ACCUPLACER Advanced Algebra & Functions Quiz: Factoring To Solve Equations
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Factoring To Solve EquationsQuestion 1 of 20

What is the complete set of real solutions for the equation 8x3+27=08x^3 + 27 = 0?

{−32}\{-\frac{3}{2}\}
{32}\{\frac{3}{2}\}
{−32,32}\{-\frac{3}{2}, \frac{3}{2}\}
{0,−32,32}\{0, -\frac{3}{2}, \frac{3}{2}\}
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ACCUPLACER Advanced Algebra & Functions Quiz

ACCUPLACER Advanced Algebra & Functions Quiz: Factoring To Solve Equations

Practice Factoring To Solve Equations 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 Factoring To Solve Equations, 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

What is the complete set of real solutions for the equation 8x3+27=08x^3 + 27 = 0?

  1. {−32}\{-\frac{3}{2}\} (correct answer)
  2. {32}\{\frac{3}{2}\}
  3. {−32,32}\{-\frac{3}{2}, \frac{3}{2}\}
  4. {0,−32,32}\{0, -\frac{3}{2}, \frac{3}{2}\}
Explanation: This equation is a sum of cubes: (2x)3+33=0(2x)^3 + 3^3 = 0. The sum of cubes formula is a3+b3=(a+b)(a2−ab+b2)a^3+b^3 = (a+b)(a^2-ab+b^2). Applying this gives (2x+3)(4x2−6x+9)=0(2x+3)(4x^2 - 6x + 9) = 0. The first factor, 2x+3=02x+3=0, yields the real solution x=−32x = -\frac{3}{2}. The second factor, 4x2−6x+9=04x^2 - 6x + 9 = 0, has a discriminant of b2−4ac=(−6)2−4(4)(9)=36−144=−108b^2-4ac = (-6)^2 - 4(4)(9) = 36 - 144 = -108. Since the discriminant is negative, there are no other real solutions.

Question 2

What is the smallest solution to the equation 30+5x−5x2=030 + 5x - 5x^2 = 0?

  1. −3-3
  2. −2-2 (correct answer)
  3. 22
  4. 33
Explanation: First, rearrange the equation into standard form and factor out the greatest common factor, -5: −5x2+5x+30=0-5x^2 + 5x + 30 = 0 becomes −5(x2−x−6)=0-5(x^2 - x - 6) = 0. Divide by -5 to get x2−x−6=0x^2 - x - 6 = 0. Factoring the trinomial gives (x−3)(x+2)=0(x-3)(x+2) = 0. The solutions are x=3x=3 and x=−2x=-2. The smallest of these solutions is −2-2.

Question 3

The equation 4x2+kx+9=04x^2 + kx + 9 = 0 has exactly one real solution. If k>0k > 0, what is the value of the solution?

  1. −32-\frac{3}{2} (correct answer)
  2. −23-\frac{2}{3}
  3. 23\frac{2}{3}
  4. 32\frac{3}{2}
Explanation: For a quadratic equation to have exactly one real solution, it must be a perfect square trinomial. The form is (ax+b)2=a2x2+2abx+b2(ax+b)^2 = a^2x^2 + 2abx + b^2. Here, a2=4a^2=4 so a=2a=2, and b2=9b^2=9 so b=3b=3. The middle term kxkx must be 2abx=2(2)(3)x=12x2abx = 2(2)(3)x = 12x. So, k=12k=12. The equation is 4x2+12x+9=04x^2 + 12x + 9 = 0, which factors to (2x+3)2=0(2x+3)^2=0. Solving for xx gives 2x+3=02x+3=0, so x=−32x = -\frac{3}{2}.

Question 4

One of the factors of x4−81x^4 - 81 is x2+9x^2 + 9. What is the sum of the real solutions to the equation x4−81=0x^4 - 81 = 0?

  1. −6-6
  2. 00 (correct answer)
  3. 66
  4. 1818
Explanation: The expression x4−81x^4 - 81 is a difference of squares, (x2)2−92(x^2)^2 - 9^2, which factors to (x2−9)(x2+9)=0(x^2 - 9)(x^2 + 9) = 0. The factor x2+9=0x^2 + 9 = 0 has no real solutions. The factor x2−9=0x^2 - 9 = 0 is another difference of squares, (x−3)(x+3)=0(x-3)(x+3)=0. This gives the real solutions x=3x=3 and x=−3x=-3. The sum of these solutions is 3+(−3)=03 + (-3) = 0.

Question 5

What is the positive solution to the equation (x+3)(x−3)=40(x+3)(x-3) = 40?

  1. 55
  2. 77 (correct answer)
  3. 88
  4. 3737
Explanation: First, expand the left side of the equation, which is a difference of squares: x2−9=40x^2 - 9 = 40. Next, set the equation to zero: x2−49=0x^2 - 49 = 0. Factor the new difference of squares: (x−7)(x+7)=0(x-7)(x+7) = 0. The solutions are x=7x=7 and x=−7x=-7. The positive solution is 7.

Question 6

If 2x3−x2=18x−92x^3 - x^2 = 18x - 9, which of the following is NOT a solution?

  1. −3-3
  2. 12\frac{1}{2}
  3. 33
  4. 99 (correct answer)
Explanation: First, set the equation to zero: 2x3−x2−18x+9=02x^3 - x^2 - 18x + 9 = 0. Factor by grouping: x2(2x−1)−9(2x−1)=0x^2(2x - 1) - 9(2x - 1) = 0. This gives (x2−9)(2x−1)=0(x^2 - 9)(2x - 1) = 0, which further factors to (x−3)(x+3)(2x−1)=0(x - 3)(x + 3)(2x - 1) = 0. The solutions are x=3x = 3, x=−3x = -3, and x=12x = \frac{1}{2}. The value that is NOT a solution is 9.

Question 7

What is the sum of all solutions to the equation x−8x+15=0x - 8\sqrt{x} + 15 = 0?

  1. 88
  2. 1515
  3. 1616
  4. 3434 (correct answer)
Explanation: This equation is quadratic in form. Let u=xu = \sqrt{x}, so u2=xu^2 = x. The equation becomes u2−8u+15=0u^2 - 8u + 15 = 0. Factoring gives (u−3)(u−5)=0(u-3)(u-5)=0, so u=3u=3 or u=5u=5. Substitute back: x=3\sqrt{x}=3 or x=5\sqrt{x}=5. Squaring both sides of each equation gives x=9x=9 and x=25x=25. Both solutions are valid. The sum of the solutions is 9+25=349 + 25 = 34.

Question 8

If 6x2−13x+5=06x^2 - 13x + 5 = 0, what is the larger of the two solutions?

  1. 12\frac{1}{2}
  2. 23\frac{2}{3}
  3. 53\frac{5}{3} (correct answer)
  4. 52\frac{5}{2}
Explanation: To solve the equation 6x2−13x+5=06x^2 - 13x + 5 = 0, we can factor it. The factored form is (3x−5)(2x−1)=0(3x - 5)(2x - 1) = 0. Setting the first factor to zero, 3x−5=03x - 5 = 0, gives x=53x = \frac{5}{3}. Setting the second factor to zero, 2x−1=02x - 1 = 0, gives x=12x = \frac{1}{2}. Comparing the two solutions, 53≈1.67\frac{5}{3} \approx 1.67 and 12=0.5\frac{1}{2} = 0.5. The larger solution is 53\frac{5}{3}.

Question 9

The solutions to the equation x2−8x+12=0x^2 - 8x + 12 = 0 are x=2x=2 and x=6x=6. Which of the following equations has solutions that are each 3 greater than the solutions of the original equation?

  1. x2−14x+45=0x^2 - 14x + 45 = 0 (correct answer)
  2. x2−11x+18=0x^2 - 11x + 18 = 0
  3. x2−8x+15=0x^2 - 8x + 15 = 0
  4. x2−2x−3=0x^2 - 2x - 3 = 0
Explanation: The original solutions are 2 and 6. The new solutions must be 3 greater than these, so they are 2+3=52+3=5 and 6+3=96+3=9. To find the equation with these solutions, use the factors (x−5)(x-5) and (x−9)(x-9). The equation is (x−5)(x−9)=0(x-5)(x-9) = 0. Expanding this gives x2−9x−5x+45=0x^2 - 9x - 5x + 45 = 0, which simplifies to x2−14x+45=0x^2 - 14x + 45 = 0.

Question 10

Which of the following is a solution to the equation (2x−1)2=49(2x - 1)^2 = 49?

  1. −4-4
  2. −3-3 (correct answer)
  3. 33
  4. 2525
Explanation: The equation can be solved by setting it to zero and factoring as a difference of squares: (2x−1)2−49=0(2x - 1)^2 - 49 = 0, which is (2x−1−7)(2x−1+7)=0(2x - 1 - 7)(2x - 1 + 7) = 0. This simplifies to (2x−8)(2x+6)=0(2x - 8)(2x + 6) = 0. The solutions are from 2x−8=02x - 8 = 0 (which gives x=4x = 4) and 2x+6=02x + 6 = 0 (which gives x=−3x = -3). Of the choices provided, −3-3 is a solution.

Question 11

What is the complete solution set for the equation 3x3−12x2+9x=03x^3 - 12x^2 + 9x = 0?

  1. {1,3}\{1, 3\}
  2. {−1,−3}\{-1, -3\}
  3. {0,1,3}\{0, 1, 3\} (correct answer)
  4. {0,−1,−3}\{0, -1, -3\}
Explanation: First, factor out the greatest common factor, 3x3x: 3x(x2−4x+3)=03x(x^2 - 4x + 3) = 0. Then, factor the trinomial: 3x(x−1)(x−3)=03x(x - 1)(x - 3) = 0. By the zero product property, the solutions are found by setting each factor to zero: 3x=03x = 0 gives x=0x=0, x−1=0x - 1 = 0 gives x=1x=1, and x−3=0x - 3 = 0 gives x=3x=3. The complete solution set is {0,1,3}\{0, 1, 3\}.

Question 12

For which value of xx does (x−2)2−9=0(x - 2)^2 - 9 = 0?

  1. x=5x = 5 only
  2. x=−1x = -1 only
  3. x=5x = 5 or x=−1x = -1 (correct answer)
  4. x=2x = 2 or x=3x = 3
Explanation: Recognize this as a difference of squares: (x−2)2−9=(x−2)2−32=[(x−2)−3][(x−2)+3]=(x−5)(x+1)=0(x - 2)^2 - 9 = (x - 2)^2 - 3^2 = [(x - 2) - 3][(x - 2) + 3] = (x - 5)(x + 1) = 0. This gives x=5x = 5 or x=−1x = -1. Choices A and B identify only one solution each. Choice D comes from incorrectly expanding or factoring the expression.

Question 13

The equation 4x2−4x−8=04x^2 - 4x - 8 = 0 can be solved by factoring. What is the positive solution?

  1. x=1x = 1
  2. x=2x = 2 (correct answer)
  3. x=4x = 4
  4. x=8x = 8
Explanation: First factor out the GCF: 4(x2−x−2)=04(x^2 - x - 2) = 0, so x2−x−2=0x^2 - x - 2 = 0. Factor the quadratic: (x−2)(x+1)=0(x - 2)(x + 1) = 0. The solutions are x=2x = 2 and x=−1x = -1. The positive solution is x=2x = 2. Choice A comes from incorrectly factoring or solving. Choice C uses the leading coefficient. Choice D uses the constant term from the original equation.

Question 14

If (x+3)(2x−1)=x+3(x + 3)(2x - 1) = x + 3, what are all possible values of xx?

  1. x=−3x = -3 only
  2. x=1x = 1 only
  3. x=−3x = -3 or x=12x = \frac{1}{2}
  4. x=−3x = -3 or x=1x = 1 (correct answer)
Explanation: When you encounter an equation where one side is a factor of the other, like (x+3)(2x−1)=x+3(x + 3)(2x - 1) = x + 3, you're dealing with a factoring situation that requires careful algebraic manipulation. To solve this equation, start by moving everything to one side: (x+3)(2x−1)−(x+3)=0(x + 3)(2x - 1) - (x + 3) = 0. Now you can factor out the common term (x+3)(x + 3): (x+3)[(2x−1)−1]=0(x + 3)[(2x - 1) - 1] = 0, which simplifies to (x+3)(2x−2)=0(x + 3)(2x - 2) = 0, or (x+3)⋅2(x−1)=0(x + 3) \cdot 2(x - 1) = 0. Using the zero product property, either x+3=0x + 3 = 0 or x−1=0x - 1 = 0. This gives you x=−3x = -3 or x=1x = 1. You can verify: when x=−3x = -3, both sides equal 00; when x=1x = 1, both sides equal 44. Looking at the wrong answers: Choice A (x=−3x = -3 only) misses the second solution by failing to apply the zero product property completely. Choice B (x=1x = 1 only) makes the same error in reverse. Choice C (x=−3x = -3 or x=12x = \frac{1}{2}) likely results from an arithmetic error when solving 2x−1=12x - 1 = 1, incorrectly getting 2x=122x = \frac{1}{2} instead of 2x=22x = 2. The correct answer is D. Remember: when you have a product equal to a single term, factor out common expressions first, then use the zero product property. Always check both solutions by substituting back into the original equation.

Question 15

What is the largest real solution to the equation x4−13x2+36=0x^4 - 13x^2 + 36 = 0?

  1. 22
  2. 33 (correct answer)
  3. 44
  4. 99
Explanation: This equation is quadratic in form. Let u=x2u = x^2. The equation becomes u2−13u+36=0u^2 - 13u + 36 = 0. Factoring gives (u−9)(u−4)=0(u - 9)(u - 4) = 0, so u=9u = 9 or u=4u = 4. Substituting back x2x^2 for uu gives x2=9x^2 = 9 or x2=4x^2 = 4. The solutions for xx are x=±3x = \pm 3 and x=±2x = \pm 2. The largest of these solutions is 3.

Question 16

The area of a rectangle is 54 square inches. The length is 3 inches more than the width. What is the perimeter of the rectangle in inches?

  1. 1515
  2. 2727
  3. 3030 (correct answer)
  4. 4242
Explanation: Let ww be the width. Then the length l=w+3l = w + 3. The area is l×w=(w+3)w=54l \times w = (w+3)w = 54. This gives the quadratic equation w2+3w−54=0w^2 + 3w - 54 = 0. Factoring yields (w+9)(w−6)=0(w+9)(w-6) = 0. Since the width must be positive, w=6w=6. The length is l=6+3=9l = 6 + 3 = 9. The perimeter is 2(l+w)=2(9+6)=2(15)=302(l+w) = 2(9+6) = 2(15) = 30 inches.

Question 17

What is the solution set for the equation x−6=16xx - 6 = \frac{16}{x}?

  1. {−8,2}\{-8, 2\}
  2. {−4,4}\{-4, 4\}
  3. {−2,8}\{-2, 8\} (correct answer)
  4. {2,8}\{2, 8\}
Explanation: Assuming x≠0x \neq 0, multiply both sides by xx to clear the denominator: x(x−6)=16x(x-6) = 16, which simplifies to x2−6x=16x^2 - 6x = 16. Set the equation to zero: x2−6x−16=0x^2 - 6x - 16 = 0. Factoring the quadratic gives (x−8)(x+2)=0(x-8)(x+2) = 0. The solutions are x=8x = 8 and x=−2x = -2.

Question 18

If x2−3x=10x^2 - 3x = 10, what is the sum of the possible values of xx?

  1. −7-7
  2. −3-3
  3. 33 (correct answer)
  4. 77
Explanation: To solve the equation, first set it to zero: x2−3x−10=0x^2 - 3x - 10 = 0. Factoring the quadratic gives (x−5)(x+2)=0(x - 5)(x + 2) = 0. The solutions are x=5x = 5 and x=−2x = -2. The sum of these values is 5+(−2)=35 + (-2) = 3.

Question 19

The length of a rectangular garden is 1 foot less than twice its width. Its area is 120 square feet. What is its width in feet?

  1. 7.57.5
  2. 88 (correct answer)
  3. 1010
  4. 1515
Explanation: Let ww be the width. The length ll is 2w−12w-1. The area is w(2w−1)=120w(2w-1) = 120. This gives the equation 2w2−w−120=02w^2 - w - 120 = 0. Factoring this quadratic (by grouping or trial and error) gives (2w+15)(w−8)=0(2w+15)(w-8) = 0. The possible solutions are w=−152w = -\frac{15}{2} or w=8w=8. Since width must be positive, the width is 8 feet.

Question 20

How many distinct real solutions does the equation 16x4−1=016x^4 - 1 = 0 have?

  1. 22 (correct answer)
  2. 11
  3. 00
  4. 44
Explanation: The equation is a difference of squares: (4x2)2−12=0(4x^2)^2 - 1^2 = 0. It factors into (4x2−1)(4x2+1)=0(4x^2 - 1)(4x^2 + 1) = 0. The second factor, 4x2+1=04x^2 + 1 = 0, has no real solutions because x2x^2 cannot be negative. The first factor, 4x2−1=04x^2 - 1 = 0, is also a difference of squares, factoring into (2x−1)(2x+1)=0(2x - 1)(2x + 1) = 0. This yields two distinct real solutions: x=12x = \frac{1}{2} and x=−12x = -\frac{1}{2}.