Use these CLEP College Algebra sample questions to review algebraic operations, equations and inequalities, functions, real and complex numbers, and the binomial theorem. The questions include both routine calculations and concept-based applications.
Topics Covered
- Properties of logarithms
- Factoring polynomials
- Quadratic equations
- Absolute-value inequalities
- Systems of linear equations
- Domains of functions
- Inverse functions
- Quadratic graph transformations
- Complex-number operations
- The binomial theorem
Sample Questions
- A. logb((x + 2y)/z)
- B. logb(xy2z)
- C. logb(xy2/z)
- D. logb(x2y/z)
- E. logb(xz/y2)
Show Answer
Use the power rule to rewrite 2logb(y) as logb(y2).
The product rule combines addition inside one logarithm, and the quotient rule converts subtraction into division. Therefore, the expression becomes logb(xy2/z).
- A. (x − 4)(x − 1)(x + 1)
- B. (x + 4)(x − 1)(x + 1)
- C. (x − 4)(x2 + 1)
- D. (x + 4)(x2 − 1)
- E. (x − 2)2(x + 1)
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Factor by grouping: x2(x − 4) − 1(x − 4) = (x − 4)(x2 − 1).
Then factor the difference of squares: x2 − 1 = (x − 1)(x + 1).
- A. x = −7 and x = 2
- B. x = −2 and x = 7
- C. x = 2 and x = 7
- D. x = −14 and x = 1
- E. x = 5 ± √14
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Factor the quadratic as (x − 7)(x + 2) = 0.
Set each factor equal to zero. This gives x = 7 or x = −2.
- A. x < −1 or x > 4
- B. −4 < x < 1
- C. −1 ≤ x ≤ 4
- D. −1 < x < 4
- E. x > 4
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Rewrite the absolute-value inequality as −5 < 2x − 3 < 5.
Add 3 throughout and divide by 2: −2 < 2x < 8, so −1 < x < 4. The endpoints are excluded because the original inequality is strict.
- A. (1, 5)
- B. (2, 3)
- C. (3, 1)
- D. (4, −1)
- E. (5, −3)
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Add the equations to eliminate y: 3x = 9, so x = 3.
Substitute into x − y = 2: 3 − y = 2, which gives y = 1.
- A. (−∞, 4) ∪ (4, ∞)
- B. [1, ∞)
- C. [1, 4) ∪ (4, ∞)
- D. (1, 4) ∪ (4, ∞)
- E. (−∞, 1] ∪ (4, ∞)
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The expression under the square root requires x − 1 ≥ 0, so x ≥ 1.
The denominator cannot equal zero, so x ≠ 4. Combining these restrictions gives [1, 4) ∪ (4, ∞).
- A. (x − 5)/3
- B. (x + 5)/3
- C. 3x + 5
- D. 5 − 3x
- E. 1/(3x − 5)
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Write y = 3x − 5, then interchange x and y: x = 3y − 5.
Solve for y: x + 5 = 3y, so y = (x + 5)/3.
- A. It has vertex (2, 3) and opens upward.
- B. It has vertex (−2, 3) and opens upward.
- C. It has vertex (2, −3) and opens downward.
- D. It has vertex (−2, 3) and opens downward.
- E. It has vertex (−3, 2) and opens downward.
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The vertex form is a(x − h)2 + k. Here, h = −2 and k = 3, so the vertex is (−2, 3).
Because the coefficient a = −1 is negative, the parabola opens downward.
- A. −5 − 10i
- B. 5 − 10i
- C. 11 − 10i
- D. 11 + 10i
- E. −11 + 10i
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Multiply using distribution: 3 − 12i + 2i − 8i2.
Because i2 = −1, the last term becomes +8. Combine terms to obtain 11 − 10i.
- A. 10
- B. 20
- C. 32
- D. 40
- E. 80
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To obtain x3, choose three of the five factors to contribute x and the remaining two to contribute 2.
The coefficient is C(5, 3) · 22 = 10 · 4 = 40.
How to Use These Questions
Identify the algebraic structure before performing calculations. Look for logarithm rules, common factors, recognizable quadratic forms, domain restrictions, function transformations, or binomial coefficients.
Check solutions in the original equation or system whenever possible. For function questions, distinguish restrictions on inputs from properties of the graph, and remember that inverse functions reverse the original input-output relationship.
These practice questions are not official CLEP questions and are not endorsed by the College Board.
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