College Algebra Quiz: Polynomial Long Division And Synthetic Division
3 questions · exam conditions
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Polynomial Long Division And Synthetic DivisionQuestion 1 of 3

When dividing 3x4−5x3+2x2−7x+43x^4 - 5x^3 + 2x^2 - 7x + 4 by x2−2x+1x^2 - 2x + 1, what is the remainder?

−3x+2-3x + 2
−5x+6-5x + 6
2x−12x - 1
x+3x + 3
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College Algebra Quiz

College Algebra Quiz: Polynomial Long Division And Synthetic Division

Practice Polynomial Long Division And Synthetic Division in College Algebra 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 Long Division And Synthetic Division, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

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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

When dividing 3x4−5x3+2x2−7x+43x^4 - 5x^3 + 2x^2 - 7x + 4 by x2−2x+1x^2 - 2x + 1, what is the remainder?

  1. −3x+2-3x + 2 (correct answer)
  2. −5x+6-5x + 6
  3. 2x−12x - 1
  4. x+3x + 3
Explanation: Using polynomial long division: 3x4−5x3+2x2−7x+4=(x2−2x+1)(3x2+x+2)+(−3x+2)3x^4 - 5x^3 + 2x^2 - 7x + 4 = (x^2 - 2x + 1)(3x^2 + x + 2) + (-3x + 2). The remainder is −3x+2-3x + 2. Choice B results from sign errors during division. Choice C comes from incorrect handling of the constant term. Choice D represents an incomplete calculation stopping too early.

Question 2

If synthetic division of 2x3−7x2+kx−152x^3 - 7x^2 + kx - 15 by (x−3)(x - 3) results in a remainder of 0, what is the value of kk?

  1. k=2k = 2
  2. k=5k = 5
  3. k=8k = 8 (correct answer)
  4. k=11k = 11
Explanation: Since the remainder is 0, (x−3)(x - 3) is a factor, so f(3)=0f(3) = 0. Substituting: 2(3)3−7(3)2+k(3)−15=02(3)^3 - 7(3)^2 + k(3) - 15 = 0, which gives 54−63+3k−15=054 - 63 + 3k - 15 = 0, so 3k=243k = 24 and k=8k = 8. Choice A comes from using x=−3x = -3 instead of x=3x = 3. Choice B results from arithmetic errors in the calculation. Choice D comes from forgetting the negative sign on the constant term.

Question 3

When 6x3−13x2+px+q6x^3 - 13x^2 + px + q is divided by (x−2)(x+1)(x - 2)(x + 1), the remainder is 5x−35x - 3. What are the values of pp and qq?

  1. p=7,q=6p = 7, q = 6
  2. p=−2,q=9p = -2, q = 9 (correct answer)
  3. p=4,q=−8p = 4, q = -8
  4. p=1,q=12p = 1, q = 12
Explanation: Since the divisor has degree 2, we can write 6x3−13x2+px+q=(x−2)(x+1)⋅Q(x)+5x−36x^3 - 13x^2 + px + q = (x - 2)(x + 1) \cdot Q(x) + 5x - 3. At x=2x = 2: 48−52+2p+q=10−348 - 52 + 2p + q = 10 - 3, so 2p+q=112p + q = 11. At x=−1x = -1: −6−13−p+q=−5−3-6 - 13 - p + q = -5 - 3, so −p+q=11-p + q = 11. Solving: p=−2,q=9p = -2, q = 9. The other choices result from sign errors or incorrect substitution in the system of equations.