separately, the first two and last two terms:
6 x 2 − 15 x − 8 x + 20 = 3 x ( 2 x − 5) − 4( 2 x − 5)
Step 6 – Factor out the common binomial factor.
3 x ( 2 x − 5) − 4 ( 2 x − 5) = ( 2 x − 5)( 3 x − 4 )
41
FACTORING – EXERCISES
1. 3 x 2 − 10 x − 8
2. 2 x 2 (x − 3) − 5 x( x − 3) − 3( x − 3)
3. x 3 + 2 x 2 − 9 x − 18
4. x 2 − 10 x + 25 − y 2
5. x 6 − y 6
6. a 3 −
1
8
7. ( x + 5) 2 + 4( x + 5) + 4
8. 1 − 2t 3 + t 6
9. 3 x 4 − 18 x 3 + 27 x 2
10. 100 x 2 − 100 x − 600
11. 8 x 2 − 2 x − 15
12. x 2 + 3 xy + 2 y 2
13. 2 x 2 ( x + 2) + 13x( x + 2) + 15( x + 2)
14. x 2 + x − 12
15. x 2 y + 3 y + 2 x 2 + 6
16. 18a 2 − 50
17. x 2 + 8 x + 15
18. 16 x 3 − 20 x 2 − 50 x
19. 9 x 2 + 33 x + 30
20. 12 x 2 + 2 x − 2
42
FACTORING – ANSWER TO EXERCISES
2. 2 x 2 ( x − 3) − 5 x ( x − 3) − 3 ( x − 3)
1. 3 x 2 − 10 x − 8
3 x 2 − 12 x + 2 x − 8
3x ( x − 4 ) + 2 ( x − 4 )
( x − 3)(2 x 2 − 5 x − 3)
( x − 3)(2 x 2 − 6 x + x − 3)
( 3x + 2 )( x − 4 )
( x − 3)(2 x( x − 3) + 1( x − 3))
( x − 3)( 2 x + 1)( x − 3) = ( x − 3) ( 2 x + 1)
2
3. x 3 + 2 x 2 − 9 x − 18
4. x 2 − 10 x + 25 − y 2
( x 2 − 10 x + 25) − y 2
( x − 5) 2 − y 2
Difference of squares: (a 2 − b 2 ) = (a − b)(a + b)
x 2 ( x + 2) − 9( x + 2)
( x 2 − 9)( x + 2)
x 2 − 9 = ( x + 3)( x − 3)
where a =
y
( x − 5) and b =
( x − 5 − y )( x − 5 + y )
( x − 3)( x + 3)( x + 2)
6. a 3 −
5. x 6 − y 6
x 6 − y 6=
(x ) −( y )
3
2
3
2
Difference of cubes:
Difference of squares: (a 2 − b 2 ) = (a − b)(a + b)
3
where
=
a x=
and b y 3
(
x −y = x −y
6
6
3
3
)( x
3
+y
3
)
(
= (a − b) (a
a 3 + b3 = ( a + b ) a 2 − ab + b 2
a −b
3
2
+ a +bb
2
)
)
= ( x + y )( x 2 − xy + y 2 )( x − y )( x 2 + xy + y 2 )
= ( x + y )( x − y )( x 2 + xy + y 2 )( x 2 − xy + y 2 )
43
(
a 3 − b3 = ( a − b ) a 2 + a +bb 2
a = a and b =
)
1
2
2
1  2

1 1 
=
 a −   a + ( a )   +   
2 

2 2 
Sum & Difference of Cubes:
3
1
8
1 
a 1

=  a −  a 2 + + 
2 
2 4

7. ( x + 5 ) + 4 ( x + 5 ) + 4 =
( x + 7 )( x + 7 ) = ( x + 7 )
2
8. 1 − 2t 3 + t 6 =
(t
3
)(
) ( t − 1) ( t
2
−1 t3 −1 =
(
2
)
+ t +1
)
2
2
2
9. 3 x 4 − 18 x 3 + 27 x=
3x 2 x 2 − 6 x + 9= 3x 2 ( x − 3)
2
10. 100 x 2 − 100 x − 600= 100 ( x − 3)( x + 2 )
11. 8 x 2 − 2 x − 15 =
( 4 x + 5)( 2 x − 3)
12. x 2 + 3 xy + 2 y 2 =( x + 2 y )( x + y )
13. 2 x 2 ( x + 2 ) + 13 x ( x + 2 ) + 15 ( x + 2 ) = ( x + 2 )( 2 x + 3)( x + 5 )
14. x 2 + x − 12 =
( x + 4 )( x − 3)
15. x 2 y + 3 y + 2 x 2 + 6=
(x
2
)
+ 3 ( y + 2)
16. 18a 2 − 50= 2 ( 3a + 5 )( 3a − 5 )
17. x 2 + 8 x + 15 = ( x + 5 )( x + 3)
18. 16 x 3 − 20 x 2 − 50 x= 2 x ( 2 x − 5 )( 4 x + 5 )
19. 9 x 2 + 33 x + 30 = 3 ( x + 2 )( 3 x + 5 )
2 2 ( 3 x − 1)( 2 x + 1)
20. 12 x 2 + 2 x −=
44