Home

Tutoring

Subjects

Live Classes

Study Coach

Essay Review

On-Demand Courses

Colleges

Games


Sign up

Log in

Opening subject page...

Loading your content

Practice

  • All Subjects
  • Algebra Flashcards
  • SAT Math Practice Tests
  • Math Question of the Day
  • Live Classes
  • On-Demand Courses

Varsity Tutors

  • Find a Tutor
  • Test Prep
  • Online Classes
  • K-12 Learning
  • College Search
  • VarsityTutors.com

© 2026 Varsity Tutors. All rights reserved.

← Back to quizzes

College Algebra Quiz

College Algebra Quiz: Polynomial Operations Add Multiply And Factor

Practice Polynomial Operations Add Multiply And Factor in College Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 7

0 of 7 answered

If f(x)=x2−4x+3f(x) = x^2 - 4x + 3f(x)=x2−4x+3 and g(x)=2x+1g(x) = 2x + 1g(x)=2x+1, what is the leading coefficient of f(x)⋅g(x)f(x) \cdot g(x)f(x)⋅g(x)?

Select an answer to continue

What this quiz covers

This quiz focuses on Polynomial Operations Add Multiply And Factor, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

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

If f(x)=x2−4x+3f(x) = x^2 - 4x + 3f(x)=x2−4x+3 and g(x)=2x+1g(x) = 2x + 1g(x)=2x+1, what is the leading coefficient of f(x)⋅g(x)f(x) \cdot g(x)f(x)⋅g(x)?

  1. 3
  2. 2 (correct answer)
  3. 1
  4. -8

Explanation: When you see a question asking for the leading coefficient of a product of polynomials, you're working with the concept that the leading term of the product comes from multiplying the leading terms of each factor. To find the leading coefficient of f(x)⋅g(x)f(x) \cdot g(x)f(x)⋅g(x), you need to identify the leading term of each polynomial first. For f(x)=x2−4x+3f(x) = x^2 - 4x + 3f(x)=x2−4x+3, the leading term is x2x^2x2 (coefficient = 1). For g(x)=2x+1g(x) = 2x + 1g(x)=2x+1, the leading term is 2x2x2x (coefficient = 2). When you multiply these polynomials, the highest degree term in the product comes from multiplying the leading terms: x2⋅2x=2x3x^2 \cdot 2x = 2x^3x2⋅2x=2x3. Therefore, the leading coefficient is 2, making B the correct answer. Let's examine why the other options are incorrect. Choice A (3) might tempt you if you mistakenly think the constant term from f(x)f(x)f(x) determines the leading coefficient. Choice C (1) would be your answer if you forgot that g(x)g(x)g(x) has a coefficient of 2 on its leading term and only considered the coefficient from f(x)f(x)f(x). Choice D (-8) appears to come from multiplying coefficients incorrectly, perhaps −4×2=−8-4 \times 2 = -8−4×2=−8, but this involves the wrong terms entirely. Remember this key principle: when multiplying polynomials, you only need to multiply the leading terms to find the leading coefficient of the product. This saves time and avoids the need to multiply out the entire expression.

Question 2

Factor completely: 6x3−15x2−9x6x^3 - 15x^2 - 9x6x3−15x2−9x

  1. 3x(2x2−5x−3)3x(2x^2 - 5x - 3)3x(2x2−5x−3)
  2. 3x(2x−3)(x+1)3x(2x - 3)(x + 1)3x(2x−3)(x+1)
  3. 3x(2x+1)(x−3)3x(2x + 1)(x - 3)3x(2x+1)(x−3) (correct answer)
  4. x(6x2−15x−9)x(6x^2 - 15x - 9)x(6x2−15x−9)

Explanation: First, factor out the greatest common factor: 6x3−15x2−9x=3x(2x2−5x−3)6x^3 - 15x^2 - 9x = 3x(2x^2 - 5x - 3)6x3−15x2−9x=3x(2x2−5x−3). Next, we need to factor 2x2−5x−32x^2 - 5x - 32x2−5x−3. We look for two numbers that multiply to (2)(−3)=−6(2)(-3) = -6(2)(−3)=−6 and add to −5-5−5. These numbers are −6-6−6 and 111. We rewrite: 2x2−5x−3=2x2−6x+x−3=2x(x−3)+1(x−3)=(2x+1)(x−3)2x^2 - 5x - 3 = 2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)2x2−5x−3=2x2−6x+x−3=2x(x−3)+1(x−3)=(2x+1)(x−3). Therefore, 6x3−15x2−9x=3x(2x+1)(x−3)6x^3 - 15x^2 - 9x = 3x(2x + 1)(x - 3)6x3−15x2−9x=3x(2x+1)(x−3). Choice A stops after factoring out the GCF. Choice B has the wrong signs in the factorization of the quadratic. Choice D only factors out xxx instead of the complete GCF 3x3x3x.

Question 3

When 2x3−5x2+x+32x^3 - 5x^2 + x + 32x3−5x2+x+3 is divided by x−2x - 2x−2, what is the remainder?

  1. 1 (correct answer)
  2. -3
  3. 5
  4. -1

Explanation: By the Remainder Theorem, when a polynomial P(x)P(x)P(x) is divided by (x−a)(x - a)(x−a), the remainder is P(a)P(a)P(a). Here, we need to find P(2)P(2)P(2) where P(x)=2x3−5x2+x+3P(x) = 2x^3 - 5x^2 + x + 3P(x)=2x3−5x2+x+3. Substituting x=2x = 2x=2: P(2)=2(2)3−5(2)2+(2)+3=2(8)−5(4)+2+3=16−20+2+3=1P(2) = 2(2)^3 - 5(2)^2 + (2) + 3 = 2(8) - 5(4) + 2 + 3 = 16 - 20 + 2 + 3 = 1P(2)=2(2)3−5(2)2+(2)+3=2(8)−5(4)+2+3=16−20+2+3=1. Choice B results from a sign error in the calculation. Choice C comes from forgetting to include the constant term. Choice D results from an arithmetic error in combining the terms.

Question 4

Which expression is equivalent to (2x−3)2−(x+1)2(2x - 3)^2 - (x + 1)^2(2x−3)2−(x+1)2?

  1. x2−14x+8x^2 - 14x + 8x2−14x+8
  2. 3x2−14x+103x^2 - 14x + 103x2−14x+10
  3. 3x−43x - 43x−4
  4. 3x2−14x+83x^2 - 14x + 83x2−14x+8 (correct answer)

Explanation: When you encounter the difference of two squares or need to expand multiple binomial expressions, you have two main approaches: expand each term separately or look for algebraic patterns that can simplify your work. Let's expand each squared term individually. For (2x−3)2(2x - 3)^2(2x−3)2, we get: (2x−3)2=(2x)2−2(2x)(3)+32=4x2−12x+9(2x - 3)^2 = (2x)^2 - 2(2x)(3) + 3^2 = 4x^2 - 12x + 9(2x−3)2=(2x)2−2(2x)(3)+32=4x2−12x+9 For (x+1)2(x + 1)^2(x+1)2: (x+1)2=x2+2x+1(x + 1)^2 = x^2 + 2x + 1(x+1)2=x2+2x+1 Now substituting back into the original expression: (2x−3)2−(x+1)2=(4x2−12x+9)−(x2+2x+1)(2x - 3)^2 - (x + 1)^2 = (4x^2 - 12x + 9) - (x^2 + 2x + 1)(2x−3)2−(x+1)2=(4x2−12x+9)−(x2+2x+1) Distributing the negative sign and combining like terms: =4x2−12x+9−x2−2x−1=3x2−14x+8= 4x^2 - 12x + 9 - x^2 - 2x - 1 = 3x^2 - 14x + 8=4x2−12x+9−x2−2x−1=3x2−14x+8 This confirms answer choice D is correct. Answer A gives x2−14x+8x^2 - 14x + 8x2−14x+8, which has the wrong coefficient on x2x^2x2 — likely from incorrectly calculating 4x2−x2=x24x^2 - x^2 = x^24x2−x2=x2 instead of 3x23x^23x2. Answer B shows 3x2−14x+103x^2 - 14x + 103x2−14x+10, which has the correct variable terms but the wrong constant (101010 instead of 888) — probably from sign errors when subtracting the constant terms. Answer C gives 3x−43x - 43x−4, suggesting someone incorrectly tried to factor this as a difference of squares pattern. When expanding binomial differences, always distribute negative signs carefully and double-check your arithmetic when combining like terms — coefficient errors are the most common mistakes on these problems.

Question 5

Which of the following is equivalent to (3x2−7x+2)−(x2+4x−5)+(2x2−x+3)(3x^2 - 7x + 2) - (x^2 + 4x - 5) + (2x^2 - x + 3)(3x2−7x+2)−(x2+4x−5)+(2x2−x+3)?

  1. 4x2−12x+104x^2 - 12x + 104x2−12x+10 (correct answer)
  2. 4x2−12x4x^2 - 12x4x2−12x
  3. 6x2−12x+106x^2 - 12x + 106x2−12x+10
  4. 4x2−2x+104x^2 - 2x + 104x2−2x+10

Explanation: We combine like terms step by step: (3x2−7x+2)−(x2+4x−5)+(2x2−x+3)=3x2−7x+2−x2−4x+5+2x2−x+3(3x^2 - 7x + 2) - (x^2 + 4x - 5) + (2x^2 - x + 3) = 3x^2 - 7x + 2 - x^2 - 4x + 5 + 2x^2 - x + 3(3x2−7x+2)−(x2+4x−5)+(2x2−x+3)=3x2−7x+2−x2−4x+5+2x2−x+3. Collecting like terms: x2x^2x2 terms: 3x2−x2+2x2=4x23x^2 - x^2 + 2x^2 = 4x^23x2−x2+2x2=4x2. xxx terms: −7x−4x−x=−12x-7x - 4x - x = -12x−7x−4x−x=−12x. Constant terms: 2+5+3=102 + 5 + 3 = 102+5+3=10. Therefore, the result is 4x2−12x+104x^2 - 12x + 104x2−12x+10. Choice B incorrectly omits the constant term. Choice C incorrectly adds the x2x^2x2 coefficients as 3+1+2=63 + 1 + 2 = 63+1+2=6 instead of 3−1+2=43 - 1 + 2 = 43−1+2=4. Choice D incorrectly computes the xxx coefficient as −7+4−1=−4-7 + 4 - 1 = -4−7+4−1=−4 instead of −7−4−1=−12-7 - 4 - 1 = -12−7−4−1=−12.

Question 6

What is the result when 4x3−6x2+8x4x^3 - 6x^2 + 8x4x3−6x2+8x is divided by 2x2x2x?

  1. 2x2−3x+4+02x2x^2 - 3x + 4 + \frac{0}{2x}2x2−3x+4+2x0​
  2. 2x2−3x+42x^2 - 3x + 42x2−3x+4 (correct answer)
  3. 4x2−6x+84x^2 - 6x + 84x2−6x+8
  4. 2x3−3x2+4x2x^3 - 3x^2 + 4x2x3−3x2+4x

Explanation: When you encounter polynomial division problems, you're essentially distributing the division across each term in the polynomial. Think of this as the reverse of factoring out a common factor. To divide 4x3−6x2+8x4x^3 - 6x^2 + 8x4x3−6x2+8x by 2x2x2x, you divide each term separately: 4x32x−6x22x+8x2x\frac{4x^3}{2x} - \frac{6x^2}{2x} + \frac{8x}{2x}2x4x3​−2x6x2​+2x8x​. For each term, divide the coefficients and subtract the exponents of like variables. The first term: 4x32x=42⋅x3x=2x2\frac{4x^3}{2x} = \frac{4}{2} \cdot \frac{x^3}{x} = 2x^22x4x3​=24​⋅xx3​=2x2. The second term: 6x22x=62⋅x2x=3x\frac{6x^2}{2x} = \frac{6}{2} \cdot \frac{x^2}{x} = 3x2x6x2​=26​⋅xx2​=3x. The third term: 8x2x=82⋅xx=4\frac{8x}{2x} = \frac{8}{2} \cdot \frac{x}{x} = 42x8x​=28​⋅xx​=4. This gives you 2x2−3x+42x^2 - 3x + 42x2−3x+4. Choice A incorrectly adds a remainder term 02x\frac{0}{2x}2x0​, but since every term in the original polynomial contains at least one factor of xxx, there's no remainder when dividing by 2x2x2x. Choice C shows 4x2−6x+84x^2 - 6x + 84x2−6x+8, which incorrectly only divides the coefficients by 2 but leaves the exponents unchanged. Choice D gives 2x3−3x2+4x2x^3 - 3x^2 + 4x2x3−3x2+4x, which correctly halves the coefficients but fails to reduce the exponents at all. The answer is B: 2x2−3x+42x^2 - 3x + 42x2−3x+4. Remember: when dividing polynomials by monomials, always check that every term in the dividend is divisible by the divisor. If not, you'll have a remainder.

Question 7

The polynomial P(x)=x3+2x2−5x−6P(x) = x^3 + 2x^2 - 5x - 6P(x)=x3+2x2−5x−6 can be written as (x−a)(x+b)(x+c)(x - a)(x + b)(x + c)(x−a)(x+b)(x+c) where aaa, bbb, and ccc are positive integers. What is the value of a+b+ca + b + ca+b+c?

  1. 7
  2. 8
  3. 5
  4. 6 (correct answer)

Explanation: When you encounter a polynomial that needs to be factored into linear terms, you're looking for the roots (zeros) of the polynomial. These roots will give you the values that make each factor equal to zero. To factor P(x)=x3+2x2−5x−6P(x) = x^3 + 2x^2 - 5x - 6P(x)=x3+2x2−5x−6, start by testing possible rational roots using the Rational Root Theorem. The possible rational roots are the factors of the constant term (-6) divided by the factors of the leading coefficient (1): ±1, ±2, ±3, ±6. Testing x=1x = 1x=1: P(1)=1+2−5−6=−8P(1) = 1 + 2 - 5 - 6 = -8P(1)=1+2−5−6=−8 (not a root) Testing x=−1x = -1x=−1: P(−1)=−1+2+5−6=0P(-1) = -1 + 2 + 5 - 6 = 0P(−1)=−1+2+5−6=0 (this is a root!) Since (x+1)(x + 1)(x+1) is a factor, use polynomial division to find: P(x)=(x+1)(x2+x−6)P(x) = (x + 1)(x^2 + x - 6)P(x)=(x+1)(x2+x−6) Factor the quadratic: x2+x−6=(x+3)(x−2)x^2 + x - 6 = (x + 3)(x - 2)x2+x−6=(x+3)(x−2) Therefore: P(x)=(x+1)(x+3)(x−2)=(x−2)(x+1)(x+3)P(x) = (x + 1)(x + 3)(x - 2) = (x - 2)(x + 1)(x + 3)P(x)=(x+1)(x+3)(x−2)=(x−2)(x+1)(x+3) In the form (x−a)(x+b)(x+c)(x - a)(x + b)(x + c)(x−a)(x+b)(x+c), we have a=2a = 2a=2, b=1b = 1b=1, and c=3c = 3c=3, so a+b+c=6a + b + c = 6a+b+c=6. Choice A (7) might result from incorrectly including negative signs in your sum. Choice B (8) could come from adding the actual roots (-1, -3, 2) and getting their absolute values confused. Choice C (5) might result from computational errors during factoring. Study tip: Always verify your factorization by expanding it back to the original polynomial. This catches sign errors and ensures your factors are correct.