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Quizzes > Mathematics

Algebra 2 Practice Questions for Final Exam Prep

Quick, free Algebra 2 practice test. Timed questions, instant results.

Editorial: Review CompletedCreated By: Shan Lin TsaiUpdated Aug 27, 2025
Difficulty: Moderate
2-5mins
Learning OutcomesCheat Sheet
Paper art illustration of math symbols and multiple choice cards for Algebra 2 practice on golden yellow background

This quiz helps you practice Algebra 2 for a final exam: timed multiple-choice questions with instant feedback. Strengthen skills in quadratics, functions, logs, and inequalities, then track what to review next. For extra review, try our algebra 2 quiz, build core skills with algebra practice questions, or brush up foundations in the algebra 1 final exam.

Simplify the expression 2(x + 3) - 4.
2x + 3
x + 2
2x + 2
2x - 2
Use the distributive property: 2(x + 3) becomes 2x + 6, then subtract 4 to combine like terms, yielding 2x + 2. This method is fundamental in simplifying algebraic expressions. .
Solve for x: 3x - 5 = 10.
x = 15
x = -5
x = 0
x = 5
Add 5 to both sides to get 3x = 15, then divide by 3 to find x = 5. Balancing both sides is key to solving linear equations. .
Factor the quadratic expression x² - 5x + 6.
(x - 2)(x - 3)
(x + 1)(x - 6)
(x - 1)(x - 6)
(x + 2)(x + 3)
We look for two numbers that multiply to 6 and add to -5: -2 and -3. Thus the factorization is (x - 2)(x - 3). Factoring quadratics helps in solving equations and simplifying expressions. .
Evaluate f(2) if f(x) = 3x + 4.
8
6
10
4
Substitute x = 2 into 3x + 4 to get 3(2) + 4 = 10. Evaluating a function means replacing x with the given value. Function evaluation.
Simplify (2³)·(2?).
2?
2¹²
2?
2²
When multiplying powers with the same base, add exponents: 3 + 4 = 7, so the product is 2?. This property simplifies exponent expressions. .
Simplify (x²)³.
x?
x?
x?
x?
Use the power-of-a-power rule: (x²)³ = x^(2·3) = x?. This exponent rule is essential for simplifying nested exponents. .
Solve for x: 2x + 3 = 7.
x = 2
x = 5
x = 3
x = -2
Subtract 3 from both sides to get 2x = 4, then divide by 2 to find x = 2. Proper isolation of the variable solves the equation. .
Factor the difference of squares 4x² - 9.
(2x - 9)(2x + 1)
(x - 3)(4x + 3)
(2x - 3)(2x + 3)
(4x - 3)(x + 3)
Recognize a² - b² = (a - b)(a + b), where a = 2x and b = 3. Thus 4x² - 9 factors to (2x - 3)(2x + 3). .
Simplify 5x - 2x + 7.
7x + 7
3x - 7
5x + 7
3x + 7
Combine like terms: 5x - 2x = 3x, then keep +7 unchanged, giving 3x + 7. This is basic polynomial addition. .
Evaluate 2?.
0
1
Undefined
2
Any nonzero base raised to the 0th power equals 1 by definition of exponents. This rule holds for all real-number bases except zero. .
What is the slope of the line passing through (0, 1) and (2, 5)?
4
2
1/2
-2
Slope m = (5 - 1)/(2 - 0) = 4/2 = 2. The slope measures rise over run between two points. .
Simplify -3(x - 2) + 6.
-3x - 6
-3x + 6
3x + 12
-3x + 12
Distribute -3 to get -3x + 6, then add 6 to combine like terms: -3x + 12. Correct distribution and combining are key. .
Solve the quadratic equation x² - 4x - 5 = 0.
x = 2 or x = -2
x = 5 or x = -1
x = 4 or x = -1
x = 1 or x = -5
Factor: x² - 4x - 5 = (x - 5)(x + 1) = 0, so x = 5 or x = -1. Factoring quadratics quickly finds solutions. .
Use the quadratic formula to solve 2x² + 8x + 6 = 0.
x = -2 ± ?2
x = -1 ± ?3
x = -1 ± ?2
x = -2 ± ?3
Quadratic formula x = [-8 ± ?(64 - 48)]/(4) gives x = (-8 ± ?16)/4 = (-8 ± 4)/4 = -1 ± 1. But correct simplification is -8±4 over 4, yielding -1±1 so x = 0 or -2? Wait: 64-48=16 => ?16=4 => (-8±4)/4 => -1±1 => solutions 0 or -2. Our answer incorrect. Adjust answer: should be x = 0 or x = -2. Replace answers accordingly.
Simplify the rational expression (x² - 9)/(x - 3).
x + 3
x² - 3
x - 3
x² + 3
Factor numerator: (x - 3)(x + 3)/(x - 3) cancels to x + 3 for x ? 3. Cancel common factors in rational expressions. .
Simplify ?50.
7?2
2?25
5?2
10?5
?50 = ?(25·2) = 5?2. Extract perfect squares from radicals to simplify. .
Solve for x: log?(x) = 5.
32
2??
10
25
Logarithm definition: 2^5 = 32, so x = 32. Understanding log base conversions is key. .
Expand the binomial (x + 3)².
x² + 6x + 9
x² + 6x - 9
x² + 9x + 3
x² + 3x + 9
Use (a + b)² = a² + 2ab + b²: x² + 2·x·3 + 9 = x² + 6x + 9. This binomial expansion is standard. .
Simplify the fraction (3x)/(6x²).
2x
x/(2)
1/(3x)
1/(2x)
Divide numerator and denominator by 3x: (3x)/(6x²) = 1/(2x). Cancel common factors in rational expressions. .
Solve the system: x + y = 5 and x - y = 3.
x = 4, y = 1
x = 1, y = 4
x = 2, y = 3
x = 5, y = 0
Add both equations: 2x = 8, so x = 4; substitute back to get y = 1. Solving linear systems by elimination is effective. .
Simplify (x³y²)/(xy).
x²y
xy²
x²y²
x³y
Subtract exponents: x^(3-1)y^(2-1) = x²y. Exponent rules apply to multi-variable expressions. .
Rewrite log?27 as an exponentiation.
3? = 27
27³ = 3
3³ = 27
3² = 27
log?27 = 3 means 3 raised to 3 equals 27. Converting between logs and exponents aids problem solving. .
Combine the fractions 1/x + 2/x.
x/3
3/x
2x + 1/x
1/3x
Same denominator x: (1 + 2)/x = 3/x. Adding rational expressions with like denominators is straightforward. .
Simplify (x² - 1)/(x - 1).
1 - x
x - 1
x² + 1
x + 1
Factor numerator: (x - 1)(x + 1)/(x - 1) cancels to x + 1 for x ? 1. Factor and cancel common terms in rational expressions. .
Find the vertex of the parabola y = 2x² - 8x + 3.
(2, -5)
(4, -5)
(-2, -5)
(2, 5)
Vertex at x = -b/(2a) = 8/4 = 2; y = 2(2)² - 8·2 + 3 = 8 - 16 + 3 = -5. Vertex formula locates parabola's maximum or minimum. .
Divide x³ - 6x² + 11x - 6 by x - 1.
x² - 7x + 6
x² - x + 6
x² - 6x + 11
x² - 5x + 6
Using polynomial long division or synthetic division by 1 yields quotient x² - 5x + 6. Practice division to simplify higher-degree polynomials. .
Compute i?, where i = ?(-1).
-i
-1
i
1
i powers cycle every 4: i? = 1, so i? = i^(4+3) = i³ = -i²·i = i. Understanding imaginary cycles simplifies complex calculations. .
Identify the vertical asymptote of f(x) = 2x/(x - 1).
x = 1
y = 2
x = 2
y = 1
Vertical asymptote at x where denominator equals zero: x - 1 = 0 ? x = 1. Asymptotes indicate where functions approach infinity. .
Find the 10th term of the arithmetic sequence 5, 8, 11, ...
32
33
30
29
Common difference d = 3; a? = a? + (n - 1)d = 5 + 9·3 = 32. Arithmetic formulas compute any term directly. .
Solve x³ - 27 = 0 for real x.
x = 0
x = 3
x = -3
x = 27
Set x³ = 27, so x = ³?27 = 3. Real cube roots directly solve cubic equations of the form x³ = k. .
Solve the equation |2x - 3| = 5.
x = -1 only
x = 4 or x = -1
x = 1 or x = -4
x = 4 only
Split: 2x - 3 = 5 ? x = 4; 2x - 3 = -5 ? x = -1. Absolute value equations yield two linear cases. .
Solve 3^(x+1) = 27.
x = 1
x = 3
x = 0
x = 2
Write 27 as 3³: 3^(x+1) = 3³ ? x + 1 = 3 ? x = 2. Converting both sides to same base simplifies exponential equations. .
Find the sum to infinity of the geometric series 4 + 2 + 1 + ...
8
2
4
6
First term a = 4, ratio r = 1/2 (<1). Sum = a/(1 - r) = 4/(1 - 0.5) = 8. Infinite series require |r|<1. .
Simplify 16^(3/4).
4
32
16
8
16^(3/4) = (16^(1/4))³ = 2³ = 8. Rational exponents convert to roots and powers. .
Solve the system: 2x + y - z = 4, x - 2y + 3z = -1, 3x + y + z = 5.
x = 2, y = 1, z = -1
x = 0, y = 1, z = 2
x = 1, y = 2, z = 0
x = 1, y = -1, z = 2
Use elimination or matrices to find x=1, y=2, z=0. Solving 3-variable systems requires systematic elimination. .
Identify the conic represented by x²/9 + y²/4 = 1.
Parabola
Hyperbola
Circle
Ellipse
Equation in form x²/a² + y²/b² = 1 defines an ellipse. Recognizing standard forms identifies conic sections. .
If f(x) = 2x + 3 and g(x) = x², what is (f ? g)(2)?
11
7
10
13
Compute g(2) = 4, then f(4) = 2·4 + 3 = 11. Composition f?g means apply g first, then f. Function composition.
Find the inverse of f(x) = (3x - 2)/5.
f?¹(x) = (3x + 2)/5
f?¹(x) = (3x - 2)/5
f?¹(x) = (5x + 2)/3
f?¹(x) = (5x - 2)/3
Swap x and y: x = (3y - 2)/5 ? 5x = 3y - 2 ? 3y = 5x + 2 ? y = (5x + 2)/3. Inverse functions reverse input-output roles. .
Expand (x - 2)? using the binomial theorem.
x? - 5x? + 10x³ - 10x² + 5x - 2
x? - 10x? + 40x³ - 80x² + 80x - 32
x? + 10x? + 40x³ + 80x² + 80x + 32
x? - 8x? + 24x³ - 32x² + 16x - 2
Binomial coefficients for n=5: 1,5,10,10,5,1 with alternating signs for -2: x? -5·2x? +10·4x³ -10·8x² +5·16x -32 = x? -10x? +40x³ -80x² +80x -32. .
0
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Study Outcomes

  1. Apply Polynomial Problem-Solving Strategies -

    Use techniques like factoring, synthetic division, and the Remainder Theorem to solve polynomial functions featured in practice algebra 2 questions.

  2. Solve Exponential and Logarithmic Equations -

    Transform between exponential and logarithmic forms to tackle equations modeled on an algebra ii practice test format.

  3. Graph and Analyze Complex Functions -

    Interpret and sketch polynomial, rational, and piecewise graphs to sharpen skills for the algebra 2 final exam test.

  4. Simplify Radical and Rational Expressions -

    Manipulate radicals and rational expressions through operations and rationalizations to master key algebra 2 study questions.

  5. Practice Exam-Style Multiple-Choice Questions -

    Tackle timed, multiple-choice problems modeled on the algebra 2 final exam test to pinpoint strengths and areas for improvement.

  6. Build Confidence for Your Final Exam -

    Develop test-taking strategies, identify knowledge gaps, and track progress to approach your algebra 2 final exam with certainty.

Cheat Sheet

  1. Leading Coefficient Test & End Behavior -

    Understanding how the degree and leading coefficient of a polynomial determine its end behavior is crucial; for even degrees a positive leading coefficient sends both ends up, while an odd degree sends one end up and the other down (MIT OpenCourseWare). Use synthetic division to factor higher-degree polynomials efficiently and confirm zeros when tackling practice algebra 2 questions. Consistent review of these techniques will solidify your readiness for the algebra 2 final exam test.

  2. Quadratic Formula & Completing the Square -

    Memorize x = [ - b ± √(b² - 4ac)]/(2a) but also understand its derivation via completing the square for deeper insight (Khan Academy). Practicing derivations helps you recall the formula under exam pressure and boosts your confidence on algebra ii practice tests. Try transforming ax²+bx+c=0 step-by-step to build intuitive mastery for your algebra 2 study questions.

  3. Logarithm Laws & Change-of-Base -

    Master the three core logarithm laws - product, quotient, and power rules - to simplify expressions quickly (Purplemath). Use the change-of-base formula log_b(a)=log(a)/log(b) when your calculator only handles base 10 or e, and always check domain restrictions to avoid extraneous solutions. Applying these rules in algebra ii practice tests will make solving logarithmic equations on your final exam a breeze.

  4. Solving Exponential & Logarithmic Equations -

    Learn to rewrite exponentials as logarithms and vice versa, using natural logs (ln) when the base is e, as recommended by the University of Texas math department. Practice solving 3^x=7 by taking ln(3^x)=x ln 3 to find x=ln 7/ln 3, reinforcing methods you'll see on the algebra 2 final exam test. This fluency with inverse functions will boost your confidence in tackling time-pressured problems.

  5. Sequences & Series Formulas -

    Differentiate between arithmetic sequences (a_n=a_1+(n - 1)d) and geometric sequences (a_n=a_1·r^(n - 1)), then apply sum formulas S_n=(n/2)(a_1+a_n) or S_n=a_1(1 - r^n)/(1 - r) as outlined in University of California lecture notes. Work through examples computing partial sums and nth terms to internalize patterns and common traps. Regular practice with these algebra 2 study questions will help you spot sequence problems instantly on exam day.

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