ACCUPLACER Advanced Algebra & Functions Quiz: Solving Polynomial Equations
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Solving Polynomial EquationsQuestion 1 of 20

The volume of a rectangular prism is represented by the polynomial V(x)=x3+5x2−6xV(x) = x^3 + 5x^2 - 6x, where xx is a positive dimension. For which value of xx is the volume equal to zero?

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ACCUPLACER Advanced Algebra & Functions Quiz

ACCUPLACER Advanced Algebra & Functions Quiz: Solving Polynomial Equations

Practice Solving Polynomial 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 Solving Polynomial 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

The volume of a rectangular prism is represented by the polynomial V(x)=x3+5x2−6xV(x) = x^3 + 5x^2 - 6x, where xx is a positive dimension. For which value of xx is the volume equal to zero?

  1. 1 (correct answer)
  2. 2
  3. 3
  4. 6
Explanation: Set the volume polynomial equal to zero and solve for xx.
[x^3 + 5x^2 - 6x = 0\]
Factor out the greatest common factor, xx:
[x(x2x^2 + 5x - 6) = 0\]
Factor the quadratic trinomial:
[x(x+6)(x-1) = 0\]
The solutions are x=0x=0, x=−6x=-6, and x=1x=1. Since the problem states that xx is a positive dimension, the only valid answer is x=1x=1.

Question 2

What is the sum of the real solutions for the equation 4x4−9x2=04x^4 - 9x^2 = 0?

  1. -9/4
  2. 3
  3. 3/2
  4. 0 (correct answer)
Explanation: First, factor out the greatest common factor, x2x^2.
[x^2(4x24x^2 - 9) = 0\]
The second factor is a difference of squares, (2x)2−32(2x)^2 - 3^2.
[x^2(2x - 3)(2x + 3) = 0\]
Set each factor to zero to find the solutions: x2=0  ⟹  x=0x^2=0 \implies x=0 (a repeated root), 2x−3=0  ⟹  x=3/22x-3=0 \implies x=3/2, and 2x+3=0  ⟹  x=−3/22x+3=0 \implies x=-3/2.
The distinct real solutions are 0,3/2,−3/20, 3/2, -3/2.
The sum of the solutions is 0+3/2+(−3/2)=00 + 3/2 + (-3/2) = 0.

Question 3

What is the complete solution set for the equation (x−5)(x+2)=x−5(x-5)(x+2) = x-5?

  1. {-1}
  2. {5}
  3. {-1, 5} (correct answer)
  4. {-2, 5}
Explanation: To solve, first set the equation to zero. Do not divide by (x−5)(x-5) as this would eliminate a solution.
[(x-5)(x+2) - (x-5) = 0\]
Factor out the common term (x−5)(x-5):
[(x-5)((x+2) - 1) = 0\]
Simplify the expression in the second parenthesis:
[(x-5)(x+1) = 0\]
Set each factor to zero: x−5=0x-5=0 gives x=5x=5, and x+1=0x+1=0 gives x=−1x=-1.
The complete solution set is {−1,5}\{-1, 5\}.

Question 4

If x>0x > 0, what is the solution to x3+3x2=10xx^3 + 3x^2 = 10x?

  1. 2 (correct answer)
  2. 5
  3. 7
  4. 10
Explanation: First, set the equation to zero by subtracting 10x10x from both sides.
[x^3 + 3x^2 - 10x = 0\]
Next, factor out the greatest common factor, xx:
[x(x2x^2 + 3x - 10) = 0\]
Factor the quadratic trinomial:
[x(x+5)(x-2) = 0\]
The solutions are x=0x=0, x=−5x=-5, and x=2x=2.
The question specifies that x>0x > 0, so the only valid solution is x=2x=2.

Question 5

The x-intercepts of the graph of a function f(x)f(x) are the real solutions to f(x)=0f(x)=0. How many distinct x-intercepts does the graph of f(x)=x4+6x3+9x2f(x) = x^4 + 6x^3 + 9x^2 have?

  1. 1
  2. 2 (correct answer)
  3. 3
  4. 4
Explanation: To find the x-intercepts, set f(x)=0f(x) = 0 and solve for xx.
[x^4 + 6x^3 + 9x^2 = 0\]
Factor out the greatest common factor, x2x^2:
[x^2(x2x^2 + 6x + 9) = 0\]
The expression in the parentheses is a perfect square trinomial.
[x^2(x+3)^2 = 0\]
The solutions are found by setting the factors to zero: x2=0  ⟹  x=0x^2 = 0 \implies x=0 and (x+3)2=0  ⟹  x=−3(x+3)^2 = 0 \implies x=-3.
Even though these roots have multiplicity 2, they correspond to only two distinct points on the x-axis. Thus, there are 2 distinct x-intercepts.

Question 6

What is the smallest integer solution to the equation x3−2x2=9x−18x^3 - 2x^2 = 9x - 18?

  1. -3 (correct answer)
  2. -2
  3. 2
  4. 3
Explanation: First, set the equation to zero.
[x^3 - 2x^2 - 9x + 18 = 0\]
Now, use factoring by grouping.
[(x3x^3 - 2x22x^2) - (9x - 18) = 0\]
[x^2(x - 2) - 9(x - 2) = 0\]
[(x2x^2 - 9)(x - 2) = 0\]
[(x-3)(x+3)(x-2) = 0\]
The solutions are x=3x=3, x=−3x=-3, and x=2x=2. All are integers. The smallest of these is -3.

Question 7

Which of the following is the complete solution set for the equation 5x3−20x=05x^3 - 20x = 0?

  1. {-2, 2}
  2. {-2, 0, 2} (correct answer)
  3. {0, 4}
  4. {-4, 4}
Explanation: First, factor out the greatest common factor, 5x5x.
[5x(x2x^2 - 4) = 0\]
Next, factor the difference of squares, x2−4x^2 - 4.
[5x(x - 2)(x + 2) = 0\]
Using the Zero Product Property, set each factor equal to zero: 5x=05x=0, x−2=0x-2=0, and x+2=0x+2=0.
This gives the solutions x=0x=0, x=2x=2, and x=−2x=-2. The complete solution set is {−2,0,2}\{-2, 0, 2\}.

Question 8

The equation x3−4x2+4x=0x^3 - 4x^2 + 4x = 0 has a real solution with a multiplicity of 2. What is the value of this solution?

  1. -2
  2. 0
  3. 2 (correct answer)
  4. 4
Explanation: First, factor the polynomial. The greatest common factor is xx.
[x(x2x^2 - 4x + 4) = 0\]
The quadratic factor is a perfect square trinomial.
[x(x-2)^2 = 0\]
The solutions are found by setting each factor to zero. x=0x=0 is one solution. (x−2)2=0(x-2)^2=0 gives x−2=0x-2=0, so x=2x=2. Because the factor (x−2)(x-2) is squared, the root x=2x=2 has a multiplicity of 2. Thus, 2 is the repeated solution.

Question 9

Given that x=3x=3 is one solution to the equation x3−x2−17x+33=0x^3 - x^2 - 17x + 33 = 0, what is the sum of the other two solutions?

  1. -11
  2. -2 (correct answer)
  3. 1
  4. 2
Explanation: If x=3x=3 is a solution, then (x−3)(x-3) is a factor of the polynomial. We can use polynomial division or Vieta's formulas. Using Vieta's formulas, for a cubic equation ax3+bx2+cx+d=0ax^3+bx^2+cx+d=0, the sum of the roots r1+r2+r3r_1+r_2+r_3 is −b/a-b/a.
In this equation, a=1a=1 and b=−1b=-1, so the sum of all three roots is −(−1)/1=1-(-1)/1 = 1.
Let the roots be r1,r2,r3r_1, r_2, r_3. We are given r1=3r_1 = 3.
So, 3+r2+r3=13 + r_2 + r_3 = 1.
Subtracting 3 from both sides gives r2+r3=1−3=−2r_2 + r_3 = 1 - 3 = -2. The sum of the other two solutions is -2.

Question 10

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

  1. 1
  2. 2 (correct answer)
  3. 3
  4. 4
Explanation: The expression is a difference of squares, (x2)2−42(x^2)^2 - 4^2.
[(x2x^2 - 4)(x2x^2 + 4) = 0\]
The first factor, x2−4x^2 - 4, is also a difference of squares.
[(x - 2)(x + 2)(x2x^2 + 4) = 0\]
Setting each factor to zero gives x−2=0  ⟹  x=2x-2=0 \implies x=2, x+2=0  ⟹  x=−2x+2=0 \implies x=-2, and x2+4=0  ⟹  x2=−4x^2+4=0 \implies x^2=-4.
The first two factors give the distinct real solutions 2 and -2. The equation x2=−4x^2=-4 has no real solutions. Therefore, there are exactly 2 distinct real solutions.

Question 11

The equation x3−8=0x^3 - 8 = 0 has one real solution and two non-real complex solutions. What is the product of the two non-real solutions?

  1. -4
  2. 2
  3. 4 (correct answer)
  4. 8
Explanation: Factor the difference of cubes: x3−23=(x−2)(x2+2x+4)=0x^3 - 2^3 = (x-2)(x^2 + 2x + 4) = 0.
The real solution comes from x−2=0x-2=0, which is x=2x=2.
The two non-real solutions come from the quadratic factor x2+2x+4=0x^2 + 2x + 4 = 0.
For a quadratic equation ax2+bx+c=0ax^2+bx+c=0, the product of the roots is given by c/ac/a.
In this case, a=1a=1 and c=4c=4. So, the product of the two non-real roots is 4/1=44/1 = 4.

Question 12

What is the sum of the distinct rational solutions to the equation 3x3−7x2+2x=03x^3 - 7x^2 + 2x = 0?

  1. 1/3
  2. 2
  3. 7/3 (correct answer)
  4. 3
Explanation: First, factor out the greatest common factor, xx:
[x(3x23x^2 - 7x + 2) = 0\]
Next, factor the quadratic trinomial 3x2−7x+23x^2 - 7x + 2. We look for two numbers that multiply to 3×2=63 \times 2 = 6 and add to -7. These are -1 and -6.
[x(3x-1)(x-2) = 0\]
Set each factor to zero to find the solutions: x=0x=0, 3x−1=0  ⟹  x=1/33x-1=0 \implies x=1/3, and x−2=0  ⟹  x=2x-2=0 \implies x=2.
All three solutions (0, 1/3, 2) are rational.
The sum of these distinct solutions is 0+1/3+2=2+1/3=7/30 + 1/3 + 2 = 2 + 1/3 = 7/3.

Question 13

What is the largest real solution to the equation 2x3−x2+8x−4=02x^3 - x^2 + 8x - 4 = 0?

  1. -2
  2. 4
  3. 2
  4. 1/2 (correct answer)
Explanation: This four-term polynomial can be solved by factoring by grouping.
[(2x32x^3 - x2x^2) + (8x - 4) = 0\]
Factor out the GCF from each pair of terms:
[x^2(2x - 1) + 4(2x - 1) = 0\]
Factor out the common binomial factor (2x−1)(2x - 1):
[(x2x^2 + 4)(2x - 1) = 0\]
This gives two possibilities: x2+4=0x^2+4=0 or 2x−1=02x-1=0.
The equation x2+4=0x^2+4=0 means x2=−4x^2=-4, which has no real solutions.
The equation 2x−1=02x-1=0 gives 2x=12x=1, so x=1/2x=1/2.
Since this is the only real solution, it is also the largest real solution.

Question 14

What is the product of the real solutions to the equation (x2−5)(x2+2)=0(x^2 - 5)(x^2 + 2) = 0?

  1. -10
  2. -5 (correct answer)
  3. 5
  4. 10
Explanation: Using the Zero Product Property, we set each factor to zero.
Case 1: x2−5=0  ⟹  x2=5  ⟹  x=±5x^2 - 5 = 0 \implies x^2 = 5 \implies x = \pm\sqrt{5}. These are two real solutions.
Case 2: x2+2=0  ⟹  x2=−2x^2 + 2 = 0 \implies x^2 = -2. This equation has no real solutions, as the square of a real number cannot be negative.
The only real solutions are 5\sqrt{5} and −5-\sqrt{5}.
The product of these solutions is (5)(−5)=−5(\sqrt{5})(-\sqrt{5}) = -5.

Question 15

How many distinct real solutions does the equation x3+27=0x^3 + 27 = 0 have?

  1. 0
  2. 1 (correct answer)
  3. 2
  4. 3
Explanation: The equation is a sum of cubes, which factors as a3+b3=(a+b)(a2−ab+b2)a^3+b^3 = (a+b)(a^2-ab+b^2).
[x^3 + 3^3 = (x+3)(x2x^2 - 3x + 9) = 0\]
The first factor, x+3=0x+3=0, gives one real solution, x=−3x=-3.
For the second factor, x2−3x+9=0x^2 - 3x + 9 = 0, we check the discriminant, b2−4acb^2-4ac:
[(-3)^2 - 4(1)(9) = 9 - 36 = -27\]
Since the discriminant is negative, this quadratic factor has no real solutions (it has two complex solutions). Therefore, the equation has only one distinct real solution.

Question 16

What is the sum of the positive real solutions to the equation x4−5x2+4=0x^4 - 5x^2 + 4 = 0?

  1. 0
  2. 2
  3. 3 (correct answer)
  4. 5
Explanation: This equation is in quadratic form. Let u=x2u = x^2. The equation becomes:
[u^2 - 5u + 4 = 0\]
Factor the quadratic:
[(u-4)(u-1) = 0\]
This gives u=4u=4 or u=1u=1. Substitute back x2x^2 for uu:
[x^2 = 4 \quad \text{or} \quad x^2 = 1\]
Solving for xx gives x=±2x = \pm 2 and x=±1x = \pm 1. The positive real solutions are 1 and 2. Their sum is 1+2=31 + 2 = 3.

Question 17

What is the product of the non-zero solutions for 2x3+5x2−3x=02x^3 + 5x^2 - 3x = 0?

  1. -3
  2. 1/2
  3. 0
  4. -3/2 (correct answer)
Explanation: First, factor out the greatest common factor, xx.
[x(2x22x^2 + 5x - 3) = 0\]
Next, factor the quadratic trinomial.
[x(2x-1)(x+3) = 0\]
The solutions are x=0x=0, x=1/2x=1/2, and x=−3x=-3.
The non-zero solutions are 1/21/2 and −3-3.
Their product is (1/2)×(−3)=−3/2(1/2) \times (-3) = -3/2.

Question 18

What is the largest solution to the equation x(x+1)=12x(x+1) = 12?

  1. 3 (correct answer)
  2. 4
  3. 11
  4. 12
Explanation: To solve the equation, first set it equal to zero.
[x(x+1) = 12\]
[x^2 + x = 12\]
[x^2 + x - 12 = 0\]
Now, factor the quadratic trinomial.
[(x+4)(x-3) = 0\]
The solutions are x=−4x = -4 and x=3x = 3. The largest of these solutions is 3.

Question 19

For the polynomial equation ax2−bx−c=0ax^2 - bx - c = 0, the solutions are x=2x=2 and x=−1/3x=-1/3. Which of the following could be the factored form of the polynomial ax2−bx−cax^2 - bx - c?

  1. (x+2)(3x−1)(x+2)(3x-1)
  2. (x−2)(x+1/3)(x-2)(x+1/3)
  3. (x+2)(x−1/3)(x+2)(x-1/3)
  4. (x−2)(3x+1)(x-2)(3x+1) (correct answer)
Explanation: If the solutions (roots) of a polynomial equation are r1r_1 and r2r_2, then the factored form of the polynomial is a(x−r1)(x−r2)a(x-r_1)(x-r_2).
Given the roots r1=2r_1 = 2 and r2=−1/3r_2 = -1/3, the factors are (x−2)(x-2) and (x−(−1/3))(x - (-1/3)), which simplifies to (x+1/3)(x+1/3).
So the factored form is a(x−2)(x+1/3)a(x-2)(x+1/3). To eliminate the fraction and match the answer choices, we can choose a value for aa, such as a=3a=3.
[3(x-2)(x+1/3) = (x-2) \cdot 3(x+1/3) = (x-2)(3x+1)\]
This corresponds to the polynomial 3x2−5x−23x^2 - 5x - 2, which has the form ax2−bx−cax^2 - bx - c (with a=3, b=5, c=2). The factored form is (x−2)(3x+1)(x-2)(3x+1).

Question 20

For what values of kk does the equation x3−7x2+kx−12=0x^3 - 7x^2 + kx - 12 = 0 have x=3x = 3 as a solution?

  1. k=8k = 8 only
  2. k=16k = 16 only (correct answer)
  3. k=8k = 8 or k=16k = 16
  4. k=12k = 12 only
Explanation: If x=3x = 3 is a solution, then substituting gives: 33−7(32)+k(3)−12=03^3 - 7(3^2) + k(3) - 12 = 0, so 27−63+3k−12=027 - 63 + 3k - 12 = 0, which simplifies to 3k−48=03k - 48 = 0, giving k=16k = 16. Choice A comes from incorrectly calculating 7×9=567 \times 9 = 56 instead of 6363. Choice C suggests multiple values exist. Choice D confuses the constant term with the parameter.