ACCUPLACER Advanced Algebra & Functions Quiz: Solving Quadratics By Factoring
20 questions · exam conditions
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Solving Quadratics By FactoringQuestion 1 of 20

A rectangle has dimensions where the length is 3 units more than the width. If the area is 28 square units, what is the width of the rectangle?

8 units
5 units
7 units
4 units
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ACCUPLACER Advanced Algebra & Functions Quiz

ACCUPLACER Advanced Algebra & Functions Quiz: Solving Quadratics By Factoring

Practice Solving Quadratics By Factoring in ACCUPLACER Advanced Algebra & Functions with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Solving Quadratics By Factoring, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Advanced Algebra & Functions.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A rectangle has dimensions where the length is 3 units more than the width. If the area is 28 square units, what is the width of the rectangle?

  1. 8 units
  2. 5 units
  3. 7 units
  4. 4 units (correct answer)
Explanation: When you encounter word problems involving rectangles with related dimensions, you're dealing with quadratic equations. The key is translating the relationship between dimensions into algebraic expressions, then using the area formula to solve. Let's define the width as ww units. Since the length is 3 units more than the width, the length is (w+3)(w + 3) units. The area of a rectangle equals length times width, so: w(w+3)=28w(w + 3) = 28 Expanding this equation: w2+3w=28w^2 + 3w = 28 Rearranging to standard form: w2+3w−28=0w^2 + 3w - 28 = 0 To factor this quadratic, we need two numbers that multiply to -28 and add to 3. Those numbers are 7 and -4: (w+7)(w−4)=0(w + 7)(w - 4) = 0 This gives us w=−7w = -7 or w=4w = 4. Since width cannot be negative, w=4w = 4 units. Let's verify: if width = 4, then length = 7, and area = 4×7=284 \times 7 = 28 ✓ Now for the wrong answers: Choice A (8 units) would give a length of 11 and area of 88—far too large. Choice B (5 units) would create a length of 8 and area of 40—also too large. Choice C (7 units) would make the length 10 and area 70—again, much larger than needed. For rectangle problems with related dimensions, always set up your variable for one dimension, express the other in terms of that variable, then use the area formula to create your equation. Remember to check that your solution makes physical sense!

Question 2

The length of a rectangular garden is 5 meters greater than its width. If the area of the garden is 36 square meters, what is the width of the garden in meters?

  1. -9
  2. 4 (correct answer)
  3. 5
  4. 9
Explanation: Let ww be the width of the garden. The length is w+5w + 5. The area is length times width, so w(w+5)=36w(w + 5) = 36. This gives the quadratic equation w2+5w−36=0w^2 + 5w - 36 = 0, which factors to (w−4)(w+9)=0(w - 4)(w + 9) = 0. The possible solutions are w=4w = 4 and w=−9w = -9. Since width must be a positive value, the width is 4 meters. Distractor D is the length of the garden.

Question 3

Which of the following are the solutions to the equation 3x2+15x−42=03x^2 + 15x - 42 = 0?

  1. x=7x = 7 and x=−2x = -2
  2. x=−7x = -7 and x=2x = 2 (correct answer)
  3. x=−3x = -3 and x=14x = 14
  4. x=3x = 3 and x=−14x = -14
Explanation: First, factor out the greatest common factor, 3: 3(x2+5x−14)=03(x^2 + 5x - 14) = 0. Then, factor the quadratic expression inside the parentheses: 3(x+7)(x−2)=03(x + 7)(x - 2) = 0. The solutions are found by setting each factor containing a variable to zero: x+7=0  ⟹  x=−7x + 7 = 0 \implies x = -7 and x−2=0  ⟹  x=2x - 2 = 0 \implies x = 2. Distractor A has incorrect signs for both solutions. Distractors C and D result from incorrect factoring attempts.

Question 4

The equation 4x2−12x+9=04x^2 - 12x + 9 = 0 has exactly one real solution. What is the value of this solution?

  1. −32-\frac{3}{2}
  2. 23\frac{2}{3}
  3. 32\frac{3}{2} (correct answer)
  4. 3
Explanation: The expression 4x2−12x+94x^2 - 12x + 9 is a perfect square trinomial, which factors to (2x−3)2=0(2x - 3)^2 = 0. Setting the factor to zero gives 2x−3=02x - 3 = 0. Solving for xx yields 2x=32x = 3, so x=32x = \frac{3}{2}. This is a 'double root,' meaning there is only one distinct solution.

Question 5

What is the product of the solutions for the equation x2+16x−13=0x^2 + \frac{1}{6}x - \frac{1}{3} = 0?

  1. −13-\frac{1}{3} (correct answer)
  2. −16-\frac{1}{6}
  3. 16\frac{1}{6}
  4. 13\frac{1}{3}
Explanation: For any quadratic equation in the form ax2+bx+c=0ax^2 + bx + c = 0, the product of the solutions is ca\frac{c}{a}. In this equation, a=1a = 1, b=16b = \frac{1}{6}, and c=−13c = -\frac{1}{3}. Therefore, the product of the solutions is −1/31=−13\frac{-1/3}{1} = -\frac{1}{3}. Alternatively, multiplying the equation by 6 to clear fractions gives 6x2+x−2=06x^2 + x - 2 = 0, which factors to (2x−1)(3x+2)=0(2x-1)(3x+2)=0. The solutions are x=1/2x=1/2 and x=−2/3x=-2/3, and their product is (1/2)(−2/3)=−1/3(1/2)(-2/3) = -1/3.

Question 6

The product of two consecutive positive odd integers is 143. What is the value of the larger integer?

  1. -13
  2. 11
  3. 13 (correct answer)
  4. 15
Explanation: Let the two consecutive positive odd integers be xx and x+2x + 2. Their product is x(x+2)=143x(x + 2) = 143. This gives the equation x2+2x−143=0x^2 + 2x - 143 = 0, which factors to (x+13)(x−11)=0(x + 13)(x - 11) = 0. The solutions for xx are x=−13x = -13 and x=11x = 11. Since the integers must be positive, the smaller integer is 11. The larger integer is 11+2=1311 + 2 = 13.

Question 7

If one solution to the equation 2x2+kx−15=02x^2 + kx - 15 = 0 is 3, what is the value of the constant kk?

  1. -5
  2. -1 (correct answer)
  3. 1
  4. 5
Explanation: Since x=3x = 3 is a solution, it must satisfy the equation. Substitute x=3x = 3 into the equation: 2(3)2+k(3)−15=02(3)^2 + k(3) - 15 = 0. This simplifies to 2(9)+3k−15=02(9) + 3k - 15 = 0, then 18+3k−15=018 + 3k - 15 = 0, and then 3+3k=03 + 3k = 0. Solving for kk gives 3k=−33k = -3, so k=−1k = -1.

Question 8

What is the complete set of solutions for the equation 5x2=10x5x^2 = 10x?

  1. {2}\{2\}
  2. {−2,2}\{-2, 2\}
  3. {−2,0}\{-2, 0\}
  4. {0,2}\{0, 2\} (correct answer)
Explanation: To solve, first set the equation to zero: 5x2−10x=05x^2 - 10x = 0. It is incorrect to divide by xx as this would lose a solution. Instead, factor out the greatest common factor, 5x5x, to get 5x(x−2)=05x(x - 2) = 0. The solutions are found by setting each factor to zero: 5x=0  ⟹  x=05x = 0 \implies x=0 and x−2=0  ⟹  x=2x-2 = 0 \implies x=2. The complete solution set is {0,2}\{0, 2\}.

Question 9

What is the positive solution to the equation (x−5)2+3(x−5)−10=0(x-5)^2 + 3(x-5) - 10 = 0?

  1. 0
  2. 2
  3. 5
  4. 7 (correct answer)
Explanation: This equation is quadratic in form. Let u=x−5u = x-5. The equation becomes u2+3u−10=0u^2 + 3u - 10 = 0. Factoring this gives (u+5)(u−2)=0(u+5)(u-2)=0, so u=−5u=-5 or u=2u=2. Substitute back to solve for xx. Case 1: x−5=−5x-5=-5, which gives x=0x=0. Case 2: x−5=2x-5=2, which gives x=7x=7. The positive solution is 7.

Question 10

What is the sum of the solutions to the equation 6x2−x=356x^2 - x = 35?

  1. −16-\frac{1}{6}
  2. 6
  3. 356\frac{35}{6}
  4. 16\frac{1}{6} (correct answer)
Explanation: First, write the equation in standard form: 6x2−x−35=06x^2 - x - 35 = 0. For a quadratic equation ax2+bx+c=0ax^2+bx+c=0, the sum of the solutions is given by −ba-\frac{b}{a}. Here, a=6a=6, b=−1b=-1, and c=−35c=-35. Therefore, the sum of the solutions is −−16=16-\frac{-1}{6} = \frac{1}{6}. Factoring the equation to (2x−5)(3x+7)=0(2x-5)(3x+7)=0 yields solutions x=5/2x=5/2 and x=−7/3x=-7/3, and their sum is 52−73=156−146=16\frac{5}{2} - \frac{7}{3} = \frac{15}{6} - \frac{14}{6} = \frac{1}{6}.

Question 11

A quadratic equation with integer coefficients has solutions at x=−4x = -4 and x=52x = \frac{5}{2}. Which of the following could be the equation?

  1. 2x2−3x−20=02x^2 - 3x - 20 = 0
  2. 2x2+13x+20=02x^2 + 13x + 20 = 0
  3. 2x2−13x−20=02x^2 - 13x - 20 = 0
  4. 2x2+3x−20=02x^2 + 3x - 20 = 0 (correct answer)
Explanation: If the solutions are x=−4x = -4 and x=52x = \frac{5}{2}, then the corresponding factors are (x−(−4))=(x+4)(x - (-4)) = (x + 4) and (x−52)(x - \frac{5}{2}). To obtain integer coefficients, we can use the factor (2x−5)(2x - 5) which also gives the root x=52x = \frac{5}{2}. The equation is formed by the product of the factors: (x+4)(2x−5)=0(x + 4)(2x - 5) = 0. Expanding this gives 2x2−5x+8x−20=02x^2 - 5x + 8x - 20 = 0, which simplifies to 2x2+3x−20=02x^2 + 3x - 20 = 0.

Question 12

What is the sum of the solutions to the equation x(x−3)=18x(x - 3) = 18?

  1. -3
  2. 3 (correct answer)
  3. 9
  4. 18
Explanation: First, rewrite the equation in the standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0. Distribute the xx to get x2−3x=18x^2 - 3x = 18. Then, subtract 18 from both sides to get x2−3x−18=0x^2 - 3x - 18 = 0. Factor the quadratic expression: (x−6)(x+3)=0(x - 6)(x + 3) = 0. The solutions are x=6x = 6 and x=−3x = -3. The sum of the solutions is 6+(−3)=36 + (-3) = 3.

Question 13

What is the distance between the two x-intercepts of the parabola defined by the equation y=2x2−8x+6y = 2x^2 - 8x + 6?

  1. 1
  2. 2 (correct answer)
  3. 3
  4. 4
Explanation: The x-intercepts occur when y=0y = 0. So we must solve the equation 2x2−8x+6=02x^2 - 8x + 6 = 0. First, factor out the common factor of 2: 2(x2−4x+3)=02(x^2 - 4x + 3) = 0. Factoring the trinomial gives 2(x−1)(x−3)=02(x-1)(x-3) = 0. The x-intercepts are at x=1x=1 and x=3x=3. The distance between these two points on the x-axis is the absolute difference of their coordinates: ∣3−1∣=2|3 - 1| = 2.

Question 14

If x2−5x=6x^2 - 5x = 6, what is the positive difference between the two solutions?

  1. 1
  2. 5
  3. 7 (correct answer)
  4. 11
Explanation: First, rearrange to standard form: x2−5x−6=0x^2 - 5x - 6 = 0. Factor: (x−6)(x+1)=0(x - 6)(x + 1) = 0. Solutions are x=6x = 6 and x=−1x = -1. The positive difference is 6−(−1)=76 - (-1) = 7. Choice A represents the difference if one solution was missed. Choice B is the coefficient of the middle term. Choice D would result from adding the solutions instead of finding their difference.

Question 15

Which of the following quadratic equations has solutions x=23x = \frac{2}{3} and x=−14x = -\frac{1}{4}?

  1. 12x2+5x−2=012x^2 + 5x - 2 = 0
  2. 12x2−5x−2=012x^2 - 5x - 2 = 0 (correct answer)
  3. 12x2−5x+2=012x^2 - 5x + 2 = 0
  4. 12x2+5x+2=012x^2 + 5x + 2 = 0
Explanation: Working backwards from the solutions, the factored form is (x−23)(x+14)=0(x - \frac{2}{3})(x + \frac{1}{4}) = 0. Expanding: x2+14x−23x−212=0x^2 + \frac{1}{4}x - \frac{2}{3}x - \frac{2}{12} = 0, which simplifies to x2−512x−16=0x^2 - \frac{5}{12}x - \frac{1}{6} = 0. Multiplying by 12: 12x2−5x−2=012x^2 - 5x - 2 = 0. Choice A has the wrong sign on the middle term. Choices C and D have incorrect signs that would yield different solutions.

Question 16

What is the smaller solution to the equation x2+2x=4x+3x^2 + 2x = 4x + 3?

  1. -3
  2. -1 (correct answer)
  3. 1
  4. 3
Explanation: To solve, first set the equation to zero by moving all terms to one side: x2+2x−4x−3=0x^2 + 2x - 4x - 3 = 0, which simplifies to x2−2x−3=0x^2 - 2x - 3 = 0. This factors into (x−3)(x+1)=0(x - 3)(x + 1) = 0. The solutions are x=3x = 3 and x=−1x = -1. The smaller of these two solutions is -1.

Question 17

If aa is a constant, which of the following is a solution for xx in the equation x2−(a+3)x+3a=0x^2 - (a+3)x + 3a = 0?

  1. x=−ax = -a
  2. x=−3x = -3
  3. x=3x = 3 (correct answer)
  4. x=a+3x = a+3
Explanation: This is a quadratic equation in the variable xx. We need to find two expressions that multiply to 3a3a and add to −(a+3)-(a+3). These are −a-a and −3-3. Thus, the equation factors to (x−a)(x−3)=0(x - a)(x - 3) = 0. The solutions are found by setting each factor to zero: x−a=0  ⟹  x=ax - a = 0 \implies x = a and x−3=0  ⟹  x=3x - 3 = 0 \implies x = 3. Of the choices provided, x=3x = 3 is a solution.

Question 18

What is the positive solution to the equation 3x2−11x=43x^2 - 11x = 4?

  1. 13\frac{1}{3}
  2. −13-\frac{1}{3}
  3. 4 (correct answer)
  4. -4
Explanation: To solve the quadratic equation, first set it to zero: 3x2−11x−4=03x^2 - 11x - 4 = 0. This can be factored as (3x+1)(x−4)=0(3x + 1)(x - 4) = 0. Setting each factor to zero gives the solutions 3x+1=0  ⟹  x=−133x + 1 = 0 \implies x = -\frac{1}{3} and x−4=0  ⟹  x=4x - 4 = 0 \implies x = 4. The question asks for the positive solution, which is 4.

Question 19

What is the product of the solutions to the equation 4x2=814x^2 = 81?

  1. −814-\frac{81}{4} (correct answer)
  2. 0
  3. 814\frac{81}{4}
  4. 81
Explanation: First, set the equation to standard form: 4x2−81=04x^2 - 81 = 0. This is a difference of squares, which factors to (2x−9)(2x+9)=0(2x - 9)(2x + 9) = 0. The solutions are x=92x = \frac{9}{2} and x=−92x = -\frac{9}{2}. The product of these solutions is (92)×(−92)=−814\left(\frac{9}{2}\right) \times \left(-\frac{9}{2}\right) = -\frac{81}{4}. Alternatively, for a quadratic ax2+bx+c=0ax^2+bx+c=0, the product of the roots is c/ac/a. Here, a=4,b=0,c=−81a=4, b=0, c=-81, so the product is −81/4-81/4.

Question 20

What is the sum of the solutions to the equation (x−1)2=3x−5(x-1)^2 = 3x - 5?

  1. -5
  2. 1
  3. 5 (correct answer)
  4. 6
Explanation: First, expand the left side of the equation: x2−2x+1=3x−5x^2 - 2x + 1 = 3x - 5. Next, set the equation to zero by moving all terms to one side: x2−2x−3x+1+5=0x^2 - 2x - 3x + 1 + 5 = 0, which simplifies to x2−5x+6=0x^2 - 5x + 6 = 0. This factors into (x−2)(x−3)=0(x-2)(x-3)=0, so the solutions are x=2x=2 and x=3x=3. Their sum is 2+3=52+3=5.