Math 3 Quiz: Factor Theorem
13 questions · exam conditions
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Factor TheoremQuestion 1 of 13

The polynomial f(x)=x3+ax2+bx+cf(x) = x^3 + ax^2 + bx + c has zeros at x=−2x = -2, x=1x = 1, and x=5x = 5. If g(x)=x4+ax3+bx2+cx+dg(x) = x^4 + ax^3 + bx^2 + cx + d and g(−2)=0g(-2) = 0, g(1)=0g(1) = 0, g(5)=0g(5) = 0, what can be concluded about the factorization of g(x)g(x)?

g(x)=(x+2)(x−1)(x−5)(x−k)g(x) = (x + 2)(x - 1)(x - 5)(x - k) for some value kk
g(x)=(x+2)(x−1)(x−5)(x+k)g(x) = (x + 2)(x - 1)(x - 5)(x + k) for some positive value kk
g(x)=(x+2)2(x−1)(x−5)g(x) = (x + 2)^2(x - 1)(x - 5) or g(x)=(x+2)(x−1)2(x−5)g(x) = (x + 2)(x - 1)^2(x - 5)
g(x)=f(x)⋅(x−k)g(x) = f(x) \cdot (x - k) where kk can be determined uniquely
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Math 3 Quiz

Math 3 Quiz: Factor Theorem

Practice Factor Theorem in Math 3 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 Factor Theorem, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 3.

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 polynomial f(x)=x3+ax2+bx+cf(x) = x^3 + ax^2 + bx + c has zeros at x=−2x = -2, x=1x = 1, and x=5x = 5. If g(x)=x4+ax3+bx2+cx+dg(x) = x^4 + ax^3 + bx^2 + cx + d and g(−2)=0g(-2) = 0, g(1)=0g(1) = 0, g(5)=0g(5) = 0, what can be concluded about the factorization of g(x)g(x)?

  1. g(x)=(x+2)(x−1)(x−5)(x−k)g(x) = (x + 2)(x - 1)(x - 5)(x - k) for some value kk (correct answer)
  2. g(x)=(x+2)(x−1)(x−5)(x+k)g(x) = (x + 2)(x - 1)(x - 5)(x + k) for some positive value kk
  3. g(x)=(x+2)2(x−1)(x−5)g(x) = (x + 2)^2(x - 1)(x - 5) or g(x)=(x+2)(x−1)2(x−5)g(x) = (x + 2)(x - 1)^2(x - 5)
  4. g(x)=f(x)⋅(x−k)g(x) = f(x) \cdot (x - k) where kk can be determined uniquely
Explanation: Since g(−2)=g(1)=g(5)=0g(-2) = g(1) = g(5) = 0, by the Factor Theorem, (x+2)(x + 2), (x−1)(x - 1), and (x−5)(x - 5) are all factors of g(x)g(x). Since g(x)g(x) is degree 4, there must be exactly one more linear factor (x−k)(x - k) for some value kk. Choice B incorrectly restricts kk to be positive. Choice C assumes repeated roots without justification. Choice D is incorrect because g(x)≠f(x)⋅(x−k)g(x) ≠ f(x) \cdot (x - k) since the coefficients don't match this pattern.

Question 2

Let Q(x)=x4−10x3+35x2−50x+24Q(x) = x^4 - 10x^3 + 35x^2 - 50x + 24. If the Factor Theorem is used to test whether (x−2)(x - 2) is a factor, and it is found that Q(2)=0Q(2) = 0, what additional information is needed to determine if (x−2)2(x - 2)^2 is also a factor?

  1. Calculate Q′(2)Q'(2) and check if it equals zero for multiplicity determination (correct answer)
  2. Perform polynomial division of Q(x)Q(x) by (x−2)(x - 2) completely
  3. Evaluate Q(4)Q(4) to determine if there are other related factors
  4. Find the value of Q′′(2)Q''(2) to verify the degree of multiplicity
Explanation: The Factor Theorem only tells us that (x−2)(x - 2) is a factor when Q(2)=0Q(2) = 0. To determine if (x−2)2(x - 2)^2 is a factor (i.e., if x=2x = 2 is a root of multiplicity at least 2), we need Q′(2)=0Q'(2) = 0. If Q′(2)=0Q'(2) = 0, then x=2x = 2 has multiplicity ≥ 2. Choice B would work but is unnecessarily complicated. Choice C is irrelevant. Choice D tests for multiplicity ≥ 3, which goes beyond what's asked.

Question 3

The polynomial K(x)=x5−3x4−13x3+39x2+36x−108K(x) = x^5 - 3x^4 - 13x^3 + 39x^2 + 36x - 108 has (x−3)(x - 3) as a factor. After factoring out (x−3)(x - 3), the resulting quotient polynomial Q(x)Q(x) also has (x−3)(x - 3) as a factor. What is the minimum degree of multiplicity of the root x=3x = 3 in K(x)K(x)?

  1. The root x=3x = 3 has multiplicity exactly 2 based on the given information alone
  2. The root x=3x = 3 has multiplicity at least 2, but could be higher without further testing (correct answer)
  3. The root x=3x = 3 has multiplicity exactly 3 since K(x)K(x) is degree 5 with this pattern
  4. The multiplicity cannot be determined from the given information about factorization alone
Explanation: The given information tells us that K(x)=(x−3)⋅Q(x)K(x) = (x-3) \cdot Q(x) and that Q(x)Q(x) also has (x−3)(x-3) as a factor, meaning Q(x)=(x−3)⋅R(x)Q(x) = (x-3) \cdot R(x) for some polynomial R(x)R(x). Therefore K(x)=(x−3)2⋅R(x)K(x) = (x-3)^2 \cdot R(x), so the multiplicity is at least 2. However, without additional information (such as testing whether R(3)=0R(3) = 0), we cannot determine if the multiplicity is exactly 2 or higher. Choice A assumes exactly 2. Choice C assumes exactly 3 without justification. Choice D is incorrect since we can determine it's at least 2.

Question 4

Consider two polynomials: F(x)=x3−6x2+11x−6F(x) = x^3 - 6x^2 + 11x - 6 and G(x)=x4−6x3+11x2−6xG(x) = x^4 - 6x^3 + 11x^2 - 6x. If F(1)=0F(1) = 0, F(2)=0F(2) = 0, and F(3)=0F(3) = 0, what can be concluded about the relationship between F(x)F(x) and G(x)G(x) using the Factor Theorem?

  1. G(x)=x⋅F(x)G(x) = x \cdot F(x) and they share exactly the same non-zero roots
  2. G(x)=F(x)+xG(x) = F(x) + x and they have completely different factorizations
  3. G(x)=x⋅F(x)G(x) = x \cdot F(x) but G(x)G(x) has an additional root at x=0x = 0 (correct answer)
  4. F(x)F(x) and G(x)G(x) are unrelated since they have different degrees
Explanation: Since F(1)=F(2)=F(3)=0F(1) = F(2) = F(3) = 0 and F(x)F(x) is cubic, we have F(x)=(x−1)(x−2)(x−3)F(x) = (x-1)(x-2)(x-3). Looking at G(x)=x4−6x3+11x2−6x=x(x3−6x2+11x−6)=x⋅F(x)G(x) = x^4 - 6x^3 + 11x^2 - 6x = x(x^3 - 6x^2 + 11x - 6) = x \cdot F(x). Therefore G(x)G(x) has all the roots of F(x)F(x) (namely 1, 2, 3) plus an additional root at x=0x = 0. Choice A is incomplete as it doesn't mention the additional root. Choice B gives the wrong relationship. Choice D is incorrect as the polynomials are clearly related.

Question 5

The polynomial P(x)=2x3+px2+qx−12P(x) = 2x^3 + px^2 + qx - 12 has (x−2)(x - 2) as a factor. Additionally, when P(x)P(x) is divided by (x+1)(x + 1), the remainder is −15-15. What is the value of p+qp + q?

  1. p+q=−7p + q = -7 based on the system of equations from both conditions
  2. p+q=3p + q = 3 after applying the Factor and Remainder Theorems correctly
  3. p+q=−1p + q = -1 using the given constraints and polynomial division (correct answer)
  4. p+q=5p + q = 5 from solving the resulting linear system of equations
Explanation: Since (x−2)(x - 2) is a factor, P(2)=0P(2) = 0: 16+4p+2q−12=016 + 4p + 2q - 12 = 0, so 4p+2q=−44p + 2q = -4 or 2p+q=−22p + q = -2. Since the remainder when dividing by (x+1)(x + 1) is −15-15, P(−1)=−15P(-1) = -15: −2+p−q−12=−15-2 + p - q - 12 = -15, so p−q=−1p - q = -1. Solving the system: 2p+q=−22p + q = -2 and p−q=−1p - q = -1 gives p=−1,q=0p = -1, q = 0. Therefore p+q=−1p + q = -1.

Question 6

Consider the polynomial h(x)=x4−6x3+9x2+6x−10h(x) = x^4 - 6x^3 + 9x^2 + 6x - 10. If it is known that h(1)=0h(1) = 0 and h(5)=0h(5) = 0, which statement about the complete factorization is most likely correct?

  1. h(x)=(x−1)(x−5)(x2+2)h(x) = (x - 1)(x - 5)(x^2 + 2) with no real linear factors remaining (correct answer)
  2. h(x)=(x−1)(x−5)(x−2)(x+1)h(x) = (x - 1)(x - 5)(x - 2)(x + 1) with all factors being linear and real
  3. h(x)=(x−1)(x−5)(x2−2x+2)h(x) = (x - 1)(x - 5)(x^2 - 2x + 2) with the quadratic having complex roots
  4. h(x)=(x−1)2(x−5)(x+2)h(x) = (x - 1)^2(x - 5)(x + 2) with x=1x = 1 being a repeated root
Explanation: Using the Factor Theorem and polynomial division: h(x)÷(x−1)(x−5)=h(x)÷(x2−6x+5)=x2+2h(x) ÷ (x - 1)(x - 5) = h(x) ÷ (x^2 - 6x + 5) = x^2 + 2. Therefore h(x)=(x−1)(x−5)(x2+2)h(x) = (x - 1)(x - 5)(x^2 + 2). The quadratic factor x2+2x^2 + 2 has no real roots. Choice B gives an incorrect factorization. Choice C has the wrong quadratic factor. Choice D assumes x=1x = 1 is a repeated root, but h′(1)≠0h'(1) ≠ 0.

Question 7

Given that P(x)=x4+mx3+nx2+px+qP(x) = x^4 + mx^3 + nx^2 + px + q and that P(x)−12=(x−2)2(x+1)(x−3)P(x) - 12 = (x-2)^2(x+1)(x-3), what is the value of P(2)P(2)?

  1. P(2)=0P(2) = 0 because (x−2)(x-2) is a factor of P(x)−12P(x) - 12
  2. P(2)=12P(2) = 12 since P(x)−12P(x) - 12 has (x−2)2(x-2)^2 as a factor (correct answer)
  3. P(2)=24P(2) = 24 due to the multiplicity of the root at x=2x = 2
  4. P(2)=−12P(2) = -12 from the relationship between P(x)P(x) and the factored form
Explanation: Since P(x)−12=(x−2)2(x+1)(x−3)P(x) - 12 = (x-2)^2(x+1)(x-3), we have P(x)=(x−2)2(x+1)(x−3)+12P(x) = (x-2)^2(x+1)(x-3) + 12. To find P(2)P(2), substitute x=2x = 2: P(2)=(2−2)2(2+1)(2−3)+12=0⋅3⋅(−1)+12=0+12=12P(2) = (2-2)^2(2+1)(2-3) + 12 = 0 \cdot 3 \cdot (-1) + 12 = 0 + 12 = 12. Choice A incorrectly assumes P(2)=0P(2) = 0. Choice C and D give incorrect values based on misunderstanding the relationship.

Question 8

The polynomial W(x)=x3+2x2−5x−6W(x) = x^3 + 2x^2 - 5x - 6 can be factored as (x+1)(x+3)(x−2)(x + 1)(x + 3)(x - 2). If V(x)=W(2x)V(x) = W(2x), which of the following represents the correct factorization of V(x)V(x)?

  1. V(x)=(x+1)(x+3)(x−2)V(x) = (x + 1)(x + 3)(x - 2) since factorization remains unchanged
  2. V(x)=(2x+1)(2x+3)(2x−2)V(x) = (2x + 1)(2x + 3)(2x - 2) by direct substitution method
  3. V(x)=8(x+2)(x+6)(x−4)V(x) = 8(x + 2)(x + 6)(x - 4) using horizontal compression rules
  4. V(x)=8(x+12)(x+32)(x−1)V(x) = 8(x + \frac{1}{2})(x + \frac{3}{2})(x - 1) after scaling transformation (correct answer)
Explanation: When you encounter polynomial transformations like V(x)=W(2x)V(x) = W(2x), you're dealing with horizontal scaling that affects both the roots and the leading coefficient of the factored form. To find V(x)=W(2x)V(x) = W(2x), substitute 2x2x into the original polynomial: V(x)=(2x)3+2(2x)2−5(2x)−6=8x3+8x2−10x−6V(x) = (2x)^3 + 2(2x)^2 - 5(2x) - 6 = 8x^3 + 8x^2 - 10x - 6. You can factor out 8: V(x)=8(x3+x2−54x−34)V(x) = 8(x^3 + x^2 - \frac{5}{4}x - \frac{3}{4}). The key insight is understanding how the transformation affects the roots. If W(x)W(x) has roots at x=−1,−3,2x = -1, -3, 2, then V(x)=W(2x)V(x) = W(2x) has roots where 2x=−1,−3,22x = -1, -3, 2, giving us x=−12,−32,1x = -\frac{1}{2}, -\frac{3}{2}, 1. The horizontal compression by factor 2 shrinks all root positions by half. Additionally, since the leading term becomes 8x38x^3, we get the factor of 8. Therefore, V(x)=8(x+12)(x+32)(x−1)V(x) = 8(x + \frac{1}{2})(x + \frac{3}{2})(x - 1). Choice A ignores the transformation entirely. Choice B incorrectly substitutes 2x2x directly into each factor without considering how transformations actually affect roots. Choice C makes an error in the root calculations—these would correspond to V(x)=W(x2)V(x) = W(\frac{x}{2}), not V(x)=W(2x)V(x) = W(2x). Remember: for f(ax)f(ax) transformations, divide the original roots by aa to find the new roots, and account for how the leading coefficient changes through the substitution.

Question 9

Given that P(x)=2x4−5x3−23x2+60xP(x) = 2x^4 - 5x^3 - 23x^2 + 60x and P(4)=0P(4) = 0, which of the following statements must be true?

  1. (x−4)(x - 4) is a factor of P(x)P(x), and P(x)=(x−4)(2x3+3x2−11x−15)P(x) = (x - 4)(2x^3 + 3x^2 - 11x - 15)
  2. (x+4)(x + 4) is a factor of P(x)P(x), and P(x)=(x+4)(2x3−13x2+29x−15)P(x) = (x + 4)(2x^3 - 13x^2 + 29x - 15)
  3. (x−4)(x - 4) is a factor of P(x)P(x), and P(x)=(x−4)(2x3−13x2+29x−15)P(x) = (x - 4)(2x^3 - 13x^2 + 29x - 15) (correct answer)
  4. (4x−1)(4x - 1) is a factor of P(x)P(x), and P(x)=(4x−1)(x3−x2−6x+15)P(x) = (4x - 1)(x^3 - x^2 - 6x + 15)
Explanation: By the Factor Theorem, if P(4)=0P(4) = 0, then (x−4)(x - 4) is a factor of P(x)P(x). Using polynomial long division or synthetic division: P(x)=(x−4)(2x3−13x2+29x−15)P(x) = (x - 4)(2x^3 - 13x^2 + 29x - 15). Choice A has the correct factor but wrong quotient. Choice B uses the wrong factor (x+4)(x + 4). Choice D incorrectly assumes (4x−1)(4x - 1) is a factor, which would require P(1/4)=0P(1/4) = 0, not P(4)=0P(4) = 0.

Question 10

The polynomial T(x)=2x4+bx3+cx2+dx+eT(x) = 2x^4 + bx^3 + cx^2 + dx + e has the property that T(1)=T(−1)=T(2)=T(−2)=0T(1) = T(-1) = T(2) = T(-2) = 0. Which statement about the coefficients must be true?

  1. The coefficient b=0b = 0 and d=0d = 0 due to symmetry properties
  2. All coefficients bb, cc, dd, and ee are uniquely determined by the given zeros
  3. The coefficient e=8e = 8 and the polynomial is completely determined
  4. The coefficients satisfy b=d=0b = d = 0 and c=−10,e=8c = -10, e = 8 (correct answer)
Explanation: Since T(x)T(x) has zeros at ±1±1 and ±2±2, we have T(x)=2(x−1)(x+1)(x−2)(x+2)=2(x2−1)(x2−4)=2(x4−5x2+4)=2x4−10x2+8T(x) = 2(x-1)(x+1)(x-2)(x+2) = 2(x^2-1)(x^2-4) = 2(x^4-5x^2+4) = 2x^4-10x^2+8. Comparing with T(x)=2x4+bx3+cx2+dx+eT(x) = 2x^4 + bx^3 + cx^2 + dx + e, we get b=0b = 0, c=−10c = -10, d=0d = 0, and e=8e = 8. Choice A is partially correct but incomplete. Choice B is correct that they're uniquely determined but doesn't give the values. Choice C gives only ee correctly.

Question 11

Given that T(x)=2x3−7x2+4x+kT(x) = 2x^3 - 7x^2 + 4x + k where kk is a constant, and (x−2)(x-2) is a factor of T(x)T(x), what is the remainder when T(x+1)T(x+1) is divided by (x−1)(x-1)?

  1. The remainder is 00 because the translation preserves the factor relationship (correct answer)
  2. The remainder is −3-3 from evaluating the shifted polynomial at the division point
  3. The remainder is 33 due to the constant term adjustment in the translation
  4. The remainder is kk since the parameter directly affects the remainder value
Explanation: First, since (x−2)(x-2) is a factor of T(x)T(x), we have T(2)=0T(2) = 0. This gives us T(2)=2(8)−7(4)+4(2)+k=16−28+8+k=−4+k=0T(2) = 2(8) - 7(4) + 4(2) + k = 16 - 28 + 8 + k = -4 + k = 0, so k=4k = 4. Therefore, T(x)=2x3−7x2+4x+4T(x) = 2x^3 - 7x^2 + 4x + 4. Now, T(x+1)=2(x+1)3−7(x+1)2+4(x+1)+4T(x+1) = 2(x+1)^3 - 7(x+1)^2 + 4(x+1) + 4. By the Remainder Theorem, the remainder when T(x+1)T(x+1) is divided by (x−1)(x-1) is T(1+1)=T(2)=0T(1+1) = T(2) = 0. This makes sense because if (x−2)(x-2) is a factor of T(x)T(x), then T(x+1)T(x+1) has (x−1)(x-1) as a factor (since when x=1x = 1, we get x+1=2x+1 = 2, which is a zero of TT).

Question 12

If p(x)=x4−5x3+7x2−5x+6p(x) = x^4 - 5x^3 + 7x^2 - 5x + 6 and p(2)=0p(2) = 0, which statement about the polynomial q(x)=x3−3x2+x−3q(x) = x^3 - 3x^2 + x - 3 is necessarily true?

  1. q(x)q(x) is the complete factorization of p(x)p(x) divided by (x−2)(x-2)
  2. q(x)q(x) represents the quotient when p(x)p(x) is divided by (x−2)(x-2) (correct answer)
  3. q(x)q(x) has the same zeros as p(x)p(x) except for x=2x = 2
  4. q(x)q(x) must have (x−2)(x-2) as one of its factors
Explanation: Since p(2)=0p(2) = 0, by the Factor Theorem, (x−2)(x-2) is a factor of p(x)p(x). This means p(x)=(x−2)⋅q(x)p(x) = (x-2) \cdot q(x) where q(x)q(x) is the quotient polynomial. Performing polynomial division of p(x)p(x) by (x−2)(x-2) yields q(x)=x3−3x2+x−3q(x) = x^3 - 3x^2 + x - 3. Choice A is incorrect because q(x)q(x) may factor further. Choice C is incorrect because q(x)q(x) could have additional zeros or share some zeros with the original polynomial. Choice D is incorrect because there's no reason q(x)q(x) would contain (x−2)(x-2) as a factor.

Question 13

If F(x)=x4−10x2+9F(x) = x^4 - 10x^2 + 9 and we know that F(1)=0F(1) = 0, which statement about the complete factorization of F(x)F(x) is correct?

  1. F(x)=(x−1)(x+1)(x2−9)F(x) = (x-1)(x+1)(x^2-9) showing two linear factors and one quadratic
  2. F(x)=(x−1)(x+1)(x−3)(x+3)F(x) = (x-1)(x+1)(x-3)(x+3) revealing four distinct linear factors (correct answer)
  3. F(x)=(x2−1)(x2−9)F(x) = (x^2-1)(x^2-9) expressing the polynomial as a product of two quadratics
  4. F(x)=(x−1)2(x2+9)F(x) = (x-1)^2(x^2+9) indicating a repeated linear factor and irreducible quadratic
Explanation: We can verify F(1)=1−10+9=0F(1) = 1 - 10 + 9 = 0, confirming (x−1)(x-1) is a factor. Notice that F(x)=x4−10x2+9F(x) = x^4 - 10x^2 + 9 is actually a quadratic in x2x^2. Let u=x2u = x^2, then F(x)=u2−10u+9=(u−1)(u−9)=(x2−1)(x2−9)F(x) = u^2 - 10u + 9 = (u-1)(u-9) = (x^2-1)(x^2-9). This can be factored further: (x2−1)(x2−9)=(x−1)(x+1)(x−3)(x+3)(x^2-1)(x^2-9) = (x-1)(x+1)(x-3)(x+3). So the complete linear factorization has four factors, and we can verify that x=±1,±3x = \pm 1, \pm 3 are all zeros. Choice A is incomplete, Choice C doesn't show complete linear factorization, and Choice D is incorrect as it suggests (x−1)2(x-1)^2 which would mean F′(1)=0F'(1) = 0, but F′(x)=4x3−20xF'(x) = 4x^3 - 20x and F′(1)=4−20=−16≠0F'(1) = 4 - 20 = -16 \neq 0.