Free Algebra 2 Practice Test Questions

Use these Algebra 2 practice questions to review systems of equations, work-rate problems, rational equations, factoring, quadratic equations, completing the square, rational functions, linear functions, and logarithms. After answering each question, open the explanation to see the step-by-step solution.

Algebra 2 Topics Covered

  • Solving systems of equations
  • Working with rates and time
  • Solving rational equations
  • Factoring polynomials
  • Using the quadratic formula
  • Completing the square
  • Finding vertical asymptotes
  • Finding the domain of a function
  • Solving logarithmic equations

Algebra 2 Practice Questions

1. Solve the system of equations.

y = −3x + 4

x + 4y = −6

  1. x = −2, y = −1
  2. x = −2, y = 10
  3. x = 2, y = −2
  4. x = 3, y = −5
  5. x = 4, y = −8
Show Answer

Answer: C. x = 2, y = −2

Since the first equation already has y isolated, use substitution.

x + 4y = −6

Substitute −3x + 4 for y:

x + 4(−3x + 4) = −6

x − 12x + 16 = −6

−11x = −22

x = 2

Now substitute 2 for x in the first equation:

y = −3(2) + 4

y = −6 + 4

y = −2

The solution is x = 2, y = −2.

2. John can mow his lawn in 3 hours, and his sister Julie can mow it in 2 hours. How long will it take them to mow the lawn if they work together?

  1. 1 hour 12 minutes
  2. 1 hour 15 minutes
  3. 1 hour 20 minutes
  4. 1 hour 30 minutes
  5. 1 hour 35 minutes
Show Answer

Answer: A. 1 hour 12 minutes

John mows at a rate of:

13 lawn per hour

Julie mows at a rate of:

12 lawn per hour

Add their rates:


13
+
12
=
56

Together, they mow 56 of the lawn per hour.

To mow 1 whole lawn:


56
t = 1

t = 65 hours

65 hours is 1.2 hours, which is 1 hour 12 minutes.

3. Solve the equation.


5x
−
3x + 4
= 2

  1. −5
  2. −5 and 2
  3. 2
  4. 2 and 4
  5. 4
Show Answer

Answer: B. −5 and 2

Multiply both sides by x(x + 4) to eliminate the denominators.

5(x + 4) − 3x = 2x(x + 4)

5x + 20 − 3x = 2x2 + 8x

2x + 20 = 2x2 + 8x

0 = 2x2 + 6x − 20

0 = x2 + 3x − 10

0 = (x + 5)(x − 2)

x = −5 or x = 2

Neither value makes a denominator equal zero, so both solutions are valid.

4. Factor the expression completely.

6x3 − 4x2 − 16x

  1. 0
  2. 2x(3x2 − 2x − 8)
  3. 2x(3x + 4)(x − 2)
  4. 4x(2x + 1)(x − 4)
  5. 2x(2x2 + 7x − 4)
Show Answer

Answer: C. 2x(3x + 4)(x − 2)

First, factor out the greatest common factor.

6x3 − 4x2 − 16x = 2x(3x2 − 2x − 8)

Now factor the trinomial:

3x2 − 2x − 8 = (3x + 4)(x − 2)

So the complete factorization is:

2x(3x + 4)(x − 2)

5. Solve the equation for x.

5x2 + 6x = 3

  1. −6 ± √225
  2. −3 ± √225
  3. −3 ± 2√65
  4. 3 ± 2√65
  5. 6 ± 2√65
Show Answer

Answer: C. −3 ± 2√65

First, move all terms to one side:

5x2 + 6x − 3 = 0

Use the quadratic formula:

x =

−b ± √(b2 − 4ac)
2a

Here, a = 5, b = 6, and c = −3.

x =

−6 ± √(62 − 4(5)(−3))
2(5)

x =

−6 ± √96
10

Since √96 = 4√6, simplify:

x =

−6 ± 4√6
10

=

−3 ± 2√6
5

6. What should be added to both sides of the equation x2 − 12x = 5 in order to solve it by completing the square?

  1. −36
  2. −12x
  3. −6
  4. 12x
  5. 36
Show Answer

Answer: E. 36

To complete the square, take half of the coefficient of x, then square the result.

The coefficient of x is −12.

Half of −12 is −6.

(−6)2 = 36

So 36 should be added to both sides.

x2 − 12x + 36 = 5 + 36

The left side becomes a perfect square trinomial:

(x − 6)2 = 41

7. Find the vertical asymptotes of the function.


y =

x2 − 36
x2 − 8x + 15

  1. x = −5 and x = −3
  2. x = −5, x = −3, and x = 6
  3. x = 3 and x = 5
  4. x = 3 and x = 6
  5. x = 6
Show Answer

Answer: C. x = 3 and x = 5

For a rational function, vertical asymptotes occur where the denominator equals zero, as long as the factor does not cancel with the numerator.

Set the denominator equal to zero:

x2 − 8x + 15 = 0

Factor the denominator:

(x − 3)(x − 5) = 0

x = 3 or x = 5

The numerator factors as (x − 6)(x + 6), so no factors cancel with the denominator.

Therefore, the vertical asymptotes are x = 3 and x = 5.

8. Two cars are traveling north along a highway. The first drives at 40 mph, and the second, which leaves 3 hours later, travels at 60 mph. How long after the second car leaves will it take for the second car to catch the first?

  1. 1 hour 12 minutes
  2. 2 hours
  3. 5 hours
  4. 6 hours
  5. 6 hours 40 minutes
Show Answer

Answer: D. 6 hours

Let t represent the number of hours the second car travels.

The first car has been driving for 3 hours longer, so its time is t + 3.

Distance = rate × time

First car’s distance: 40(t + 3)

Second car’s distance: 60t

The second car catches the first when the distances are equal:

40(t + 3) = 60t

40t + 120 = 60t

120 = 20t

t = 6

The second car catches the first 6 hours after it leaves.

9. What is the domain of the function f(x) = 2x − 4?

  1. x ≥ −12
  2. x < −12
  3. x > 0
  4. x ≥ 12
  5. All real numbers
Show Answer

Answer: E. All real numbers

The domain is the set of all possible input values.

The function f(x) = 2x − 4 is linear.

It does not contain a square root, logarithm, or denominator that could restrict the value of x.

Therefore, the domain is all real numbers.

10. Solve the equation for x.


log2

7x + 3
x

= 3

  1. −3
  2. −12
  3. 2
  4. 3
  5. 5
Show Answer

Answer: D. 3

A logarithm can be rewritten in exponential form.

Since log2 of the expression equals 3, the expression must equal 23.



7x + 3
x

= 8

7x + 3 = 8x

3 = x

So x = 3.

How to Use These Algebra 2 Practice Questions

Start by answering each question before opening the explanation. Then compare your work to the step-by-step solution. If you miss a question, review the related Algebra 2 skill before moving on.

For extra review, focus on the topics you miss most often. Algebra 2 questions often depend on recognizing the correct setup, such as substitution, factoring, the quadratic formula, completing the square, or rewriting logarithmic equations.

 

Last Updated: July 3, 2026