ACCUPLACER Advanced Algebra & Functions Quiz: Polynomial Zeros
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Polynomial ZerosQuestion 1 of 20

A cubic polynomial has integer coefficients. If two of its zeros are x=5x=5 and x=2x=\sqrt{2}, which of the following must also be a zero of the polynomial?

−2-\sqrt{2}
5−25-\sqrt{2}
15\frac{1}{5}
i2i\sqrt{2}
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ACCUPLACER Advanced Algebra & Functions Quiz

ACCUPLACER Advanced Algebra & Functions Quiz: Polynomial Zeros

Practice Polynomial Zeros 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 Polynomial Zeros, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Advanced Algebra & Functions.

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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.

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Question 1

A cubic polynomial has integer coefficients. If two of its zeros are x=5x=5 and x=2x=\sqrt{2}, which of the following must also be a zero of the polynomial?

  1. −2-\sqrt{2} (correct answer)
  2. 5−25-\sqrt{2}
  3. 15\frac{1}{5}
  4. i2i\sqrt{2}
Explanation: The Irrational Conjugate Root Theorem states that if a polynomial with rational (and thus integer) coefficients has an irrational zero of the form a+bca + b\sqrt{c}, then its conjugate a−bca - b\sqrt{c} must also be a zero. In this case, the zero 2\sqrt{2} can be written as 0+20 + \sqrt{2}. Its conjugate is 0−20 - \sqrt{2}, or −2-\sqrt{2}.

Question 2

If x2−4x^2 - 4 is a factor of the polynomial P(x)=x3−kx2−4x+8P(x) = x^3 - kx^2 - 4x + 8, what is the value of kk?

  1. -4
  2. -2
  3. 2 (correct answer)
  4. 4
Explanation: If x2−4x^2 - 4 is a factor, then its roots, x=2x=2 and x=−2x=-2, must be zeros of P(x)P(x). We can use either zero to find kk. Using x=2x=2, we set P(2)=0P(2) = 0: P(2)=(2)3−k(2)2−4(2)+8=8−4k−8+8=0P(2) = (2)^3 - k(2)^2 - 4(2) + 8 = 8 - 4k - 8 + 8 = 0. This simplifies to 8−4k=08 - 4k = 0, which gives 4k=84k=8 and k=2k=2.

Question 3

A polynomial of degree 4 with real coefficients has zeros at 3i3i and 1−i1-i. What is the product of all the zeros of this polynomial?

  1. -18
  2. 0
  3. 18 (correct answer)
  4. 20
Explanation: Since the polynomial has real coefficients, complex zeros must come in conjugate pairs. If 3i3i is a zero, then −3i-3i must also be a zero. If 1−i1-i is a zero, then 1+i1+i must also be a zero. The four zeros are 3i,−3i,1−i,1+i3i, -3i, 1-i, 1+i. The product is (3i)(−3i)(1−i)(1+i)(3i)(-3i)(1-i)(1+i). This simplifies to (−9i2)(12−i2)=(9)(1−(−1))=(9)(2)=18(-9i^2)(1^2-i^2) = (9)(1 - (-1)) = (9)(2) = 18.

Question 4

What is the product of the real zeros of the equation x4−13x2+36=0x^4 - 13x^2 + 36 = 0?

  1. -36
  2. 0
  3. 13
  4. 36 (correct answer)
Explanation: This equation is quadratic in form. Let u=x2u = x^2. The equation becomes u2−13u+36=0u^2 - 13u + 36 = 0, which factors as (u−9)(u−4)=0(u-9)(u-4) = 0. So, u=9u=9 or u=4u=4. Substituting back, we have x2=9x^2 = 9 or x2=4x^2 = 4. The solutions are x=±3x = \pm 3 and x=±2x = \pm 2. All four zeros are real. Their product is (3)(−3)(2)(−2)=(−9)(4)=36(3)(-3)(2)(-2) = (-9)(4) = 36.

Question 5

A polynomial P(x)P(x) has P(−2)=0P(-2) = 0 and P(5)=0P(5) = 0. Which of the following quadratic expressions must be a factor of P(x)P(x)?

  1. x2+3x−10x^2 + 3x - 10
  2. x2−3x−10x^2 - 3x - 10 (correct answer)
  3. x2+7x+10x^2 + 7x + 10
  4. x2−7x+10x^2 - 7x + 10
Explanation: If P(−2)=0P(-2)=0, then by the Factor Theorem, (x−(−2))(x - (-2)) or (x+2)(x+2) is a factor. If P(5)=0P(5)=0, then (x−5)(x-5) is a factor. Since both are factors, their product must also be a factor. The product is (x+2)(x−5)=x2−5x+2x−10=x2−3x−10(x+2)(x-5) = x^2 - 5x + 2x - 10 = x^2 - 3x - 10.

Question 6

The polynomial f(x)=x3−2x2+kx+18f(x) = x^3 - 2x^2 + kx + 18 has a zero at x=−3x=-3. What is the value of kk?

  1. -9 (correct answer)
  2. -3
  3. 3
  4. 9
Explanation: If x=−3x=-3 is a zero of the polynomial, then f(−3)=0f(-3) = 0. Substitute x=−3x=-3 into the function: f(−3)=(−3)3−2(−3)2+k(−3)+18=0f(-3) = (-3)^3 - 2(-3)^2 + k(-3) + 18 = 0. This becomes −27−2(9)−3k+18=0-27 - 2(9) - 3k + 18 = 0, which simplifies to −27−18−3k+18=0-27 - 18 - 3k + 18 = 0, or −27−3k=0-27 - 3k = 0. Solving for kk gives −3k=27-3k = 27, so k=−9k = -9.

Question 7

The graph of the polynomial P(x)=(x−a)2(x−1)P(x) = (x-a)^2 (x-1) passes through the point (0,−16)(0, -16). If a>0a > 0, what is the value of aa?

  1. 2
  2. 4 (correct answer)
  3. 8
  4. 16
Explanation: The point (0,−16)(0, -16) is the y-intercept, which means that P(0)=−16P(0) = -16. Substitute x=0x=0 into the polynomial's equation: P(0)=(0−a)2(0−1)=(−a)2(−1)=−a2P(0) = (0-a)^2(0-1) = (-a)^2(-1) = -a^2. Setting this equal to -16 gives −a2=−16-a^2 = -16, which simplifies to a2=16a^2 = 16. The possible values for aa are 4 and -4. Since the problem states that a>0a > 0, the value of aa must be 4.

Question 8

The graph of a polynomial function y=f(x)y = f(x) is tangent to the x-axis at x=−4x=-4 and crosses the x-axis at x=1x=1. Which of the following could be the equation for f(x)f(x)?

  1. f(x)=(x−4)2(x+1)f(x) = (x-4)^2(x+1)
  2. f(x)=(x+4)(x−1)2f(x) = (x+4)(x-1)^2
  3. f(x)=(x−4)(x+1)2f(x) = (x-4)(x+1)^2
  4. f(x)=(x+4)2(x−1)f(x) = (x+4)^2(x-1) (correct answer)
Explanation: If the graph is tangent to the x-axis at a zero, that zero has an even multiplicity (e.g., 2, 4, ...). Tangency at x=−4x=-4 implies a factor of (x+4)k(x+4)^k where kk is even. If the graph crosses the x-axis, the zero has an odd multiplicity (e.g., 1, 3, ...). Crossing at x=1x=1 implies a factor of (x−1)m(x-1)^m where mm is odd. The simplest form matching this is (x+4)2(x−1)(x+4)^2(x-1).

Question 9

If rr and ss are the zeros of a polynomial P(x)P(x), which of the following polynomials must have zeros at r−2r-2 and s−2s-2?

  1. P(x+2)P(x+2) (correct answer)
  2. P(x−2)P(x-2)
  3. P(x)+2P(x) + 2
  4. P(x)−2P(x) - 2
Explanation: Let g(x)g(x) be the new polynomial. We want the zeros of g(x)g(x) to be r−2r-2 and s−2s-2. Let zz be a zero of g(x)g(x). Then z=r−2z = r-2 or z=s−2z = s-2. This implies z+2=rz+2 = r or z+2=sz+2 = s. Since rr and ss are zeros of P(x)P(x), it means that P(r)=0P(r)=0 and P(s)=0P(s)=0. Therefore, P(z+2)=0P(z+2)=0. The new polynomial is g(x)=P(x+2)g(x) = P(x+2).

Question 10

The zeros of the polynomial P(x)=x3+kx2+11x−6P(x) = x^3 + kx^2 + 11x - 6 are r1r_1, r2r_2, and r3r_3. If r1+r2+r3=5r_1 + r_2 + r_3 = 5, what is the value of kk?

  1. 11
  2. 5
  3. -5 (correct answer)
  4. -11
Explanation: For a cubic polynomial of the form ax3+bx2+cx+dax^3 + bx^2 + cx + d, the sum of the zeros is given by the formula −ba-\frac{b}{a}. In this polynomial, a=1a=1 and b=kb=k. Therefore, the sum of the zeros is −k1=−k-\frac{k}{1} = -k. We are given that the sum of the zeros is 5. So, −k=5-k = 5, which means k=−5k = -5.

Question 11

If r(x)r(x) is a polynomial function and the equation r(x)=5r(x) = 5 has solutions x=2x = 2 and x=6x = 6, which statement must be true about r(x)−5r(x) - 5?

  1. r(x)−5r(x) - 5 has zeros at x=2x = 2 and x=6x = 6 only
  2. r(x)−5r(x) - 5 has zeros at x=2x = 2 and x=6x = 6, and possibly others (correct answer)
  3. r(x)−5r(x) - 5 has a zero at x=5x = 5 and possibly others
  4. r(x)−5r(x) - 5 has no real zeros since r(x)=5r(x) = 5 for some values
Explanation: If r(x)=5r(x) = 5 when x=2x = 2 and x=6x = 6, then r(2)−5=0r(2) - 5 = 0 and r(6)−5=0r(6) - 5 = 0. This means r(x)−5r(x) - 5 has zeros at x=2x = 2 and x=6x = 6. However, since we don't know the degree of r(x)r(x), there could be additional zeros of r(x)−5r(x) - 5. Choice A incorrectly assumes these are the only zeros. Choice C confuses the input and output values. Choice D misunderstands that r(x)=5r(x) = 5 having solutions means r(x)−5r(x) - 5 has zeros.

Question 12

A polynomial function P(x)P(x) has zeros at x=−1x=-1 and x=5x=5. Which of the following functions, g(x)g(x), must have zeros at x=2x=2 and x=8x=8?

  1. g(x)=P(x−3)g(x) = P(x-3) (correct answer)
  2. g(x)=P(x+3)g(x) = P(x+3)
  3. g(x)=P(x)−3g(x) = P(x) - 3
  4. g(x)=3P(x)g(x) = 3P(x)
Explanation: Let the new zeros be xnx_n. We want g(xn)=0g(x_n) = 0. For g(x)=P(x−3)g(x) = P(x-3), we have P(xn−3)=0P(x_n-3) = 0. This means the input to PP, which is xn−3x_n-3, must be one of the original zeros (-1 or 5). If xn−3=−1x_n-3 = -1, then xn=2x_n = 2. If xn−3=5x_n-3 = 5, then xn=8x_n = 8. These are the desired zeros.

Question 13

For what value of kk is (x−2)(x-2) a factor of the polynomial P(x)=x3−3x2+kx−10P(x) = x^3 - 3x^2 + kx - 10?

  1. -15
  2. -3
  3. 4
  4. 7 (correct answer)
Explanation: According to the Factor Theorem, if (x−2)(x-2) is a factor of P(x)P(x), then P(2)P(2) must equal 0. Substitute x=2x=2 into the polynomial: P(2)=(2)3−3(2)2+k(2)−10=8−3(4)+2k−10=8−12+2k−10P(2) = (2)^3 - 3(2)^2 + k(2) - 10 = 8 - 3(4) + 2k - 10 = 8 - 12 + 2k - 10. Simplifying gives −14+2k=0-14 + 2k = 0. Solving for kk yields 2k=142k = 14, so k=7k = 7.

Question 14

A polynomial with real coefficients has a zero of 4−5i4 - 5i. Which of the following must also be a zero of the polynomial?

  1. −4−5i-4 - 5i
  2. −4+5i-4 + 5i
  3. 4+5i4 + 5i (correct answer)
  4. 5−4i5 - 4i
Explanation: The Complex Conjugate Root Theorem states that if a polynomial has real coefficients and a+bia + bi is a zero, then its complex conjugate, a−bia - bi, must also be a zero. The complex conjugate of 4−5i4 - 5i is 4+5i4 + 5i.

Question 15

A polynomial f(x)f(x) has degree 3. The graph of y=f(x)y=f(x) has x-intercepts at x=−2x=-2 and x=4x=4. If f(x)f(x) has exactly two distinct real zeros, which of the following statements must be true?

  1. The zero at x=4x=4 has a multiplicity of 2.
  2. The zero at x=−2x=-2 has a multiplicity of 2.
  3. Either x=−2x=-2 or x=4x=4 has a multiplicity of 2. (correct answer)
  4. The polynomial has one non-real complex zero.
Explanation: A polynomial of degree 3 must have 3 zeros, counting multiplicity. We are given two distinct real zeros, -2 and 4. To have a total of 3 zeros, one of these real zeros must have a multiplicity of 2. For example, the zeros could be -2, -2, 4 or -2, 4, 4. We cannot determine which zero has the multiplicity without more information. Since the polynomial has real coefficients (implied by the graph), any non-real zeros would come in conjugate pairs, adding at least 2 zeros, which would exceed the degree of 3.

Question 16

Given that x=1x=1 is a zero of f(x)=x3−x2−9x+9f(x) = x^3 - x^2 - 9x + 9, what is the sum of the other two zeros?

  1. -1
  2. 0 (correct answer)
  3. 1
  4. 9
Explanation: For a cubic polynomial ax3+bx2+cx+dax^3+bx^2+cx+d, the sum of the three zeros is −ba-\frac{b}{a}. Here, a=1a=1 and b=−1b=-1, so the sum of all three zeros is −−11=1-\frac{-1}{1} = 1. If one zero is 1, and the other two are r2r_2 and r3r_3, then 1+r2+r3=11 + r_2 + r_3 = 1. Subtracting 1 from both sides gives r2+r3=0r_2 + r_3 = 0.

Question 17

A polynomial is defined by P(x)=x2(x−3)(x+2)3P(x) = x^2(x-3)(x+2)^3. What is the sum of the distinct real zeros of P(x)P(x)?

  1. -3
  2. -1
  3. 0
  4. 1 (correct answer)
Explanation: The zeros of a polynomial are the values of xx for which P(x)=0P(x) = 0. The factors are x2x^2, (x−3)(x-3), and (x+2)3(x+2)^3. Setting each factor to zero gives x=0x=0, x=3x=3, and x=−2x=-2. These are the distinct real zeros. The sum is 0+3+(−2)=10 + 3 + (-2) = 1.

Question 18

What is the product of the zeros of the polynomial function f(x)=2x3−5x2+8x−7f(x) = 2x^3 - 5x^2 + 8x - 7?

  1. -7
  2. −72-\frac{7}{2}
  3. 52\frac{5}{2}
  4. 72\frac{7}{2} (correct answer)
Explanation: For a polynomial of the form ax3+bx2+cx+dax^3 + bx^2 + cx + d, the product of the zeros is given by the formula −da-\frac{d}{a}. In this polynomial, a=2a = 2 and d=−7d = -7. Therefore, the product of the zeros is −−72-\frac{-7}{2} = 72\frac{7}{2}.

Question 19

A polynomial with rational coefficients has 2−32 - \sqrt{3} as one of its zeros. Which of the following must also be a zero of this polynomial?

  1. −2+3-2 + \sqrt{3}
  2. −2−3-2 - \sqrt{3}
  3. 3−2\sqrt{3} - 2
  4. 2+32 + \sqrt{3} (correct answer)
Explanation: The Irrational Conjugate Root Theorem states that if a polynomial has rational coefficients and an irrational zero of the form a+bca+b\sqrt{c}, then its conjugate a−bca-b\sqrt{c} must also be a zero. The conjugate of 2−32 - \sqrt{3} is 2+32 + \sqrt{3}.

Question 20

According to the Rational Root Theorem, which of the following is NOT a possible rational zero of the polynomial P(x)=3x4−5x3+x2−8x+4P(x) = 3x^4 - 5x^3 + x^2 - 8x + 4?

  1. 23\frac{2}{3}
  2. -4
  3. 32\frac{3}{2} (correct answer)
  4. 13\frac{1}{3}
Explanation: The Rational Root Theorem states that any possible rational zero must be of the form pq\frac{p}{q}, where pp is a factor of the constant term (4) and qq is a factor of the leading coefficient (3). The factors of 4 are ±1,±2,±4\pm 1, \pm 2, \pm 4. The factors of 3 are ±1,±3\pm 1, \pm 3. The value 32\frac{3}{2} is not possible because 3 is not a factor of 4 and 2 is not a factor of 3.