MAT0020

MAT0020/24C

Beginning Algebra

Competency Exam

Explained with Practice Problems and Sample Exams

Created by Al Groccia for use in MAT0024C Beginning Algebra at Valencia College

MAT0020/24C Beginning Algebra

State ALGEBRA Competency EXAM

Comp Exam Question	Algebra State Test Competency
1	Order of Operations (No Grouping / No Exponents)
2	Order of Operations (with Grouping and Exponents)
3	Absolute Value (with Addition and Subtraction)
4	Simplify Algebraic Expressions (Using the Distributive Property)
5	Evaluate an Algebraic Expression
6	Solve a Linear Equation
7	Solve a Linear Equation with Fraction Coefficient(s)
8	Solve a Literal Equation
9	Translate a Word Problem into an Algebraic Expression
10	Solve a Word Problem
11	Translate a Word Problem into a Proportion
12	Simplify Exponential Expressions (Positive Integer Exponents)
13	Simplify Exponential Expressions (Positive and Negative Integer Exp.)
14	Simplify Exponential Expressions (Positive, Neg. and Zero Integer Exp.)
15	Scientific Notation (To and From)
16	Subtraction of Polynomials: (polynomial) – (polynomial)
17	Multiply a Monomial and a Binomial: (monomial)(binomial)
18	Multiply Two Binomials: (binomial)(binomial)
19	Factoring a Polynomial: Greatest Common Factor (GCF)
20	Factoring a Polynomial: Difference of Two Squares
21	Factoring a Polynomial: By Grouping
22	Factoring a Polynomial: A Trinomial
23	Simplify a Rational Expression Reduce by Factoring
24	Solving Quadratic Equations by Factoring (Leading Coefficient is ONE, a=1)
25	Solving Quadratic Equations by Factoring (Leading Coefficient is not ONE)
26	Simplify Square Root of a Monomial
27	Simplify Square Root of a Polynomial Using the Distributive Property
28	Solving a Linear Inequality
29	Identify Intercepts of a Linear (ax + by = c)
30	Match Linear Equation to Graph (ax + by = c or y = mx + b)

Question 1 on the MAT0020/24C Beginning Algebra State Exam

Order of Operations (No Grouping / No Exponents)

1. Simplify:

A. 19 B. 34 C. 7 D. 0

Solution:

Simplify:

You need to know order of operations:

1) Parentheses

2) Exponents

3) Multiplication/ Division (which ever comes first left to right)

4) Addition / Subtraction (which ever comes first left to right)

In this problem, the first step would be the division:

Then complete the subtraction from left to right.

The solution is: 0

Practice Problems for Question 1

Order of Operations (No Grouping / No Exponents)

1. Simplify:

a. -29 b. 33 c. 31 d. -43

2. Simplify:

a. b. 27 c. 23 d.

3. Simplify:

a. 31 b. -72 c. d. 66

4. Simplify:

a. 11 b. 23 c. -1 d. -5

Solutions:

1. b 2. c 3. a 4. d

Question 2 on the MAT0020/24C Beginning Algebra State Exam

Order of Operations (with Grouping and Exponents)

2. Simplify:

A. -25 B. 90 C. D.

Solution:

Simplify:

You need to know order of operations:

1) Parentheses

2) Exponents

3) Multiplication/ Division (which ever comes first left to right)

4) Addition / Subtraction (which ever comes first left to right)

In this problem, the first step would be the parenthesis:

The next step would be exponents:

The next step would be division because it occurs first from left to right:

The next step would be multiplication:

The next step would be subtraction:

The solution is: -25

Practice Problems for Question 2

Order of Operations (with Grouping and Exponents)

1. Simplify:

a. 224 b. 35 c. d. 80

2. Simplify:

a. 4 b. 20 c. 80 d. 2

3. Simplify:

a. -33 b. 66 c. -44 d. 206

4. Simplify:

a. 27 b. 51 c. 87 d. 3

Solutions:

1. d 2. b 3. c 4. a

Question 3 on the MAT0020/24C Beginning Algebra State Exam

Absolute Value (with Addition and Subtraction)

3. Simplify:

A. 16 B. 31 C. 3 D. 15

Solution:

Simplify:

You need to know order of operations.

In this problem, the first step would be to complete what is inside the absolute value, the absolute value is acting similar to parenthesis.

The next step is to take the absolute value:

The next step is complete the addition:

The solution is: 15

Practice Problems for Question 3

Absolute Value (with Addition and Subtraction)

1. Simplify:

a. 37 b. 19 c. -11 d. 7

2. Simplify:

a. 15 b. 3 c. -3 d. -15

3. Simplify:

a. 25 b. 15 c. -15 d. 3

4. Simplify:

a. 15 b. 13 c. 17 d. 5

Solutions:

1. b 2. c 3. b 4. d

Question 4 on the MAT0020/24C Beginning Algebra State Exam

Simplify Algebraic Expressions (Using the Distributive Property)

4. Simplify:

A. B. C. D.

Solution:

Simplify:

In this problem, the first step would be to simplify within the brackets. Within the brackets, use the distributive property first.

The next step is to combine like terms with the brackets.

The next step is to use distributive property.

The solution is: -42x-180

Practice Problems for Question 4

Simplify Algebraic Expressions (Using the Distributive Property)

1. Simplify:

a. b. c. d.

2. Simplify:

a. b. c. d.

3. Simplify:

a. b. c. d.

4. Simplify:

a. b. c. d.

Solutions:

1. a 2. c 3. c 4. b

Question 5 on the MAT0020/24C Beginning Algebra State Exam

Evaluate an Algebraic Expression

5. Evaluate the given expression when w = - 4:

A. – 79 B. – 119 C. – 113 D. – 73

Solution:

Evaluate the given expression when w = - 4:

In this problem, the first step would be to substitute the value into the variable.

Then using order of operations, the next step would be to complete the exponent.

The next step would be to complete multiplication from left to right.

The next step is to complete subtraction then addition left to right.

The solution is: -113

Practice Problems for Question 5

Evaluate an Algebraic Expression

1. Evaluate the given expression when w = -4:

a. 35 b. 61 c. 51 d. 45

2. Evaluate the given expression when w = -3:

a. -39 b. -27 c. -45 d. -33

3. Evaluate the given expression when x = -9, y = 4, z = -7:

a. -137 b. 151 c. 137 d. -151

4. Evaluate the given expression when x = -9, y = 2, z = -2:

a. -158 b. 158 c. -160 d. -166

Solutions:

1. c 2. a 3. a 4. d

Question 6 on the MAT0020/24C Beginning Algebra State Exam

Solve a Linear Equation

6. Solve for r:

A. B. C. D.

Solution:

Solve for r:

The first step would be to simplify both sides of the equation using distributive property.

The next step would be to move the variables to the same side of the equation.

The next step is to move the variables to other side of the equation.

The next step is to solve for r and simplify solution.

The solution is:

Note: You can check your solution by substituting it back into your equation.

Practice Problems for Question 6

Solve a Linear Equation

1. Solve for r:

a. b. c. d.

2. Solve for x:

a. b. c. d.

3. Solve for y:

a. b. c. d.

4. Solve for y:

a. b. c. d.

Solutions:

1. d 2. a 3. d 4. c

Question 7 on the MAT0020/24C Beginning Algebra State Exam

Solve a Linear Equation with Fraction Coefficient(s)

7. Solve for y:

A. B. C. D.

Solution:

Solve for y:

One way to solve this problem is to eliminate the fractions by multiply the entire equation by the least common denominator of the fractions. In this case the least common denominator is 12.

The next step is to solve for y.

The solution is:

Note: You can check your solution by substituting it back into your equation.

Practice Problems for Question 7

Solve a Linear Equation with Fraction Coefficient(s)

1. Solve for x:

a. b. c. d.

2. Solve for t:

a. b. c. d.

3. Solve for r:

a. b. c. d.

4. Solve for r:

a. b. c. d.

Solutions:

1. b 2. b 3. c 4. b

Question 8 on the MAT0020/24C Beginning Algebra State Exam

Solve a Literal Equation

8. Solve for t:

A. B. C. D.

Solution:

Solve for t:

The first step to solve for t is to move the -8z:

The next step is isolate t by dividing by 7:

Since this solution is not one of the choices we can simplify further by dividing each part of the numerator by 7.

The solution is:

Practice Problems for Question 8

Solve a Literal Equation

1. Solve for x:

a. b. c. d.

2. Solve for W:

a. b. c. d.

3. Solve for y:

a. b. c. d.

4. Solve for v:

a. b. c. d.

Solutions:

1. a 2. b 3. d 4. c

Question 9 on the MAT0020/24C Beginning Algebra State Exam

Translate a Word Problem into an Algebraic Expression

9. The sum of a number and 16 is 4 less than twice the number.

Find the equation that could be used to find this number, x.

A. B. C. D.

Solution:

The sum of a number and 16 is 4 more than twice the number.

Translating each part:

The sum of a number and 16 means: x + 16

The word “is” means “=”

Then 4 less than twice the number means: 2x - 4

The solution is: x + 16 = 2x - 4

Practice Problems for Question 9

Translate a Word Problem into an Algebraic Expression

1. If 9 times a number is increased by 20, the result is 22 less than the square of the number. Choose the equation that could be used to find this number x.

a. b.

c. d.

2. The sum of a number and 2 is 3 more than twice the number.

Find the equation that could be used to find this number, x.

a. b. c. d.

3. If 3 times the sum of a number and 5 is equal to 7.

Find the equation that could be used to find this number, x.

a. b. c. d.

4. If 6 times a number is decreased by 8, the result is 5 less than twice the number. Choose the equation that could be used to find this number x.

a. b.

c. d.

Solutions:

1. d 2. d 3. a 4. c

Question 10 on the MAT0020/24C Beginning Algebra State Exam

Solve a Word Problem

10. The length of a rectangle is 2 feet more than the width. The perimeter of the rectangle is 72 feet. Find the length.

A. 19 feet B. 17 feet C. 37 feet D. 35 feet

Solution:

The first step is to translate the statements into algebraic expressions.

The length of a rectangle is 2 feet more than the width would be represented by: x + 2

The width would be represented by: x

Here is a visual representation:

The next step is to create an algebraic equation to solve for x.

Since the perimeter is the sum of the sides, the equation would be:

(x) + (x + 2) + (x) + (x + 2) = 72

The next step is to combine like terms:

4x + 4 = 72

Then solve for x:

x = 17

Since x is the width, then the width is 17. The length is x + 2, so the length is 19.

The answer is 19.

Note: You can check your solution by substituting it back into your algebraic equation.

Practice Problems for Question 10

Solve a Word Problem

Solutions explained in following pages.

1. A CD is priced at $15.00, but it is on sale for 20% off. What is the sale price of the CD?

a. $3.00 b. $10.00 c. $18.00 d. $12.00

2. If a sony play station costs $250 after a 15% discount, what was the original cost?

a. $294.12 b. $212.50 c. $287.50 d. $399.46

3. If a palm pilot costs $1300 after a 20% increase in price, what was the original cost?

a. $1625.00 b. $1083.33 c. $1560.00 d. $1040.00

4. Find the simple interest percent if you invested $1000.00 for 5 years and you received $500.00 in interest.

a. 20% b. 50% c. 10% d. 40%

5. The width of a rectangular garden is 8 meters less than its length. Its perimeter is 76 meters. Find the length of the garden.

a. 23 meters b. 76 meters c. 345 meters d. 15 meters

6. The perimeter of a triangle is 51 inches. The length of the middle side is 5 inches more than the length of the smaller side and the largest side is 4 inches less than three times the length of the smallest side. Find the length of the middle side.

a. 10 inches b. 15 inches c. 26 inches d. 5 inches

7. Two shrimp boats start from the same port at the same time, but they head in opposite directions. The slower boat travels 15 knots per hour slower than the fast boat. At the end of 12 hours, they were 600 nautical miles apart. How many nautical miles had the slow boat traveled by the end of the 12-hour period?

a. 210 nautical miles b. 17.5 nautical miles

c. 2.5 nautical miles c. 390 nautical miles

Solutions to Practice Word Problems

1. D

To find the discount you multiply the original price by the percent of the discount.

So, the discount is $15.00(0.20) = $3.00.

To find the sale price you subtract the original price from the discount.

So, to find the sale price you take 15.00 – 3.00 = 12.00. The sale price is $12.00.

________________________________________________________________________

2. A Check

Original Cost – Discount = Sale Price 294.12 - .15(294.12) = 250

X - .15X = 250 294.12 – 44.12 = 250

1X - .15X = 250 250 = 250

.85X = 250

.85 .85

X = 294.12

________________________________________________________________________

3. B

Original Cost + Increase = New Price Check: 1083.33 + .20(1083.33) = 1300

X + .20X = 1300 1083.33 + 216.67 = 1300

1X + .20X = 1300

1300 = 1300

1.20X = 1300

1.20 1.20

X = $1083.33

________________________________________________________________________

4. C

The formula to find simple interest is: Interest = Principal x Rate x Time (I = P*R*T)

The information gives would lead to: 500 = (1000)X (5)

Then simplify: 500 = X(5000)

Then solve for x by dividing both sides by 5000: 500/5000 = X(5000)/5000

X = 0.10, the simple interest percent would be 10%.

Check: 500 = 1000(0.10)(5)

500 = 500

________________________________________________________________________

5. A

W - 8 Width: W -8

Length: W

W W

Perimeter: 76

W - 8

Equation: W + W - 8 + W + W - 8 = 76 Width: 23 – 8 = 15

Combine Like Terms: 4W - 16 = 76 Length: 23

+ 16 +16

4W = 92 Check: 15 + 23 + 15 + 23 = 76

4 4 76 = 76

Solve: W = 23

________________________________________________________________________

6. B

Small: X

Middle: X + 5

Large: 3X - 4 X X + 5

Perimeter: 51

3X - 4

The equation will be: X + X + 5 + 3X - 4 = 51

Combine like terms: 5X + 1 = 51

Solve: - 1 - 1

5X = 50 Check: 10 + 10 + 5 + 3(10) - 4 = 51

5 5 10 + 15 + 26 = 51

25 + 26 = 51

X = 10 51 = 51

Small: X = 10

Middle: 10 + 5 = 15

Large: 3(10) - 4 = 26

________________________________________________________________________

7. A

Type	D	R	T
Fast Boat	12X	X	12
Slow Boat	12(X-15)	X	12
Total	600

Fill in the 4 x 4 chart with the given information.

To fill in the distance column use the formula Distance = Rate x Time

You now have the formula:

12X + 12(X - 15) = 600

Distributive Property: 12X + 12X -180 = 600

Combine Like Terms: 24X - 180 = 600

+180 +180

Solve for X: 24X = 780

24 24

X = 32.5

Substitute X into the chart and answer the question:

Type	D	R	T
Fast Boat	12(32.5) =390 miles	32.5 knots	12
Slow Boat	12(32.5-15) = 210 miles	32.5-15 = 17.5 knots	12
Total	600 miles

The answer from the chart is the slow boat traveled 210 miles.

Question 11 on the MAT0020/24C Beginning Algebra State Exam

Translate a Word Problem into a Proportion

11. Identify the proportion listed below that solves this problem:

A car can travel 500 miles on 5 gallons of gasoline.

How far can the car travel on 32 gallons of gasoline?

A. B. C. D.

Solution:

Identify the proportion listed below that solves this problem:

A car can travel 500 miles on 5 gallons of gasoline.

How far can the car travel on 32 gallons of gasoline?

The first step is to determine how to step up the way the proportion will be written.

One way is:

So the proportion set up will be:

Since that is not an option, another way to set up the proportion is:

The solution is:

Remember to solve a proportion, multiply the diagonal values.

The first step to solve: is

This is the same as: is

Practice Problems for Question 11

Translate a Word Problem into a Proportion

1. Identify the proportion listed below that solves this problem:

A car can travel 415 miles on 8 gallons of gasoline. How far can the car travel on 25 gallons?

a. b. c. d.

2. Identify the proportion listed below that solves this problem:

If 46 pounds of jelly beans cost 65 cents, how many pounds of jelly beans can be purchased for 130 cents?

a. b. c. d.

3. Identify the proportion listed below that solves this problem:

If Jim can eat 10 pies in 20 minutes, how many pies could Jim eat in 45 minutes?

a. b. c. d.

4. Identify the proportion listed below that solves this problem:

A shoe factory can produce 1000 pairs of shoes every 3 hours. How long would it take the shoe factory to produce 5000 pairs of shoes?

a. b. c. d.

Solutions:

1. a 2. b 3. a 4. d

Question 12 on the MAT0020/24C Beginning Algebra State Exam

Simplify Exponential Expressions (Positive Integer Exponents)

12. Simplify:

A. B. C. D.

Solution:

Simplify:

Know the exponent rules:

Exponent Rules:

The first step using order of operations is to complete the exponents using exponent rules.

The next step is to combine like terms using exponent rules.

The solution is:

Practice Problems for Question 12

Simplify Exponential Expressions (Positive Integer Exponents)

1. Simplify:

a. b. c. d.

2. Simplify:

a. b. c. d.

3. Simplify:

a. b. c. d.

4. Simplify:

a. b. c. d.

Solutions:

1. c 2. c 3. b 4. a

Question 13 on the MAT0020/24C Beginning Algebra State Exam

Simplify Exponential Expressions (Positive and Negative Integer Exp.)

13. Simplify:

A. B. C. D.

Solution:

Simplify:

Know the exponent rules:

Exponent Rules:

One way to simplify is to make all of the negative exponents positive by moving them from numerator to denominator or vice versa.

The next step is the combine like terms using the exponent rules.

The solution is:

Practice Problems for Question 13

Simplify Exponential Expressions (Positive and Negative Integer Exp.)

1. Simplify:

a. b. c. d.

2. Simplify:

a. b. c. d.

3. Simplify:

a. b. c. d.

4. Simplify:

a. b. c. d.

Solutions:

1. a 2. c 3. a 4. b

Question 14 on the MAT0020/24C Beginning Algebra State Exam

Simplify Exponential Expressions (Positive, Neg. and Zero Integer Exp.)

14. Simplify:

A. B. C. D.

Solution:

Simplify:

Know your exponent rules.

Exponent Rules:

The first step is to simplify the 0 exponent.

The next step is to simplify using the exponent rules.

The answer is:

Practice Problems for Question 14

Simplify Exponential Expressions (Positive, Neg. and Zero Integer Exp.)

1. Simplify:

a. b. c. d.

2. Simplify:

a. b. c. d.

3. Simplify:

a. b. c. d.

4. Simplify:

a. b. c. d.

Solutions:

1. c 2. a 3. d 4. c

Question 15 on the MAT0020/24C Beginning Algebra State Exam

Scientific Notation (To and From)

15. Convert to scientific notation: 0.000000195

A. B. C. D.

Solution:

Convert to scientific notation: 0.000000195

The first step will be to determine where the decimal place should be to convert this number from standard notation to scientific notation. In scientific notation the decimal should be after the first non zero from the left.

In this case the decimal should be after the 1 which is 7 movements to the right.

If the decimal moved to the right the exponent is negative, to the left the exponent is positive.

Then place in this form:

The solution is:

Note: This can be checked by converting from scientific notation to standard notation.

To convert to standard notation, start at the decimal point and the -7 means move the decimal point 7 movements to the left (due to the negative). A positive exponent means move the decimal to the right, a negative means move to the left.

Practice Problems for Question 15

Scientific Notation (To and From)

1. Convert to standard form:

a. b. c. d.

2. Convert to standard form:

a. b. c. d.

3. Convert to scientific notation:

a. b. c. d.

4. Convert to scientific notation:

a. b. c. d.

Solutions:

1. b 2. d 3. a 4. a

Question 16 on the MAT0020/24C Beginning Algebra State Exam

Addition/Subtraction of Polynomials

16. Simplify:

A. B. C. D.

Solution:

Simplify:

One way to simplify is to rewrite this problem by distributing the subtraction sign:

The next step is to combine like terms.

The solution is:

Practice Problems for Question 16

Addition/Subtraction of Polynomials

1. Simplify:

a. b. c. d.

2. Simplify:

a. b. c. d.

3. Simplify:

a. b. c. d.

4. Simplify:

a. b. c. d.

Solutions:

1. a 2. d 3. c 4. a

Question 17 on the MAT0020/24C Beginning Algebra State Exam

Multiply a Monomial and a Binomial

17. Simplify:

A. B. C. D.

Solution:

Simplify:

The first step to simplify is to use the distributive property and simplify.

The solution is:

Practice Problems for Question 17

Multiply a Monomial and a Binomial

1. Simplify:

a. b. c. d.

2. Simplify:

a. b. c. d.

3. Simplify:

a. b. c. d.

4. Simplify:

a. b.

c. d.

Solutions:

1. d 2. a 3. c 4. a

Question 18 on the MAT0020/24C Beginning Algebra State Exam

Multiply Two Binomials

18. Simplify:

A. B. C. D.

Solution:

Simplify:

One way to simplify this problem is with the FOIL method:

First: (4x)(6x):

Outer: (4x)(-7): -28x

Inner: (-7)(6x): -42x

Last: (-7)(-7): 49

The result is:

Then combine like terms: -28x – 42x = -70x

The solution is:

Practice Problems for Question 18

Multiply Two Binomials

Solutions Below

1. Simplify:

a. b. c. d.

2. Simplify:

a. b.

c. d.

3. Simplify:

a. b. c. d.

4. Simplify:

a. b.

c. d.

Solutions:

1. c 2. a 3. d 4. a

Question 19 on the MAT0020/24C Beginning Algebra State Exam

Factoring a Polynomial: Greatest Common Factor (GCF)

19. Factor completely:

A. B.

C. D.

Solution: Factor completely:

The first step to factor this problem is to identify which method can be used to factor. In this case the factoring method is factoring by pulling out the Greatest Common Factor.

To determine the GCF of the coefficients we look at the prime factors of the coefficients:

The Greatest Common Factor of the coefficients is: 2

To determine the Greatest Common Factor of the variables, we look for the most of each variable in each term. All of the terms must have the variable for it to be factored out. It will be the smallest exponent as long as all terms have the variable.

The Greatest Common Factor for the variable y is:

The Greatest Common Factor for the variable s is:

The Greatest Common Factor is:

If we factor out from each term, we need to show what is remaining.

If we divide each term by , that will give the remaining.

The solution is:

Note: This solution can be checked by using the distributive property but be sure to

pull out the greatest common factor.

Practice Problems for Question 19

Factoring a Polynomial: Greatest Common Factor (GCF)

1. Factor:

a. b.

c. d.

2. Factor:

a. b.

c. d.

3. Factor:

a. b.

c. d.

4. Factor:

a. b.

c. d.

Solutions:

1. b 2. b 3. d 4. b

Question 20 on the MAT0020/24C Beginning Algebra State Exam

Factoring a Polynomial: Difference of Two Squares

20. Factor completely:

A. B.

C. D.

Solution: Factor completely:

The first step to factor this problem is to identify which method can be used to factor. In this case the factoring method is difference of two square because there are only 2 terms, there is a subtraction sign in the middle, and the front and back terms can be square rooted.

Make sure the expression can be factored by pulling out the Greatest Common Factor first. If you can factor by pulling out the GCF, that should be completed first.

The first step is to take the square root of the front term is:

Then take the square root of the back term is:

The solution is:

Note: This solution can be checked by simplifying through FOIL.

Practice Problems for Question 20

Factoring a Polynomial: Difference of Two Squares

1. Factor:

a. b. c. d.

2. Factor:

a. b. c. d.

3. Factor:

a. b. c. d.

4. Factor:

a. b.

c. d.

Solutions:

1. a 2. d 3. c 4. b

Question 21 on the MAT0020/24C Beginning Algebra State Exam

Factoring a Polynomial: By Grouping

21. Factor completely:

A. B. C. D.

Solution: Factor completely:

The first step to factor this problem is to identify which method can be used to factor. In this case the factoring method is grouping because there are 4 terms, the Greatest Common Factor can not be factored out of all 4 terms, but we can group terms and then factor out a Greatest Common Factor.

The next step is to group terms that have similar factors.

The next step is to factor out similar terms from each group.

The next step is to notice that both groups now have a common factor of (r + k).

Factor out the (r + k) from both group and show what remains.

The solution is:

Note: This solution can be checked by simplifying through FOIL.

Practice Problems for Question 21

Factoring a Polynomial: By Grouping

1. Factor:

a. b. c. d.

2. Factor:

a. b. c. d.

3. Factor:

a. b. c. d.

4. Factor:

a. b. c. d.

Solutions:

1. a 2. c 3. c 4. a

Question 22 on the MAT0020/24C Beginning Algebra State Exam

Factoring a Polynomial: A Trinomial

22. Identify a factor of the following trinomial:

A. B. C. D.

Solution: Identify a factor of the following trinomial:

The first step to factor this problem is to identify which method can be used to factor. In this case the factoring method is factoring a trinomial because a trinomial is given.

Make sure another factoring method such as factor out the Greatest Common factor can be completed first. If it can that should be completed first.

There are two ways to factor this polynomial: Trial and Error and Change to grouping

First Method: Trial and Error

We know that when we factor, the factored form will be:

We will first list the possible factors:

We must get the leading term of

The possibilities are:

We know that it must multiply to the last term of + 4.

The possibilities with signs are:

**Signs are VERY important**

Since the middle term is negative, and it must add to the middle term, then it must be either: or

Then try different combinations and simply each one to try get :

Ex.

The solution is:

Since the question only requires one of the factors, the solution is:

Note: This solution can be checked by simplifying through FOIL.

Second Method: Change to grouping

Solution: Identify a factor of the following trinomial:

In this method we want the trinomial to become a polynomial with four terms so we can perform grouping. There is a specific way to change the trinomial into a polynomial with four terms.

Step 1: Find the middle factors

Multiply the leading coefficient by the last constant term.

, we want to find factors that multiply to get this number (including the sign)

The factors must also add to the middle term coefficient, in this case -25.

The factors that multiply to +24 and add to -25 are: -24 and -1

Step 2: Rewrite the polynomial

We found the two coefficients to replace the middle term to form a polynomial with 4 terms.

The trinomial will now become:

It does not matter which number goes first, but since we are going to group you want to place factors next to terms they have something in common with. For example, place the “-24t” next to the “” because they have a “6t” in common.

Step 3: Use Grouping

The solution is:

Since the question only requires one of the factors, the solution is:

Note: This solution can be checked by simplifying through FOIL.

Practice Problems for Question 22

Factoring a Polynomial: A Trinomial (mixed)

1. Factor:

a. b. c. d.

2. Factor:

a. b. c. d.

3. Factor:

a. b. c. d.

4. Factor:

a. b. c. d.

Solutions:

1. a 2. d 3. b 4. c

Practice Problems for Question 22

Factoring a Polynomial: A Trinomial

1. Identify a factor of the following trinomial:

a. b. c. d.

2. Identify a factor of the following trinomial:

a. b. c. d.

3. Identify a factor of the following trinomial:

a. b. c. d.

4. Identify a factor of the following trinomial:

a. b. c. d.

Solutions:

1. d 2. a 3. c 4. a

Question 23 on the MAT0020/24C Beginning Algebra State Exam

Simplify a Rational Expression: Reduce by Factoring

23. Simplify:

A. B. C. D.

Solution:

Simplify:

The first step to simplify this expression is if possible factor the numerator and denominator.

The next step is to simplify by looking for factors that appear in the numerator and denominator that will simplify to 1.

The solution is:

Practice Problems for Question 23

Simplify a Rational Expression: Reduce by Factoring

1. Simplify:

a. b. c. d.

2. Simplify:

a. b. c. d.

3. Simplify:

a. b. c. d.

4. Simplify:

a. b. c. d.

Solutions:

1. b 2. a 3. c 4. c

Question 24 on the MAT0020/24C Beginning Algebra State Exam

Solving Quadratic Equations by Factoring

(Leading Coefficient is ONE, a=1)

24. Solve:

A. B. C. D.

Solution:

Solve:

The first step is to factor the level side of the equation.

The next step is to set each factor equal to 0.

The next step is to solve each equation.

The solutions are:

Note: The solutions can be checked by substituting them back into the original equation.

Practice Problems for Question 24

Solving Quadratic Equations by Factoring

(Leading Coefficient is ONE, a=1)

1. Solve:

a. b. c. d.

2. Solve:

a. b. c. d.

3. Solve:

a. b. c. d.

4. Solve:

a. b. c. d.

Solutions:

1. c 2. c 3. a 4. c

Question 25 on the MAT0020/24C Beginning Algebra State Exam

Solving Quadratic Equations by Factoring

(Leading Coefficient is not ONE)

25. Solve:

A. B. C. D.

Solution: Solve:

The first step is to factor the level side of the equation.

The next step is to set each factor equal to 0.

The next step is to solve each equation.

The solutions are:

Note: The solutions can be checked by substituting them back into the original equation.

Practice Problems for Question 25

Solving Quadratic Equations by Factoring

(Leading Coefficient is not ONE)

1. Solve:

a. b. c. d.

2. Solve:

a. b. c. d.

3. Solve:

a. b. c. d.

4. Solve:

a. b. c. d.

Solutions:

1. c 2. b 3. b 4. d

Question 26 on the MAT0020/24C Beginning Algebra State Exam

Simplify Square Root of a Monomial

26. Simplify completely:

A. B. C. D.

Solution:

Simplify completely:

One method of simplifying this radical expression is to break the radicand into prime factors.

The next step is for every pair, one value is brought out the radical, then simplify.

Another method is to create two radicals were one radical contains what can be square rooted (perfect squares) and the other radical contains what can not be square rooted (non perfect squares).

The next step is to take the square root of the perfect squares.

The solution is:

Practice Problems for Question 26

Simplify Square Root of a Monomial

1. Simplify completely:

a. b. c. d.

2. Simplify completely:

a. b. c. d.

3. Simplify completely:

a. b. c. d.

4. Simplify completely:

a. b. c. d.

Solutions:

1. a 2. c 3. b 4. d

Question 27 on the MAT0020/24C Beginning Algebra State Exam

Simplify Square Root of a Polynomial Using the Distributive Property

27. Simplify:

A. B. C. D.

Solution:

Simplify:

The first step to simplify this problem is to use the distributive property:

The next step is to multiply the radical expressions. Remember to multiply radical expressions, multiply the coefficients and multiply the radicands.

The next step is to simply, if needed, the radical terms.

The solution is:

Practice Problems for Question 27

Simplify Square Root of a Polynomial Using the Distributive Property

1. Simplify completely:

a. b. c. d.

2. Simplify completely:

a. b. c. d.

3. Simplify completely:

a. b. c. d.

4. Simplify completely:

a. b. c. d.

Solutions:

1. d 2. a 3. c 4. a

Question 28 on the MAT0020/24C Beginning Algebra State Exam

Solving a Linear Inequality

28. Solve the inequality:

A. B. C. D.

Solution: Solve the inequality:

The first step is to get the variables on the same side. Since this is an inequality it is preferred to have the variables on the left side because inequalities are generally written with the variable on the left so they are easier to read.

The next step is to solve for x. Remember if both sides of an inequality is multiplied or divided by a negative, the inequality changes from less than to greater than or from greater than to less than.

The solution is:

Note: This solution can be checked if you substitute a value that is such as 0 into the original equation.

Practice Problems for Question 28

Solving a Linear Inequality

1. Solve the inequality:

a. b. c. d.

2. Solve the inequality:

a. b. c. d.

3. Solve the inequality:

a. b. c. d.

4. Solve the inequality:

a. b. c. d.

Solutions:

1. c 2. d 3. c 4. b

Question 29 on the MAT0020/24C Beginning Algebra State Exam

Identify Intercepts of a Linear (ax + by = c)

29. Find the y –intercept for:

A. B. C. D.

Solution:

Find the y –intercept for:

To find the x-intercept of a linear equation make y = 0 and solve for x. To find the y-intercept of a linear equation make x = 0 and solve for y.

The first step to find the y-intercept is to substitute 0 for x and solve for y.

x	y
0	?

x	y
0	-1

The second step is to write the solution as a coordinate (x, y). The value for x is 0 and the value for y is -1.

The solution is: (0, -1)

Practice Problems for Question 29

Identify Intercepts of a Linear (ax + by = c)

1. Find the x-intercept for:

a. b. c. d.

2. Find the y-intercept for:

a. b. c. d.

3. Find the y-intercept for:

a. b. c. d.

4. Find the x-intercept and y-intercept for:

a. x-intercept: y-intercept: b. x-intercept: y-intercept:

c. x-intercept: y-intercept: d. x-intercept: y-intercept:

Solutions:

1. c 2. a 3. c 4. b

Question 30 on the MAT0020/24C Beginning Algebra State Exam

Match Linear Equation to Graph

30. Find the graph that best matches the linear equation:

A. B.

C. D.

One method is to create a table of values. Choose a value for x or y, then solve for the other variable. Three points are sufficient for a linear equation.

x	y
0	?
2	?
4	?

x	y	(x, y)
0	-4	(0, -4)
2	-3	(2, -3)
4	-2	(4, -2)

Label the points and draw a line through the points.

Another method is to use the slope intercept form (y = mx + b) to graph the equation.

The first step is to make sure that the linear equation is written as y =.

The next step is to determine the slope and y-intercept of the linear equation.

In the form: y = mx + b, the m is the slope and the b is the y intercept.

Slope (m) = and y intercept (b) = -4

To graph, first place a point on the y intercept at -4. The y intercept is the point where the graph crosses the y axis. In this case the y intercept is -4.

The next step is to use the slope to find the next point. The slope can be expressed as .

The numerator determines if you move up (positive value) or down (negative value). The denominator determines if you move left (negative value) or right (positive value).

In this case, Slope (m) = mean move up 1 unit and to the right 2 units.

Then draw a straight line connecting the points.

This is the solution.

Practice Problems for Question 30

Match Linear Equation to Graph

1. Find the graph that best matches the given linear equation:

a. b.