Mathematics Placement

Mathematics Placement

• The ACT COMPASS math test is a self-adaptive test, which potentially tests

students within four different levels of math including pre-algebra, algebra, college algebra, and trigonometry. As you answer questions correctly, you will move into more difficult levels of math. Similarly, if you answer questions incorrectly, the computerized test will begin to ask questions from a lower level of math.

• Multiple-choice items in each of the five mathematics placement areas test the following:

– basic skills—performing a sequence of basic operations

– application—applying sequences of basic operations to novel settings or in complex ways

– analysis—demonstrating conceptual understanding of principles and relationships in mathematical operations

• Students are permitted to use approved calculators when completing the COMPASS® _{mathematics placement or diagnostic tests. An online calculator is}

available for those students who wish to access it via Microsoft Windows.

• Because this is an adaptive test, you may change your answer while you are still on a problem, but once you go on to another problem, you may not go back to a

Math Diagnostics Test

• Depending on the results of you Math Placement Test, you may be required to take the Math Diagnostic Test. This math test evaluates students' skill levels in 15 subareas in Pre-Algebra and Algebra:

• Pre-Algebra

– Integers

– Decimals

– Exponents, square roots, and scientific notation

– Fractions

– Percentages

– Averages (means, medians, and modes)

• Algebra

– Substituting values

– Setting up equations

– Factoring polynomials

– Exponents and radicals

– Basic operations/polynomials

– Linear equations/one variable

– Linear equations/two variables

Mathematics Placement

Sample Questions (PreAlgebra)

•

Following are 16 sample Algebra Placement Test

Questions taken from the ACT COMPASS website.

•

First you will see the question, then the following

slide will have the answer.

•

If you need some additional refreshers, the

remainder of the slides cover the content from the

Algebra section.

Algebra Placement Test

Sample Questions

Algebra Placement Test

Sample Questions

This is an example of Substituting Values into Algebraic Expressions. The correct answer is A (-4). You would need to be familiar with Evaluating Expressions to correctly answer this question (see Evaluating Expressions slides for additional information on this topic). To solve:

Step 1: Substitute value into expression Step 2: Solve

4

2

8

2

1

9

1

3

1

)

3

(

2

Algebra Placement Test

Sample Questions

2. Doctors use the term maximum heart rate (MHR) when referring to the quantity found by starting with 220 beats per minute and subtracting 1 beat per minute for each year of a person’s age. Doctors recommend exercising 3 or 4 times each week for at least20 minutes with your heart rate increased from its resting heart rate (RHR) to its training heart rate (THR), where

THR = RHR + .65(MHR – RHR)

Which of the following is closest to the THR of a 43-year-old person whose RHR is

54 beats per minute? A. 197

B. 169 C. 162 D. 134 E. 80

Algebra Placement Test

Sample Questions

2. Doctors use the term maximum heart rate (MHR) when referring to the quantity found by starting with 220 beats per minute and subtracting 1 beat per minute for each year of a person’s age. Doctors recommend exercising 3 or 4 times each week for at least 20 minutes with your heart rate increased from its resting heart rate (RHR) to its training heart rate (THR), where

THR = RHR + .65(MHR – RHR)

Which of the following is closest to the THR of a 43-year-old person whose RHR is

54 beats per minute? A. 197

B. 169 C. 162

D. 134

E. 80

This is an example of Substituting Values into Algebraic Expressions. The correct answer is D (134). You would need to be familiar with Evaluating Expressions to correctly answer this question (see Evaluating Expressions slides for additional

information on this topic). To solve: THR = 54 + .65[(220-1(43)) – 54] = 54+.65(123) = 54+79.95 = 133.95

Algebra Placement Test

Sample Questions

3. When getting into shape by exercising, the subject’s maximum recommended number of heartbeats per minute (h) can be determined by subtracting the subject’s age (a) from 220 and then taking 75% of that value. This relation is expressed by which of the following formulas?

A. h = .75(220 – a)

B. h = .75(220) – a

C. h = 220 – .75a

D. .75h = 220 – a

Algebra Placement Test

Sample Questions

3. When getting into shape by exercising, the subject’s maximum recommended number of heartbeats per minute (h) can be determined by subtracting the subject’s age (a) from 220 and then taking 75% of that value. This relation is expressed by which of the following formulas?

A. h = .75(220 – a)

B. h = .75(220) – a

C. h = 220 – .75a

D. .75h = 220 – a

E. 220 = .75(h – a)

This is an example of Setting Up Equations for Given Situations. The correct answer is A (h = .75(220 – a)). You would need to be familiar with Writing Equations to

correctly answer this question (see Writing Equations slides for additional information on this topic). To solve: Remember to change the percent to a decimal: 75% = .75. Other key words are subtract from (-) and of (x).

Algebra Placement Test

Sample Questions

4. An airplane flew for 8 hours at an airspeed of x miles per hour (mph), and for 7 more hours at 325 mph. If the average airspeed for the entire flight was 350 mph, which of the following equations could be used to find x ?

A. x + 325 = 2(350)

B. x + 7(325) = 15(350)

C. 8x – 7(325) = 350

D. 8x + 7(325) = 2(350)

Algebra Placement Test

Sample Questions

4. An airplane flew for 8 hours at an airspeed of x miles per hour (mph), and for 7

more hours at 325 mph. If the average airspeed for the entire flight (which was 15 hours total) was 350 mph, which of the following equations could be used to find x ?

A. x + 325 = 2(350)

B. x + 7(325) = 15(350)

C. 8x – 7(325) = 350

D. 8x + 7(325) = 2(350)

E . 8x + 7(325) = 15(350)

This is an example of Setting Up Equations for Given Situations. The correct answer is E (8x + 7(325) = 15(350) ). You would need to be familiar with Writing Equations to correctly answer this question (see Writing Equations slides for additional

Algebra Placement Test

Sample Questions

5. Which of the following is equivalent to 3a + 4b – (–6a – 3b) ?

A. 16ab

B. –3a + b

C. –3a + 7b

D. 9a + b

Algebra Placement Test

Sample Questions

5. Which of the following is equivalent to 3a + 4b – (–6a – 3b) ?

A. 16ab

B. –3a + b

C. –3a + 7b

D. 9a + b

E. 9a + 7b

This is an example of Basic Operations with Polynomials. The correct answer is E (9a + 7b). You would need to be familiar with Combining Like Terms to correctly answer this question (see Basic Operations with Polynomials slides for additional information on this topic).

To solve: Distribute the negative through the parenthesis and then combine like terms: 3a + 4b – (–6a – 3b) = 3a + 4b + 6a + 3b = (3a + 6a) + (4b + 3b) = 9a + 7b

Algebra Placement Test

Sample Questions

Algebra Placement Test

Sample Questions

This is an example of Basic Operations with Polynomials. The correct answer is A (3a2_{b – ab}2 _{+ 3a}2_b2_{). You would need to be familiar with Adding Polynomials and}

Combining Like Terms to correctly answer this question (see Basic Operations with Polynomials slides for additional information on this topic).

To solve: Sum means add the two polynomials, then combine like terms: (3a2_{b + 2a}2_b2_{)+ (-ab}2_+a2_b2_{) = 3a}2_{b - ab}2 _{+ (2a}2_b2 _{+ a}2_b2_{) = 3a}2_{b – ab}2 _{+ 3a}2_b2

Algebra Placement Test

Sample Questions

7. Which of the following is a factor of the polynomial x2 _{– x – 20 ?} A. x – 5

B. x – 4

C. x + 2

D. x + 5

Algebra Placement Test

Sample Questions

7. Which of the following is a factor of the polynomial x2 _{– x – 20 ?}

A. x – 5

B. x – 4

C. x + 2

D. x + 5

E. x + 10

This is an example of Factoring Polynomials. The correct answer is A (x-5). You would need to be familiar with Factoring Polynomials to correctly answer this

question (see Factoring Polynomials slides for additional information on this topic). To solve: Factor the quadratic equation: x2 _{– x – 20 = (x – 5)(x + 4). Check using} FOIL.

Algebra Placement Test

Sample Questions

8. Which of the following is a factor of x2 _{– 5x – 6 ?} A. (x + 2)

B. (x – 6)

C. (x – 3)

D. (x – 2)

Algebra Placement Test

Sample Questions

8. Which of the following is a factor of x2 _{– 5x – 6 ?} A. (x + 2)

B. (x – 6)

C. (x – 3)

D. (x – 2)

E. (x – 1)

This is an example of Factoring Polynomials. The correct answer is B (x - 6). You would need to be familiar with Factoring Polynomials to correctly answer this

question (see Factoring Polynomials slides for additional information on this topic). To solve: Factor the quadratic equation: x2 _{– 5x – 6 = (x – 6)(x + 1). Check using} FOIL.

Algebra Placement Test

Sample Questions

Algebra Placement Test

Sample Questions

This is an example of Linear Equations in One Variable. The correct answer is E (-½).

You would need to be familiar with Solving Linear Equations to correctly answer this question (see Solving Linear Equations slides for additional information on this topic). To solve: Solve for x:

2 1 2 10 2 11 5 2 11 2 11 5 11 ) 5 ( 2 x x x

Algebra Placement Test

Sample Questions

Algebra Placement Test

Sample Questions

This is an example of Linear Equations in One Variable. The correct answer is C (-1).

You would need to be familiar with Solving Linear Equations to correctly answer this question (see Solving Linear Equations slides for additional information on this topic). To solve: Solve for x:

1 10 10 10 15 10 5 10 15 10 5 2 3 10 3 10 8 2 1 1 10 3 5 4 x x x x

Algebra Placement Test

Sample Questions

Algebra Placement Test

Sample Questions

This is an example of Exponents. The correct answer is B. You would need to be familiar with Rules for Exponents to correctly answer this question (see Rules for Exponents slides for additional information on this topic).

To solve: Apply rules for exponents:

2 3 2 3 2 2 2 5 3 3 2 3 5 3

4

4

)

)(

)(

)(

4

16

(

4

16

t

z

r

z

t

r

z

z

t

t

r

r

z

rt

tz

r

Algebra Placement Test

Sample Questions

Algebra Placement Test

Sample Questions

This is an example of Exponents. The correct answer is C. You would need to be familiar with Rules for Exponents to correctly answer this question (see Rules for Exponents slides for additional information on this topic).

To solve: First, you want to get rid of the radical signs in the denominator. To do that you would multiply the numerator and the denominator by the expression.

y x xy x y x y x y x x y x x 9 3 3 3 3 3

Algebra Placement Test

Sample Questions

Algebra Placement Test

Sample Questions

This is an example of Rational Expressions. The correct answer is B (x + 8). You would need to be familiar with Rational Expressions to correctly answer this

question (see Rational Expressions slides for additional information on this topic). To solve:

8

4

)

8

)(

4

(

4

32

12

2

x

x

x

x

x

x

x

Step 1: Factor the numerator. Step 2: Cancel common terms.

Algebra Placement Test

Sample Questions

Algebra Placement Test

Sample Questions

This is an example of Rational Expressions. The correct answer is E (-x – 3). You would need to be familiar with Rational Expressions to correctly answer this question (see Rational Expressions slides for additional information on this topic).

To solve:

3

)

3

(

3

)

3

)(

3

(

3

)

9

(

3

9

2 2

x

x

x

x

x

x

x

x

x

_{Step 1: Factor the numerator.}

Step 2: Cancel common terms.

Algebra Placement Test

Sample Questions

Algebra Placement Test

Sample Questions

This is an example of Linear Equations in Two Variables. The correct answer is D (–2/3).

You would need to be familiar with Linear Equations to correctly answer this question (see Linear Equations slides for additional information on this topic).

Use the slope intercept form: y=mx+b where m=slope and b=y-intercept 2x+3y+6 = 0 (standard form)

Solve for y: 3y = -2x – 6

Algebra Placement Test

Sample Questions

Algebra Placement Test

Sample Questions

This is an example of Linear Equations in Two Variables. The correct answer is E ((8, 1)).

You would need to be familiar with Linear Equations to correctly answer this question (see Linear Equations slides for additional information on this topic).

Graph the point and the line. A perpendicular (cross at 90 degree angle) bisector (cuts the line segment into two equal length segments). Also, for future reference, parallel lines have the same slope, and perpendicular lines have negative reciprocal slope.

Algebra Review

•

The following slides review the concepts

found on the COMPASS Algebra

Substituting Values

1. Evaluating Variable Expressions

Evaluating Variable Expressions

Section 2.1

•

Variable

- a letter that is used to stand for a

quantity that is unknown or may change

•

Variable Expression

- an expression that contains

variables

•

Terms

- addends of a variable expression

• Ex: the expression 4x

2

+ 3x - 2y + 5 has four terms that

include three

variable terms

(4x

2

, 3x, 2y) and one

constant term

(5)

•

Coefficient

is the number part of the variable term. In

the term 4x

2

_{the 4 is the coefficient.}

Note

x = 1x

-x = -1x

Evaluating the Variable Expression

•

Evaluating the Variable Expression - replace

the variable with numbers and then simplify

•

EX: xy - 3 where x = 5, y = 4

replace

5(4) - 3 = 20 - 3 = 17

Don’t forget Order of Operations - PEMDAS

Parenthesis, Exponents, Mult/Div, Add/Subtract

Practice: Evaluate

Ex1: ab

2

– a when a = 2 and b = -2

Ex2: a

2

– b

2

when a = 3 and b = -4

a - b

Ex3: a

2

+ b

2

when a = 5 and b = -3

a + b

Practice: Answers

Ex1: ab

2

– a when a = 2 and b = -2

Ex2: a

2

– b

2

when a = 3 and b = -4

a - b

Ex3: a

2

+ b

2

when a = 5 and b = -3

a + b

6

2

8

2

)

4

(

2

2

)

2

(

2

2

1

7

7

4

3

16

9

)

4

(

3

)

4

(

3

2 2

17

2

34

)

3

(

5

9

25

)

3

(

5

)

3

(

You Try: Evaluate

Ex1: 3a + 2b when a = 2 and b = 3

Ex2: 2a – (c + a)

2

when a = 2 and c = -4

Ex3: a

2

– 5a – 6 when a = -2

For Ex 4-7 use: a = -2, b = 4, c = -1, d = 3

Ex4: (b + c)

2

_{+ (a + d)}

2

Ex5: b – 2a

bc

2

_-d

Ex6: b

2

- a

ad + 3c

You Try: Answers

Ex1: 3a + 2b when a = 2 and b = 3

Ex2: 2a – (c + a)

2

when a = 2 and c = -4

Ex3: a

2

– 5a – 6 when a = -2

For Ex 4-7 use: a = -2, b = 4, c = -1, d = 3

Ex4: (b + c)

2

_{+ (a + d)}

2

Ex5: b – 2a

bc

2

_-d

Ex6: b

2

- a

ad + 3c

Ex7: 1/3(d

2

_{) – 3/8(b}

2

₎

12

6

6

)

3

(

2

)

2

(

3

0 4 4 4 ) 2 ( 2 ) 2 ( ) 2 ( 2 ) 2 4 ( ) 2 (

2 2 2

8 6 10 4 6 ) 2 ( 5 ) 2 ( 2 10 1 9 1 3 ) 3 2 ( ) 1 4

( 2 2 2 2

8

1

8

3

4

8

3

)

1

(

4

4

4

3

)

1

(

4

)

2

(

2

4

2 6 3 18 ) 3 ( 6 ) 2 ( 16 ) 1 ( 3 ) 3 ( 2 ) 2 ( 42

3

6

3

)

16

(

8

3

)

9

(

3

1

)

4

(

8

3

)

3

(

3

Setting Up Equations

1. Phrases that mean Addition

2. Phrases that mean Subtraction

3. Phrases that mean Multiplication

4. Phrases that mean Division

5. Phrases that mean Exponents

added to

6 added to y

y + 6

more than

8 more than x

x + 8

the sum of

the sum of x and z

x + z

increased by

t increased by 9

t + 9

the total of

the total of 5 and y

5 + y

plus

b plus 17

b + 17

Phrases that mean Addition

Section 2.6

minus

x minus 2

x - 2

less than

7 less than t

t - 7

less

7 less t

7 - t

subtracted from

5 subtracted from d

d - 5

decreased by

m decreased by 3

m - 3

the difference between

the difference between y and 4 y - 4

Note: The book uses less than and difference frequently. Make sure you know the difference between them!

Phrases that mean Subtraction

Section 2.6

times

10 times t

10t

of

one half of x

(1/2)x

the product of

the product of y and z

yz

multiplied by

y multiplied by 11

11y

twice

twice n

2n

Phrases that mean Multiplication

divided by

x divided by 12

x/12

the quotient of

the quotient of y and z

y/z

the ratio of

the ratio of t to 9

t/9

Phrases that mean Division

Section 2.6

the square of the square of x

x

2

the cube of

the cube of a

a

3

Phrases that mean Exponents (Powers)

Translating Phrases into Expressions

Section 2.6

Note: Remember that addition and multiplication are commutative, so order doesn’t matter. However, subtraction and division are not commutative, so order DOES matter.

Word Phrase

Variable

(can be any letter)

Expression

4 less than a number

a

a – 4

A number decreased by 9

b

b – 9

50 minus a number

z

50 – z

A number more than 5

x

5 + x

A number increased by 7

y

y + 7

the sum of 6 and a number

c

6 + c

3 plus a number

k

3 + k

The difference of a number and 8

n

n – 8

7 times a number

m

7m

The product of 12 and a number

t

12t

3/4 of a number

k

¾k

A number times 8

d

d x 8

A number divided by 2

y

y/2

6 divided by a number

s

6/s

S e c t i o n 2 . 6

•

Translate into a variable expression

–

The

total

of five

times

b and c

5b + c

Use the operations to

write the variable

expression.

Identify words that

indicate mathematical

operations.

•

Translate into a variable expression

–

The

quotient

of eight

less than

n and fourteen

n – 8

14

Use the operations to

write the variable

expression.

Identify words that

indicate mathematical

operations.

•

Translate into a variable expression

–

Thirteen

more than

the

sum

of seven and the

square

of x

(7 + x

2

) + 13

Use the operations to

write the variable

expression.

Identify words that

indicate mathematical

operations.

Your turn!

•

Translate into a variable expression

–

eighteen less than the cube of x

–

y decreased by the sum of z and nine

–

the difference between the square of q and the

sum of r and t

Your turn!

(answers)

•

Translate into a variable expression

–

eighteen less than the cube of x

x

3

- 18

–

y decreased by the sum of z and nine

y – (z + 9)

–

the difference between the square of q and the

sum of r and t

More Examples

•

Translate “a number multiplied by the total of

six and the cube of the number” into a

variable expression.

Break into parts…

–

The unknown number: n

–

The cube of a number: n

3

–

The total of six and the cube of the number: 6 +

n

3

–

Put it together

More Examples

•

Translate “the quotient of twice a number and the

difference between the number and twenty” into a

variable expression.

Break into parts…

–

The unknown number: n

–

Twice the number: 2n

–

The difference between the number and twenty: n - 20

–

Put it together

2n

n - 20

Your turn!

•

Translate “a number added to the product of five and

the square of the number” into a variable

expression.

•

Translate “the product of three and the sum of seven

and twice a number” into a variable expression.

Your turn!

(answers)

•

Translate “a number added to the product of five and

the square of the number” into a variable

expression.

5n

2

+ n

•

Translate “the product of three and the sum of seven

and twice a number” into a variable expression.

More Examples

•

After translating a verbal expression into a variable

expression, simplify the variable expression by using

the Properties of Real Numbers.

•

Translate and simplify “the total of four times an

unknown number and twice the difference between

the number and eight.”

Break into parts…

–

The unknown number: n

–

Four times the number: 4n

–

Twice the difference between the number and eight: 2(n – 8)

–

Put it together:

4n + 2(n - 8)

Your turn!

•

Translate and simplify “ a number minus the

difference between twice the number and

seventeen.”

n – (2n – 17)

n – 2n + 17

-n + 17

Factoring Polynomials

1.

Greatest Common Factor and Factoring by Grouping

2.

Factoring Trinomials: x

2

_{+ bx + c}

3.

More on Factoring Trinomials:

ax

2

_{+ bx + c}

4.

Factoring Special Products: Difference of Two Squares

and Perfect Square Trinomials

Greatest Common Factor

Factoring

is the process of breaking a product into smaller parts

called factors. Recall that the product is the answer to a

multiplication problem and the factors are the terms being

multiplied.

Examples:

For 2

∙

5 = 10, the factors are 2 and 5.

For (x - 2)(x + 4) = 0, the factors are x - 2 and x + 4.

Common Factor:

If all terms in the equation have a common factor,

you can factor it out and have a simpler equation to solve. Try to

find the greatest common factor (

GCF

) to factor out.

Greatest Common Factor

Find the GCF for each of the following:

Ex 1:

16, 40, 56

16 = 2x2x2x2 = 2

4

40 = 2x2x2x5 = 2

3

_∙5

56 = 2x2x2x7 = 2

3

_∙7

so the GCF = 2

3

_{= 8}

Ex 2: 45x

2

_y

2

_z

2

_{and 75xy}

2

_z

3

45x

2

_y

2

_z

2

_{= 3}

2

_∙5∙x

2

_∙y

2

_∙z

2

75xy

2

_z

3

_{= 3∙5}

2

_∙x∙y

2

_∙z

3

Greatest Common Factor

Example: Factor the expression 8x - 12.

• The common factors are 1, 2, and 4 because they

all divide evenly into both 8x and 12.

• The GCF is 4 and can be factored out.

• Divide 8x by 4, which is 2x.

• Divide 12 by 4, which is 3.

The factored equation is 4(2x - 3).

TIP

: Factoring out a common factor is the opposite

of applying the distributive property.

Greatest Common Factor

Factor each of the polynomials by finding the greatest common

monomial factor:

Ex 1:

14x + 21

Ex 2: -3x

2

_{+ 6x}

Ex 3:

-6ax + 9ay

Ex 4: 16x

4

_{y – 14x}

2

_y

Greatest Common Factor

Factor each of the polynomials by finding the greatest common

monomial factor:

Ex 1:

14x + 21 =

7(2x + 3)

Ex 2: -3x

2

_{+ 6x =}

_{-3x(x – 2)}

Ex 3:

-6ax + 9ay =

-3a(2x - 3y)

Ex 4: 16x

4

_{y – 14x}

2

_{y =}

_2x

2

_y(8x

2

_{– 7)}

Ex 5:

34x

4

_y

6

_{– 51x}

3

_y

5

_{+ 17x}

5

_y

4

₌

_17x

3

_y

4

_(2xy

2

_{– 3y + x}

2

₎

Factor by Grouping

Consider the expression y(x + 2) + (x + 2) as the sum of two

terms, y(x + 2) and 1(x +2). Each of these terms have a

common binomial factor of (x + 2). Factoring out this

common binomial factor by using the distributive property

gives

Factor by Grouping

Section 7.1

Factor each expression by factoring out the common binomial

factor:

Ex 1:

7y

2

_{(y + 3) + 2(y + 3)}

Ex 2: 9a(x + 1) – (x + 1)

Ex 3:

2x

2

_{(x + 5) + (x + 5)}

Ex 4: 10y(2y + 3) – 7(2y + 3)

Ex 5:

a(x – 2) – b(x – 2)

Factor by Grouping

Factor each expression by factoring out the common binomial

factor:

Ex 1:

7y

2

_{(y + 3) + 2(y + 3) =}

_(7y

2

_{+ 3)(y + 3)}

Ex 2: 9a(x + 1) – (x + 1) =

(9a – 1)(x + 1)

Ex 3:

2x

2

_{(x + 5) + (x + 5) =}

_(2x

2

_{+ 1)(x + 5)}

Ex 4: 10y(2y + 3) – 7(2y + 3) =

(10y – 7)(2y + 3)

Factor by Grouping

Factor each polynomial by grouping. If the polynomial cannot be

factored, write not factorable:

Ex 1:

bx + b + cx + c

Ex 2: x

3

_{+ 3x}

2

_{+ 6x + 18}

Ex 3:

24y – 3yz + 2xz – 16x

Ex 4: 10xy + x – y – 1

Factor by Grouping

Factor each polynomial by grouping. If the polynomial cannot be

factored, write not factorable:

Ex 1:

bx + b + cx + c = b(x + 1) + c(x + 1)

= (b + c)(x + 1)

Ex 2: x

3

_{+ 3x}

2

_{+ 6x + 18 = x}

2

_{(x + 3) + 6(x + 3)}

_{= (x}

2

_{+ 6)(x + 3)}

Ex 3:

24y – 3yz + 2xz – 16x = – 3yz + 24y + 2xz – 16x =

= -3y(z – 8) + 2x(z – 8)

= (2x – 3y)(z – 8)

Ex 4: 10xy + x – y – 1 = 10x(y + 1) – (y + 1)

= (10x – 1)(y + 1)

Factor Trinomials: x

2

_{+ bx + c}

Quadratic equations are written in this form

y = ax

2

_{+ bx + c}

where ax

2

_{is the quadratic term (to the 2}

nd

_power)

bx is the linear term (to the 1

st

_power)

c is the constant term

•

Example: Multiply

(x - 1)(x + 4)

FOIL (

First

,

Outside

,

Inside

,

Last

)

1x

2

_{+ 4x - x - 4}

_{Combine like terms.}

x

2

_{+ 3x - 4}

_{Result is a trinomial}

•

Notice that the result of multiplying binomials is a

trinomial. Factoring a trinomial requires finding the two

binomials that were originally multiplied together to give

the trinomial. The process of factoring involves working

backwards from the terms in the trinomial.

Factor Trinomials: x

2

_{+ bx + c}

To factor a trinomial with leading coefficient 1, find the two

factors of the constant term whose sum is the coefficient of

the middle term.

x

2

_{+ (}

_{a + b}

_{)x +}

_ab

Sum of constants a

and b

Product of constants a

Factor Trinomials: x

2

_{+ bx + c}

Ex: Factor: x

2

_{+ 7x + 12}

To factor x

2

_{+ 7x + 12, find the two factors of 12 whose}

sum is 7.

x

2

_{+ (}

_{3 + 4}

_{)x +}

₃₍₄₎

Therefore, x

2

_{+ 7x + 12 = (x + 3)(x + 4)}

Sum of constants a

and b

Product of constants a

Factor Trinomials: x

2

_{+ bx + c}

Ex: Factor: x

2

_{- 8x + 16}

To factor x

2

_{- 8x + 16, find the two factors of 16 whose}

sum is -8.

x

2

_{+ (}

_{-4 + -4}

_{)x +}

_-4(-4)

Therefore, x

2

_{- 8x + 16 = (x - 4)(x - 4)}

Sum of constants a

and b

Product of constants a

Factor Trinomials: x

2

_{+ bx + c}

Ex: Factor: x

2

_{+ 4x – 21}

To factor x

2

_{+ 4x – 21, find the two factors of -21}

whose sum is +4.

x

2

_{+ (}

_{7 + -3}

_{)x +}

_7(-3)

Therefore, x

2

_{+ 4x – 21 = (x +7)(x – 3)}

Sum of constants a

and b

Product of constants a

Factor Trinomials: x

2

_{+ bx + c}

We have seen from the previous 3 examples that a rule can

be formed:

To factor x

2

_{+ bx + c, if possible, find an integer pair of factors}

of c whose sum is b.

1. If

c is positive

, then both factors must have the same sign.

a. Both will be positive if b is positive

b. Both will be negative if b is negative

2. If

c is negative

, then one factor must be positive and the

other negative.

Factor Trinomials: x

2

_{+ bx + c}

Factor the trinomials, if the trinomial cannot be factored,

write not factorable:

Ex 1:

x

2

_{– x – 12}

Ex 2: y

2

_{– 3y + 2}

Ex 3:

y

2

_{+ 12y + 35}

Factor Trinomials: x

2

_{+ bx + c}

Section 7.2

Factor the trinomials, if the trinomial cannot be factored,

write not factorable:

Ex 1:

x

2

_{– x – 12 =}

_{(x + 3)(x - 4)}

Ex 2: y

2

_{– 3y + 2 =}

_{(y – 1)(y – 2)}

Ex 3:

y

2

_{+ 12y + 35 =}

_{(y + 5)(y + 7)}

Factoring Trinomials: ax

2

_{+ bx + c}

A polynomial is completely factored if none of its factors can

be factored.

To factor ax

2

_{+ bx + c, where a ≠ 1, look for a common}

monomial factor. If there is a common monomial factor,

factor out this monomial factor and factor the remaining

trinomial, if possible.

Factoring Trinomials: ax

2

_{+ bx + c}

Completely factor the polynomials:

Ex 1: 5x

2

_{– 5x – 60}

Ex 2:

7y

3

_{– 70y}

2

_{+ 168y}

Ex 3:

3x

2

_{– 18x + 30}

Ex 4:

a

4

_{+ 30a}

3

_{+ 81a}

2

Factoring Trinomials: ax

2

_{+ bx + c}

Completely factor the polynomials:

Ex 1: 5x

2

_{– 5x – 60 = 5(x}

2

_{– x – 12) =}

_{5(x – 4)(x + 3)}

Ex 2:

7y

3

_{– 70y}

2

_{+ 168y = 7y(y}

2

_{– 10y + 24) =}

_{7y(y – 4)(y – 6)}

Ex 3:

3x

2

_{– 18x + 30 =}

_3(x

2

_{– 6x + 10)}

Ex 4:

a

4

_{+ 30a}

3

_{+ 81a}

2

_{= a}

2

_(a

2

_{+ 30a + 81) =}

_a

2

_{(a + 3)(a + 27)}

Factoring Trinomials: ax

2

+ bx + c

Factoring using the ac-Method:

1. Multiply a by c

2. Find two integers whose product is ac and

whose sum b

3. Rewrite the middle term bx using the two

numbers found in Step 2 as coefficients

4. Factor by grouping the first two terms and the

last two terms

5. Factor out the common binomial factor to find

two binomial factors of the trinomial ax

2

_{+ bx + c}

Factoring Trinomials: ax

2

+ bx + c

Factoring 4x

2

_{- 5x - 6 using the ac-Method:}

1. Multiply a by c: 4(-6) = -24

2. Find two integers whose product is ac and whose sum

b: 3(-8) and 3 + -8

3. Rewrite the middle term bx using the two numbers

found in Step 2 as coefficients: 4x

2

_{- 8x + 3x - 6}

4. Factor by grouping the first two terms and the last

two terms: 4x(x - 2) + 3(x – 2)

5. Factor out the common binomial factor to find two

binomial factors of the trinomial ax

2

_{+ bx + c:}

Factoring Trinomials: ax

2

+ bx + c

Factoring using the Trial by Error Method:

1. The key is using the FOIL method (First,

Factoring Trinomials: ax

2

+ bx + c

Factor:

Ex 1: 2x

2

– 3x – 2

Ex 2: 12y

2

_{– 15y + 3}

Factoring Trinomials: ax

2

+ bx + c

Factor:

Ex 1:

2x

2

_{– 3x – 2 = 2x}

2

_{– 4x + 1x – 2}

= 2x(x – 2) + (x – 2)

=

(2x + 1)(x – 2)

Ex 2:

12y

2

_{– 15y + 3 = 3(4y}

2

_{– 5y + 1)}

= 3[4y

2

_{– 4y – 1y + 1]}

= 3[4y(y – 1) – (y – 1)]

=

3(4y – 1)(y – 1)

Difference of Squares

When factoring the difference of two squares:

x

2

_{– a}

2

_{= (x + a)(x – a)}

Ex 1: x

2

– 25 = (x + 5)(x – 5)

Ex 2: 36 – y

2

= (6 + y)(6 – y)

Perfect Square Trinomials

When factoring perfect square trinomials:

x

2

_{+ 2ax + a}

2

_{= (x + a)}

2

Ex 1: x

2

+ 20x + 100 = (x + 10)(x + 10) = (x + 10)

2

Ex 2: 49 + 14y + y

2

= (7 + y)

2

Perfect Square Trinomials

Section 7.4

When factoring perfect square trinomials:

x

2

_{– 2ax + a}

2

_{= (x - a)}

2

Ex 1: x

2

– 20x + 100 = (x – 10)(x – 10) = (x – 10)

2

Ex 2: 49 – 14y + y

2

= (7 – y)

2

Factoring Trinomials: ax

2

_{+ bx + c}

Procedures to follow when factoring Polynomials:

1. Factor out any common monomial factor.

2. Check the number of terms:

a.

Two terms:

1) Difference of two squares? – factorable 2) Sum of two squares? – not factorable

b.

Three terms:

1) Perfect square trinomial? 2) Use trial-and-error method? 3) Use the ac-method?

c.

Four terms:

1) Group terms with a common factor

Factoring Trinomials: ax

2

_{+ bx + c}

We can add to a binominal the constant term that makes it a

perfect-square trinomial ; this process is called

Completing

the Square

. In this case, (½ of our coefficient of our linear

term)

2

_{= constant term.}

To complete the squares we want: y

2

_{+ 20y +}

_{= ( )}

2

to be in the form x

2

_{+ 2ax + a}

2

_{= (x + a)}

2

_{so ½ (b) = ½ (20) = 10.}

Factoring Trinomials: ax

2

_{+ bx + c}

Complete the squares for the following:

•

Ex 1:

x

2

_{– 6x +}

_{= ( )}

2

•

Ex 2:

x

2

_{– 4x +}

_{= ( )}

2

•

Ex 3:

x

2

₊

_{+ 16 = ( )}

2

Factoring Trinomials: ax

2

_{+ bx + c}

Complete the squares for the following:

•

Ex 1:

x

2

_{– 6x + 9 = (x - 3)}

2

•

Ex 2:

x

2

_{– 4x + 4 = (x - 2)}

2

•

Ex 3:

x

2

_{+ 8x + 16 = (x + 4)}

2

Exponents and Radicals

1. Properties of Exponents

2. Simplifying Radicals

Exponents

•

Exponent:

In the expression

x

n

_(read:

_x

_{to the}

n

th power), where

x

is the base and

n

is the

exponent.

•

x

n

the base (x) is multiplied by itself

exponent (n) number of times

•

Ex:

5

2

= 5 x 5 = 25

2

3

= 2 x 2 x 2 = 8

Be careful

•

_(-3)

2

_{does not equal -3}

2

it means (-3)(-3) = 9

Exponents

•

When dealing with particular operations, there are rules for

simplifying or evaluating an exponential expression.

•

Exponent of 1:

a

1

_{= a}

•

Exponent of 0:

a

0

_{= 1, when a ≠ 0}

•

Negative Exponents:

a

-n

_{= 1 ÷ a}

n

_{= 1/a}

n

_{, where a≠0}

•

1 to a Power:

1

n

_{= 1}

•

To add and subtract exponential expressions

, like bases with

like exponents are required.

•

Example: LIKE bases with

LIKE

exponents can be simplified.

a

2

_{+ a}

2

_{= 2a}

2

Properties of Exponents

• Rule: am _{∙ a}n _{= a}m+n

• Rule: am _{÷ a}n _{= a}m_/an _{= a}m-n where a≠0

• Rule: (am₎n_{= a}m(n) • Rule: (ab)m_{= a}m_bm • Rule:

where b≠0

• To multiply two or more exponential

expressions that have the same base, add the exponents. Example: 53 _{x 5}2 _{= 5}(3+2) _{= 5}5

• To divide two or more exponential expressions that have the same base, subtract the

exponents. Example: 53 _{÷ 5}2 _{= 5}(3-2) _{= 5}1 _{= 5}

• To raise a power to a power, multiply the exponents. Example: (a6₎5_{= a}6x5 _{= a}30

• To raise a product to a power, raise each factor to that power. Example: (ab)5_{= a}5 _b5

• To raise a fraction to a power, raise both the numerator and the denominator to that power. Example:

m m m b a b a

Simplifying Radicals

2

5

2

25

50

2

3

4

3

4

3

An expression with radicals is in simplest from if the following are true.

1. No radicands (expressions under radical signs) have perfect square factors other than 1.

• Example:

2. No radicands contain fractions.

• Example:

3. No radicals appear in the denominator of a fraction.

• Example:

2

2

2

2

2

1

2

2

2

1

2

1

Note: To simplify this expression,

multiply the numerator and denominator by √2. This is

algebraically justified because it is equivalent to multiplying the original fraction by 1.

Properties of Radicals

• Rule:

• Rule:

• The square root of a product equals the product of the square roots of the factors. Example:

• The square root of a quotient equals the

quotient of the square roots of the numerator and denominator. Example:

b

a

ab

2

5

2

25

50

b

a

b

a

2

3

4

3

4

3

Examples:

1. 8

3

_{∙ 8}

0

_{= 8}

3

_{∙1 or 8}

3+0

_{= 8}

3

_{= 8∙8∙8 = 512}

2. x

3

∙ x

2

= x

3+2

= x

5

3. 3y∙ y

4

= 3y

1+4

= 3y

5

4. x

-3

= 1/ x

3

5. -10x

5

_/2x

2

_{= -5x}

5-2

_{= -5x}

3

6.

7.

2 3 10 2 3 10 2 ) 3 ( 0 ) 5 ( 5 3 5 2 0 5

4

4

4

9

36

c

b

a

c

b

a

c

b

a

b

a

c

b

a

y

xy

y

y

y

x

y

x

Basic Operations of Polynomials

1. Addition with Polynomials

2. Subtraction with Polynomials

3. Multiplication with Polynomials

Vocabulary

•

A

monomial

is an algebraic expression that contains only one term

(i.e. -2x

3

_{and 4a}

5

_).

•

A

binomial

is an algebraic expression that contains exactly two

unlike terms (i.e. 3x + 5 and a

2

_{– 3).}

•

A

trinomial

is an algebraic expression that contains exactly three

unlike terms (a

2

_{+ 6a – 7 and x}

3

_{– 8x}

2

_{+ 12x).}

•

A

polynomial

is an algebraic expression that contains one or more

unlike terms. The

degree of the polynomial

is the largest of the

degrees of its terms. The coefficient of the term of the largest

degree is called the

leading coefficient

.

•

To simplify polynomials means to add or subtract any like terms

and when possible write it in order of descending exponents.

Simplifying Polynomials

•

The sum of two or more polynomials is found by

combining like terms. Remember that like terms are

constants or terms that contain the same variable raised

to the same powers.

•

Ex: Find the sum:

(10x

2

_{+ 17x - 91) + (12x}

2

_{+ 29x - 95)}

•

To easily see the like terms, rewrite the polynomials

vertically so that the like terms line up. Then add the

coefficients of the variables.

10x

2

_{+ 17x - 91}

+12x

2

_{+ 29x - 95}

22x

2

_{+ 46x - 186}

(This is written in simplest form because it contains no

like terms.)

Simplifying Polynomials

•

Ex: Find the sum:

(7x

3

_{+ 5x}

2

_{+ x – 6) + (-3x}

2

_{+11) + (-3x}

3

_{– x}

2

_{– 5x +2)}

7x

3

_{+ 5x}

2

_{+ x – 6}

-3x

2

₊₁₁

-3x

3

_{– x}

2

_{– 5x +2}

4x

3

_{+ x}

2

_{– 4x +7}

Simplifying Polynomials

•

Remember that a negative sign written in front of a

parentheses, means to change the sign of every term within

the parenthesis. Ex:

-

(12x

2

_{+ 29x - 95) =}

_-

_12x

2

_-

_29x

₊

₉₅

•

The difference of two polynomials is found by changing the

sign of each term of the second polynomial and then

combining like terms.

•

Ex: Find the difference:

(10x

2

_{+ 17x - 91) - (12x}

2

_{+ 29x - 95)}

•

To easily see the like terms, rewrite the polynomials vertically

so that the like terms line up. Then add the coefficients of the

variables.

10x

2

_{+ 17x - 91}

-12x

2

_{- 29x + 95}

-2x

2

_{- 12x + 4}

(This is written in simplest form because it contains no like

terms.)

Simplifying Polynomials

•

Ex 1: Simplify and write in descending order:

4x + 2(x – 3) – (3x + 7)

4x + 2x – 6 – 3x – 7

Distribute

3x – 13

Combine like terms

•

Ex 2: Simplify and write in descending order:

-[6x

2

_{– 3(4 + 2x) + 9] – (x}

2

_{+ 5)}

_{Distribute inner most parenthesis}

-[6x

2

_{– 12 – 6x + 9] – x}

2

_{– 5}

_{Distribute negative through bracket}

-6x

2

_{+ 12 + 6x - 9 – x}

2

_{– 5}

_{Combine like terms}

-7x

2

_{+ 6x – 2}

Multiply Polynomials

•

To multiply a polynomial by a monomial, distribute

each term to the others by multiplication.

•

Example:

Multiply 2z(z

2

- 5 + 8).

Follow the colors.

2z(z

2

_{- 5 + 8)}

2z

3

_{- 10z}

_{+ 16z}

_{Combine like terms}

2z

3

+ 6z

Alg.: D

Multiplying Polynomials

Multiply Polynomials

•

Multiplying Binomials: Example:

(x - 1)(x + 4)

FOIL (

first

,

outside

,

inside

, last)

1x

2

_{+ 4x - x - 4}

_{Combine like terms}

x

2

+ 3x - 4

•

Notice that the result of multiplying binomials is a

trinomial.

Multiply Polynomials

•

Multiply and simplify:

8x

2

_{+ 3x – 2}

-2x + 7

56x

2

_{+21x – 14}

_{Multiply each term of the first trinomial by +7}

-16x

3

_{– 6x}

2

_{+ 4x}

_{Multiply each term of the first trinomial by -2x}

-16x

3

_{+ 50x}

2

_{+ 25x – 14}

_{Combine like terms}

Alg.: D

Multiplying Polynomials

Multiply Polynomials

•

Simplify:

(2t + 3)(2t + 3) – (t – 2)(t – 2)

4t

2

+ 6t + 6t + 9 – [t

2

– 2t – 2t + 4]

Use FOIL

4t

2

+ 12t + 9 – [t

2

– 4t + 4]

Combine like terms

4t

2

_{+ 12t + 9 – t}

2

_{+ 4t – 4}

_{Distribute negative}

3t

2

_{+ 16t + 5}

_{Combine like terms}

Alg.: D

Multiplying Polynomials

Linear Equations in One Variable

1.

Solving Linear Equations

x+b=c and ax=c

2.

Solving Linear Equations

ax+b=c

3.

More Linear Equations

ax+b=dx+c

•

Equation

– expresses the equality of two

mathematical expressions

•

Solution

– is a number when substituted

for the variable results in a true equation

Where’s the

solution?

Linear Equations

•

If a, b, and c are constants and a≠0 then a

linear equation

in x is an equation that can be

written in the form: ax + b = c.

•

(Note: A linear equation in x is also called a

first-degree equation in x because the variable

x can be written with the exponent 1. That is

x = x

1

.

Solving Linear Equations

•

_To

_{solve an equation}

_{means to find a solution}

of the equation.

•

_{The simplest equation to solve is an equation}

of the form

variable = constant

, because the

constant is the solution. (ex: x = 5)

•

The objective of solving linear (or first degree)

equations is to get the variable by itself (with

a coefficient of 1) on one side of the equation

and any constants on the other side.

Addition Property of Equations

•

The same number or variable term can be added

to each side of an equation without changing the

solution of the equation.

•

Think of an equation as a balance scale. If the

weights added to each side of the equation are

not the same, the sides no longer balance.

Whatever you do to the one side of

the equation, you have to do to the

other side to keep it balanced!

Solve equations of the form x + a = b

x + 5 = 9

-5 -5

x = 4

The goal is to rewrite the equation in the form

variable = constant.

Add the opposite of the constant term 5 to each

side.

Solve equations of the form x + a = b

Try: x + 6 = 10

4 = 10 + x

x – 3 = 12

-8 + x = 5

Solve equations of the form x + a = b

Section 3.1

Answers:

x + 6 = 10

4 = 10 + x

x = 4

-6 = x

x – 3 = 12

-8 + x = 5

x = 15

x = 13

x – 4 = 9

3 = x – 6

Multiplication Property of Equations

•

Each side of an equation can be multiplied by the same

nonzero number without changing the solution of the

equation.

•

The multiplication property is used to remove a

coefficient from a variable term in an equation by

multiplying each side of the equation by the reciprocal of

the coefficient.

•

Still think of an equation as a balance scale. What ever

you do to one side of the equation, you must do to the

other side to remain equal.

Solve equations of the form ax = b

2x = 6

½ (2x) = ½ (6)

x = 3

Try: 4x = 8

-3x = -9

3x = -27

y/8 = 2

1/3(x) = 5

x/3 = 5

Multiply each side of the equation by the

reciprocal of the coefficient.

Solve equations of the form ax = b

Section 3.1

Answers:

4x = 8

-3x = -9

x = 2

x = 3

3x = -27

y/8 = 2

x = -9

y = 16

1/3(x) = 5

x/3 = 5

Solving Application Problems

Section 3.1

1. The perimeter of a square is 4 times the

length of a side (

P = 4s

). Find the length of a

side if the perimeter of a square is 64 ½

meters.

4s = 64 ½

4

4

Solving Application Problems

2. A suit cost the store $675 and the owner said

that he wants to make a profit of $180. What

price should he mark on the suit?

Profit = Revenue – Cost

180 = R – 675

+675

+675

855 = R

Therefore, the price he would mark on the suit

would be $855.

Solving Application Problems

3. How long will a truck driver take to travel 350

miles if he averages 50 mph?

Distance = Rate x Time (d = rt)

350 = 50t

50 50

7 = t