Algebra 2 Quiz: Applying The Remainder Theorem
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Applying The Remainder TheoremQuestion 1 of 20

Use p(a)p(a) to find the remainder for division by (x−a).(x-a). If p(x)=x4−x2+2x−6,p(x)=x^4-x^2+2x-6, what is the remainder when dividing by (x−2)?(x-2)?

The remainder is 22.
The remainder is 00.
The remainder is 66.
The remainder is 1010.
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Algebra 2 Quiz

Algebra 2 Quiz: Applying The Remainder Theorem

Practice Applying The Remainder Theorem in Algebra 2 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 Applying The Remainder Theorem, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

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

Use p(a)p(a) to find the remainder for division by (x−a).(x-a). If p(x)=x4−x2+2x−6,p(x)=x^4-x^2+2x-6, what is the remainder when dividing by (x−2)?(x-2)?

  1. The remainder is 22.
  2. The remainder is 00.
  3. The remainder is 66.
  4. The remainder is 1010. (correct answer)
Explanation: This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! The Remainder Theorem states that when you divide a polynomial p(x) by (x - a), the remainder is simply p(a)—just substitute a into the polynomial and evaluate! This works because of how polynomial division works: p(x) = (x - a)·q(x) + r, where r is the remainder. Substituting x = a: p(a) = (a - a)·q(a) + r = 0 + r = r. So the remainder r equals p(a)—brilliant! For p(x) = x⁴ - x² + 2x - 6 divided by (x - 2), we need p(2): p(2) = (2)⁴ - (2)² + 2(2) - 6 = 16 - 4 + 4 - 6 = 10. Choice D correctly identifies the remainder as 10. Using the Remainder Theorem: (1) Identify the divisor (x - 2) so a = 2, (2) Substitute 2 for every x in p(x), (3) Calculate carefully: 16 - 4 + 4 - 6 = 10, (4) That result is your remainder. This method is dramatically faster than polynomial long division, especially for higher-degree polynomials!

Question 2

If the remainder when p(x)=x3−2x2+kx−4p(x)=x^3-2x^2+kx-4 is divided by (x−2)(x-2) is 66, what is the value of kk? (Use remainder =p(2)=p(2).)

  1. k=5k=5 (correct answer)
  2. k=−1k=-1
  3. k=3k=3
  4. k=1k=1
Explanation: This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! The Remainder Theorem states that when you divide a polynomial p(x) by (x - a), the remainder is simply p(a)—just substitute a into the polynomial and evaluate! This works because of how polynomial division works: p(x) = (x - a)·q(x) + r, where r is the remainder. Substituting x = a: p(a) = (a - a)·q(a) + r = 0 + r = r. So the remainder r equals p(a)—brilliant! Here, divisor (x - 2), a = 2, and remainder = 6, so p(2) = 6: (2)^3 - 2*(2)^2 + k*(2) - 4 = 8 - 2*4 + 2k - 4 = 8 - 8 + 2k - 4 = 2k - 4 = 6, so 2k = 10, k = 5. Choice D correctly finds k = 5. Choice A says k = -1, perhaps from setting 2k - 4 = 0 instead of 6 or a solving error like 2k = 2, but always solve the equation carefully! Using the Remainder Theorem: (1) Identify the divisor (x - a) and extract a (remember: (x + 3) = (x - (-3)), so a = -3), (2) Substitute a for every x in p(x), (3) Calculate carefully (use parentheses for negative values!), (4) That result is your remainder. If it equals 0, (x - a) is a factor! This method is dramatically faster than polynomial long division.

Question 3

What is the remainder when dividing p(x)=2x4−x2+5p(x)=2x^4-x^2+5 by (x+1)(x+1)? Use the Remainder Theorem (remainder =p(a)=p(a) for divisor (x−a)(x-a)).

  1. The remainder is 00.
  2. The remainder is 88.
  3. The remainder is 66. (correct answer)
  4. The remainder is 44.
Explanation: This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! The Remainder Theorem states that when you divide a polynomial p(x) by (x - a), the remainder is simply p(a)—just substitute a into the polynomial and evaluate! This works because of how polynomial division works: p(x) = (x - a)·q(x) + r, where r is the remainder. Substituting x = a: p(a) = (a - a)·q(a) + r = 0 + r = r. So the remainder r equals p(a)—brilliant! For (x + 1) = (x - (-1)), a = -1; p(-1) = 2*(-1)^4 - (-1)^2 + 5 = 21 - 1 + 5 = 2 - 1 + 5 = 6. Choice A correctly evaluates p(-1) as 6. Choice B says 4, which could happen if you mistakenly computed (-1)^4 as -1 instead of 1, leading to 2(-1) -1 +5 = -2 -1 +5 =2, but remember even powers are positive! Sign safety for negatives: when testing (x + 2) = (x - (-2)), you're evaluating at a = -2. Substitute carefully: if p(x) = x³ - 3x + 5, then p(-2) = (-2)³ - 3(-2) + 5 = -8 + 6 + 5 = 3. Use parentheses around negative values to avoid sign errors! This is where most mistakes happen with the Remainder Theorem.

Question 4

Use the Remainder Theorem to find the remainder when p(x)=x3+4x2−x−12p(x)=x^3+4x^2-x-12 is divided by (x+3).(x+3).

  1. The remainder is 00. (correct answer)
  2. The remainder is −6-6.
  3. The remainder is 66.
  4. The remainder is −24-24.
Explanation: This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! The Remainder Theorem states that when you divide a polynomial p(x) by (x - a), the remainder is simply p(a)—just substitute a into the polynomial and evaluate! This works because of how polynomial division works: p(x) = (x - a)·q(x) + r, where r is the remainder. Substituting x = a: p(a) = (a - a)·q(a) + r = 0 + r = r. So the remainder r equals p(a)—brilliant! For p(x) = x³ + 4x² - x - 12 divided by (x + 3) = (x - (-3)), we need p(-3): p(-3) = (-3)³ + 4(-3)² - (-3) - 12 = -27 + 4(9) + 3 - 12 = -27 + 36 + 3 - 12 = 0. Choice A correctly identifies the remainder as 0, which also means (x + 3) is a factor of p(x)! Sign safety for negatives: when testing (x + 3) = (x - (-3)), you're evaluating at a = -3. Substitute carefully using parentheses: (-3)³ = -27 and (-3)² = 9. The zero remainder reveals that this polynomial is exactly divisible by (x + 3)!

Question 5

If the remainder when p(x)=x4+mx2−2x+1p(x)=x^4+mx^2-2x+1 is divided by (x−2)(x-2) is 99, what is mm? (Use: remainder =p(2)=p(2).)

  1. m=0m=0
  2. m=1m=1
  3. m=2m=2
  4. m=−1m=-1 (correct answer)
Explanation: This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! The Remainder Theorem states that when you divide a polynomial p(x) by (x - a), the remainder is simply p(a)—just substitute a into the polynomial and evaluate! This works because of how polynomial division works: p(x) = (x - a)·q(x) + r, where r is the remainder. Substituting x = a: p(a) = (a - a)·q(a) + r = 0 + r = r. So the remainder r equals p(a)—brilliant! For (x-2), a=2, p(2)=2^4 +m2^2 -22 +1=16+4m-4+1=13+4m; set to 9: 13+4m=9, 4m=-4, m=-1. Choice A correctly finds m=-1. A distractor like C (1) might come from setting 16+4m-4+1=9 as 13+4m=9 but solving 4m=-4 incorrectly. Using the Remainder Theorem: (1) Identify the divisor (x - a) and extract a (remember: (x + 3) = (x - (-3)), so a = -3), (2) Substitute a for every x in p(x), (3) Calculate carefully (use parentheses for negative values!), (4) That result is your remainder. If it equals 0, (x - a) is a factor! This method is dramatically faster than polynomial long division.

Question 6

Verify that (x−3)(x-3) is a factor of p(x)=x3−6x2+11x−6p(x)=x^3-6x^2+11x-6 using the Remainder Theorem.

  1. Yes; p(3)=0p(3)=0. (correct answer)
  2. No; p(3)=6p(3)=6.
  3. Yes; p(−3)=0p(-3)=0.
  4. No; p(−3)=−6p(-3)=-6.
Explanation: This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! A special case is the Factor Theorem: (x - a) is a factor of p(x) if and only if p(a) = 0. This means the remainder is zero, so the division is exact with no remainder. We can test potential factors by just evaluating the polynomial—if p(a) = 0, we've found a factor! This beats trial-and-error factoring when testing specific values. To verify that (x - 3) is a factor of p(x) = x³ - 6x² + 11x - 6, we evaluate p(3): p(3) = (3)³ - 6(3)² + 11(3) - 6 = 27 - 6(9) + 33 - 6 = 27 - 54 + 33 - 6 = 0. Since p(3) = 0, (x - 3) is indeed a factor! Choice A correctly confirms that (x - 3) is a factor because p(3) = 0. Using the Remainder Theorem: (1) To verify a factor (x - a), check if p(a) = 0, (2) Substitute a for every x in p(x), (3) Calculate step by step, (4) If the result is 0, you've confirmed the factor! This elegant theorem connects division, remainders, and factorization.

Question 7

Is (x+2)(x+2) a factor of p(x)=x3+3x2−4x−12p(x)=x^3+3x^2-4x-12? Use the Remainder Theorem / Factor Theorem (i.e., (x−a)(x-a) is a factor iff p(a)=0p(a)=0).​

  1. Yes, because p(−2)=0p(-2)=0. (correct answer)
  2. No, because p(−2)=8p(-2)=8.
  3. Yes, because p(2)=0p(2)=0.
  4. No, because the remainder is 00.
Explanation: This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! A special case is the Factor Theorem: (x - a) is a factor of p(x) if and only if p(a) = 0. This means the remainder is zero, so the division is exact with no remainder. We can test potential factors by just evaluating the polynomial—if p(a) = 0, we've found a factor! This beats trial-and-error factoring when testing specific values. For (x + 2) = (x - (-2)), a = -2; p(-2) = (-2)^3 + 3*(-2)^2 - 4*(-2) - 12 = -8 + 3*4 + 8 - 12 = -8 + 12 + 8 - 12, which is (-8 - 12) + (12 + 8) = -20 + 20 = 0. Choice A correctly determines that yes, it is a factor because p(-2) = 0. Choice B says no because p(-2) = 8, which might come from forgetting to include the -12 or mishandling signs, but actually it's 0—keep track of all terms! Sign safety for negatives: when testing (x + 2) = (x - (-2)), you're evaluating at a = -2. Substitute carefully: if p(x) = x³ - 3x + 5, then p(-2) = (-2)³ - 3(-2) + 5 = -8 + 6 + 5 = 3. Use parentheses around negative values to avoid sign errors! This is where most mistakes happen with the Remainder Theorem.

Question 8

Use the Remainder Theorem to find the remainder when p(x)=3x3−x2−7x+2p(x)=3x^3-x^2-7x+2 is divided by (x+2)(x+2).

  1. The remainder is −12-12. (correct answer)
  2. The remainder is 1212.
  3. The remainder is −6-6.
  4. The remainder is 00.
Explanation: This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! The Remainder Theorem states that when you divide a polynomial p(x) by (x - a), the remainder is simply p(a)—just substitute a into the polynomial and evaluate! This works because of how polynomial division works: p(x) = (x - a)·q(x) + r, where r is the remainder. Substituting x = a: p(a) = (a - a)·q(a) + r = 0 + r = r. So the remainder r equals p(a)—brilliant! For (x + 2) = (x - (-2)), a = -2; p(-2) = 3*(-2)^3 - (-2)^2 - 7*(-2) + 2 = 3*(-8) - 4 + 14 + 2 = -24 - 4 + 14 + 2, which is (-24 - 4) + (14 + 2) = -28 + 16 = -12. Choice A correctly evaluates p(-2) as -12. Choice B says 12, possibly from changing all signs incorrectly or forgetting the negative in (-2)^3, but use parentheses to track signs properly! Sign safety for negatives: when testing (x + 2) = (x - (-2)), you're evaluating at a = -2. Substitute carefully: if p(x) = x³ - 3x + 5, then p(-2) = (-2)³ - 3(-2) + 5 = -8 + 6 + 5 = 3. Use parentheses around negative values to avoid sign errors! This is where most mistakes happen with the Remainder Theorem.

Question 9

Verify the claim using the Remainder Theorem: When dividing p(x)=x4−3x2+2x+1p(x)=x^4-3x^2+2x+1 by (x−1)(x-1), the remainder is 11.

  1. True, because p(1)=1p(1)=1. (correct answer)
  2. False, because p(1)=0p(1)=0 so the remainder is 00.
  3. False, because p(−1)=1p(-1)=1 so the remainder is 11.
  4. True, because p(−1)=0p(-1)=0.
Explanation: This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! The Remainder Theorem states that when you divide a polynomial p(x) by (x - a), the remainder is simply p(a)—just substitute a into the polynomial and evaluate! This works because of how polynomial division works: p(x) = (x - a)·q(x) + r, where r is the remainder. Substituting x = a: p(a) = (a - a)·q(a) + r = 0 + r = r. So the remainder r equals p(a)—brilliant! To verify, evaluate p(1): 1^4 = 1, -31^2 = -3, +21 = +2, +1; combine: 1 - 3 = -2, -2 + 2 = 0, 0 + 1 = 1. Choice A correctly verifies it's true because p(1)=1 matches the claimed remainder. Choice B incorrectly claims false with p(1)=0, perhaps from miscalculating 1 -3 +2 +1 as 0 by ignoring the last +1. Using the Remainder Theorem: (1) Identify the divisor (x - a) and extract a (remember: (x + 3) = (x - (-3)), so a = -3), (2) Substitute a for every x in p(x), (3) Calculate carefully (use parentheses for negative values!), (4) That result is your remainder. If it equals 0, (x - a) is a factor! This method is dramatically faster than polynomial long division.

Question 10

Use the Remainder Theorem to find the remainder when p(x)=x3−4x2+5x−2p(x)=x^3-4x^2+5x-2 is divided by (x−3)(x-3). (Recall: remainder =p(a)=p(a) when dividing by (x−a)(x-a).)

  1. The remainder is −2-2.
  2. The remainder is 44. (correct answer)
  3. The remainder is 00.
  4. The remainder is 1010.
Explanation: This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! The Remainder Theorem states that when you divide a polynomial p(x) by (x - a), the remainder is simply p(a)—just substitute a into the polynomial and evaluate! This works because of how polynomial division works: p(x) = (x - a)·q(x) + r, where r is the remainder. Substituting x = a: p(a) = (a - a)·q(a) + r = 0 + r = r. So the remainder r equals p(a)—brilliant! For this problem, a=3, so p(3)=3^3 -43^2 +53 -2=27-36+15-2; step by step, 27-36=-9, -9+15=6, 6-2=4. Choice C correctly finds the remainder as 4. A distractor like B (-2) might come from sign errors in subtraction, but careful calculation confirms 4. Using the Remainder Theorem: (1) Identify the divisor (x - a) and extract a (remember: (x + 3) = (x - (-3)), so a = -3), (2) Substitute a for every x in p(x), (3) Calculate carefully (use parentheses for negative values!), (4) That result is your remainder. If it equals 0, (x - a) is a factor! This method is dramatically faster than polynomial long division.

Question 11

If the remainder when p(x)=x3−2x2+kx−5p(x)=x^3-2x^2+kx-5 is divided by (x−1)(x-1) is 33, find kk. (Use remainder=p(1)\text{remainder} = p(1).)

  1. k=7k=7
  2. k=9k=9 (correct answer)
  3. k=11k=11
  4. k=3k=3
Explanation: This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! The Remainder Theorem states that when you divide a polynomial p(x) by (x - a), the remainder is simply p(a)—just substitute a into the polynomial and evaluate! This works because of how polynomial division works: p(x)=(x−a)⋅q(x)+rp(x) = (x - a) \cdot q(x) + r, where r is the remainder. Substituting x = a: p(a)=(a−a)⋅q(a)+r=0+r=rp(a) = (a - a) \cdot q(a) + r = 0 + r = r. So the remainder r equals p(a)—brilliant! Here, p(1) = 3 and we set it equal to 3: 13−2(1)2+k⋅1−5=1−2+k−5=−6+k=31^3 - 2(1)^2 + k \cdot 1 - 5 = 1 - 2 + k - 5 = -6 + k = 3, so k=9k = 9. Choice B correctly finds k=9. A distractor like k=7 in choice A might come from misadding -6 + k =3 as k=-3+6=3 or similar arithmetic error. Using the Remainder Theorem: (1) Identify the divisor (x - a) and extract a (remember: (x+3)=(x−(−3))(x + 3) = (x - (-3)), so a = -3), (2) Substitute a for every x in p(x), (3) Calculate carefully (use parentheses for negative values!), (4) That result is your remainder. If it equals 0, (x - a) is a factor! This method is dramatically faster than polynomial long division.

Question 12

For p(x)=x4−16,p(x)=x^4-16, which of the following are factors? Use the Remainder Theorem by checking whether p(a)=0p(a)=0.

I. (x−2)(x-2)

II. (x+2)(x+2)

III. (x−4)(x-4)

  1. I and II only (correct answer)
  2. II and III only
  3. I, II, and III
  4. I only
Explanation: This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! A special case is the Factor Theorem: (x - a) is a factor of p(x) if and only if p(a) = 0. This means the remainder is zero, so the division is exact with no remainder. We can test potential factors by just evaluating the polynomial—if p(a) = 0, we've found a factor! This beats trial-and-error factoring when testing specific values. For p(x) = x⁴ - 16, let's test each potential factor: For (x - 2): p(2) = 2⁴ - 16 = 16 - 16 = 0 ✓ Factor! For (x + 2) = (x - (-2)): p(-2) = (-2)⁴ - 16 = 16 - 16 = 0 ✓ Factor! For (x - 4): p(4) = 4⁴ - 16 = 256 - 16 = 240 ≠ 0 ✗ Not a factor. Choice A correctly identifies that I and II only are factors. Using the Remainder Theorem: (1) Identify the divisor and extract a (remember: (x + 2) = (x - (-2)), so a = -2), (2) Substitute a for every x in p(x), (3) Calculate carefully (use parentheses for negative values!), (4) If the result equals 0, (x - a) is a factor! This method is dramatically faster than polynomial long division.

Question 13

Is (x+2)(x+2) a factor of p(x)=x3+3x2−4x−12p(x)=x^3+3x^2-4x-12? Use the Remainder Theorem/Factor Theorem (i.e., (x−a)(x-a) is a factor iff p(a)=0p(a)=0).

  1. Yes, because p(−2)=0p(-2)=0. (correct answer)
  2. No, because p(−2)=4p(-2)=4.
  3. Yes, because p(2)=0p(2)=0.
  4. No, because p(2)=−4p(2)=-4.
Explanation: This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! A special case is the Factor Theorem: (x - a) is a factor of p(x) if and only if p(a) = 0. This means the remainder is zero, so the division is exact with no remainder. We can test potential factors by just evaluating the polynomial—if p(a) = 0, we've found a factor! This beats trial-and-error factoring when testing specific values. For (x+2) = (x - (-2)), evaluate p(-2): (-2)^3 = -8, +3*(-2)^2 = +34 = +12, -4(-2) = +8, -12; combine: -8 + 12 = 4, 4 + 8 = 12, 12 - 12 = 0. Choice A correctly determines that yes, it's a factor because p(-2)=0. Choice B says no because p(-2)=4, which might happen if you forget the -4*(-2) = +8 and stop at -8 + 12 -12 = -8. Sign safety for negatives: when testing (x + 2) = (x - (-2)), you're evaluating at a = -2. Substitute carefully: if p(x) = x³ - 3x + 5, then p(-2) = (-2)³ - 3(-2) + 5 = -8 + 6 + 5 = 3. Use parentheses around negative values to avoid sign errors! This is where most mistakes happen with the Remainder Theorem.

Question 14

A cubic polynomial f(x)f(x) has leading coefficient 1 and satisfies f(1)=f(3)=f(5)=0f(1) = f(3) = f(5) = 0. When f(x)f(x) is divided by (x−7)(x - 7), the remainder is:

  1. 24
  2. 48 (correct answer)
  3. 72
  4. 0
Explanation: Since f(x)f(x) is cubic with leading coefficient 1 and has zeros at x=1,3,5x = 1, 3, 5, we have f(x)=(x−1)(x−3)(x−5)f(x) = (x-1)(x-3)(x-5). By the Remainder Theorem, the remainder when divided by (x−7)(x-7) is f(7)=(7−1)(7−3)(7−5)=6⋅4⋅2=48f(7) = (7-1)(7-3)(7-5) = 6 \cdot 4 \cdot 2 = 48.

Question 15

A polynomial f(x)f(x) has the property that f(−3)=8f(-3) = 8 and f(1)=0f(1) = 0. When f(x)f(x) is divided by x2+2x−3x^2 + 2x - 3, what is the remainder?

  1. 4x+124x + 12
  2. 2x+22x + 2
  3. −2x+2-2x + 2 (correct answer)
  4. 3x+93x + 9
Explanation: Since x2+2x−3=(x+3)(x−1)x^2 + 2x - 3 = (x + 3)(x - 1), when f(x)f(x) is divided by this quadratic, the remainder must be linear: R(x)=ax+bR(x) = ax + b. By the Remainder Theorem, R(−3)=f(−3)=8R(-3) = f(-3) = 8 and R(1)=f(1)=0R(1) = f(1) = 0. So −3a+b=8-3a + b = 8 and a+b=0a + b = 0. Solving: b=−ab = -a and −3a+(−a)=8-3a + (-a) = 8, so −4a=8-4a = 8 and a=−2,b=2a = -2,b = 2. Therefore R(x)=−2x+2R(x) = -2x + 2. The other choices don't satisfy both conditions simultaneously.

Question 16

If g(x)=x4−3x3+2x2−5x+1g(x) = x^4 - 3x^3 + 2x^2 - 5x + 1 and the remainder when g(x)g(x) is divided by (x−c)(x - c) is 1515, which equation must be satisfied by cc?

  1. c4−3c3+2c2−5c+1=15c^4 - 3c^3 + 2c^2 - 5c + 1 = 15 (correct answer)
  2. c4−3c3+2c2−5c−14=0c^4 - 3c^3 + 2c^2 - 5c - 14 = 0
  3. c4−3c3+2c2−5c+16=0c^4 - 3c^3 + 2c^2 - 5c + 16 = 0
  4. c4−3c3+2c2−5c−1=15c^4 - 3c^3 + 2c^2 - 5c - 1 = 15
Explanation: By the Remainder Theorem, when g(x)g(x) is divided by (x−c)(x - c), the remainder equals g(c)g(c). Since the remainder is 1515, we have g(c)=15g(c) = 15. Therefore c4−3c3+2c2−5c+1=15c^4 - 3c^3 + 2c^2 - 5c + 1 = 15. Choice B incorrectly sets the expression equal to zero after subtracting 15. Choice C adds 15 instead of setting the original equal to 15. Choice D incorrectly modifies the constant term before setting equal to 15.

Question 17

Consider the polynomial R(x)=3x3−7x2+4x−12R(x) = 3x^3 - 7x^2 + 4x - 12. If (x−k)(x - k) is a factor of R(x)R(x), which of the following could be the remainder when R(x)R(x) is divided by (x−2k)(x - 2k)?

  1. Always equal to zero regardless of kk
  2. Equal to R(2k)R(2k) which depends on kk (correct answer)
  3. Always equal to −12-12 regardless of kk
  4. Equal to 2R(k)2R(k) which depends on kk
Explanation: If (x−k)(x - k) is a factor of R(x)R(x), then R(k)=0R(k) = 0. By the Remainder Theorem, the remainder when R(x)R(x) is divided by (x−2k)(x - 2k) is R(2k)R(2k). Since kk and 2k2k are generally different values, R(2k)R(2k) will depend on the specific value of kk. Choice A is wrong because there's no reason R(2k)R(2k) must equal zero. Choice C incorrectly assumes the remainder is always the constant term. Choice D confuses the relationship between R(k)R(k) and R(2k)R(2k).

Question 18

For the polynomial T(x)=2x4−3x3+x2−4x+5T(x) = 2x^4 - 3x^3 + x^2 - 4x + 5, suppose that when T(x)T(x) is divided by (x−a)(x - a), the remainder is rr. What is the remainder when T(2x)T(2x) is divided by (x−a2)(x - \frac{a}{2})?

  1. 4r4r
  2. 2r2r
  3. r2\frac{r}{2}
  4. rr (correct answer)
Explanation: When you encounter polynomial remainder problems involving transformations, think about how the Remainder Theorem connects division and function evaluation. By the Remainder Theorem, when T(x)T(x) is divided by (x−a)(x - a), the remainder equals T(a)=rT(a) = r. Now let's analyze what happens with T(2x)T(2x) divided by (x−a2)(x - \frac{a}{2}). Using the Remainder Theorem again, this remainder equals T(2x)T(2x) evaluated at x=a2x = \frac{a}{2}. Substituting: T(2⋅a2)=T(a)=rT(2 \cdot \frac{a}{2}) = T(a) = r. So the remainder is still rr, making (D) correct. Let's see why the other answers represent common misconceptions. (A) 4r4r likely comes from thinking the coefficient 2 in T(2x)T(2x) gets squared when applied to the remainder. (B) 2r2r probably results from assuming the transformation x→2xx \to 2x directly scales the remainder by the same factor of 2. **(C) r2\frac{r}{2} might come from incorrectly thinking that since we're dividing by (x−a2)(x - \frac{a}{2}) instead of (x−a)(x - a), the remainder gets scaled by 12\frac{1}{2}. The key insight is that transformations of the variable don't automatically scale the remainder in obvious ways. Instead, you must carefully apply the Remainder Theorem to the transformed situation. Remember: when working with polynomial transformations and remainders, always go back to the fundamental relationship that remainder equals the polynomial evaluated at the root of the divisor.

Question 19

The polynomial P(x)=x3+ax2+bx+cP(x) = x^3 + ax^2 + bx + c satisfies P(2)=0P(2) = 0 and P(−1)=6P(-1) = 6. If Q(x)=P(x)−3Q(x) = P(x) - 3, what is the remainder when Q(x)Q(x) is divided by (x+1)(x + 1)?

  1. −3-3
  2. 66
  3. 99
  4. 33 (correct answer)
Explanation: When you see polynomial questions involving remainders and given conditions, think about the Remainder Theorem: when a polynomial P(x)P(x) is divided by (x−a)(x - a), the remainder equals P(a)P(a). Since Q(x)=P(x)−3Q(x) = P(x) - 3, we need to find Q(−1)Q(-1) to get the remainder when Q(x)Q(x) is divided by (x+1)(x + 1) (which is (x−(−1))(x - (-1))). We know P(−1)=6P(-1) = 6 from the given information. Therefore: Q(−1)=P(−1)−3=6−3=3Q(-1) = P(-1) - 3 = 6 - 3 = 3 By the Remainder Theorem, the remainder is 33. Let's examine why the other answers are wrong: Answer A (−3-3): This would be correct if you mistakenly calculated 3−6=−33 - 6 = -3 instead of 6−3=36 - 3 = 3, reversing the subtraction order. Answer B (66): This trap answer uses P(−1)P(-1) directly, forgetting that we need Q(−1)Q(-1), not P(−1)P(-1). Remember that Q(x)=P(x)−3Q(x) = P(x) - 3, so you must subtract 33. Answer C (99): This incorrect result comes from adding instead of subtracting: P(−1)+3=6+3=9P(-1) + 3 = 6 + 3 = 9. The definition Q(x)=P(x)−3Q(x) = P(x) - 3 requires subtraction. Study tip: When working with transformed polynomials like Q(x)=P(x)+kQ(x) = P(x) + k, always apply the transformation to your known values. If you know P(a)P(a), then Q(a)=P(a)+kQ(a) = P(a) + k. Don't skip this step—it's where most errors occur on these problems.

Question 20

Given that W(x)=x5−2x4+3x3−x2+4x−6W(x) = x^5 - 2x^4 + 3x^3 - x^2 + 4x - 6 and the remainder when W(x)W(x) is divided by (x2−4)(x^2 - 4) is ax+bax + b, what is the value of a+ba + b?

  1. W(2)+W(−2)−4W(2) + W(-2) - 4
  2. W(2)+W(−2)2\frac{W(2) + W(-2)}{2}
  3. W(2)−W(−2)4+W(2)\frac{W(2) - W(-2)}{4} + W(2) (correct answer)
  4. W(2)−W(−2)2+W(2)+W(−2)2\frac{W(2) - W(-2)}{2} + \frac{W(2) + W(-2)}{2}
Explanation: Since x2−4=(x−2)(x+2)x^2 - 4 = (x-2)(x+2), the remainder when dividing by this quadratic is linear: R(x)=ax+bR(x) = ax + b. By the Remainder Theorem, R(2)=W(2)R(2) = W(2) and R(−2)=W(−2)R(-2) = W(-2). So 2a+b=W(2)2a + b = W(2) and −2a+b=W(−2)-2a + b = W(-2). Solving: 4a=W(2)−W(−2)4a = W(2) - W(-2), so a=W(2)−W(−2)4a = \frac{W(2) - W(-2)}{4}. And b=W(2)−2a=W(2)−W(2)−W(−2)2b = W(2) - 2a = W(2) - \frac{W(2) - W(-2)}{2}. Therefore a+b=W(2)−W(−2)4+W(2)−W(2)−W(−2)2=W(2)−W(−2)4+W(2)a + b = \frac{W(2) - W(-2)}{4} + W(2) - \frac{W(2) - W(-2)}{2} = \frac{W(2) - W(-2)}{4} + W(2). The other choices give incorrect combinations of these values.