College Algebra Quiz: Zeros Factors And The Factor Theorem

What this quiz covers

This quiz focuses on Zeros Factors And The Factor Theorem, 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.

Question 1

1. $(x - 3)(2x^2 - x - 1)$ (correct answer)
2. $(x - 3)(2x^2 + x - 1)$
3. $(x - 3)(2x^2 - x + 1)$
4. $(x - 3)(2x^2 + x + 1)$

Explanation: Since $(x - 3)$ is a factor, we can perform polynomial division: $f(x) = 2x^3 - 7x^2 + 2x + 3 = (x - 3)(2x^2 - x - 1)$. We can verify this by expanding or by synthetic division. Choice B has the wrong sign on the middle term of the quotient. Choice C has the wrong sign on the constant term. Choice D has wrong signs on both the middle and constant terms.

Question 2

1. $x = -4, x = -1, x = 2$
2. $x = 2, x = 5, x = 8$ (correct answer)
3. $x = -1, x = 2, x = 5$
4. $x = -2, x = 1, x = 4$

Explanation: First, $f(x) = (x + 1)(x - 2)(x - 5)$ since it has leading coefficient 1 and the given zeros. For $g(x) = f(x - 3)$, the zeros occur when $g(x) = 0$, i.e., when $f(x - 3) = 0$. This happens when $x - 3$ equals a zero of $f(x)$. So $x - 3 = -1$ gives $x = 2$; $x - 3 = 2$ gives $x = 5$; and $x - 3 = 5$ gives $x = 8$. The horizontal shift right by 3 units shifts all zeros right by 3. Choice A shifts left instead of right. Choice C gives the original zeros. Choice D uses an incorrect shift amount.

Question 3

1. $(x - 1)(x^2 - x + 6)$
2. $(x - 1)(x^2 - x - 6)$
3. $(x - 1)(x^2 + x + 6)$
4. $(x - 1)(x^2 + x - 6)$ (correct answer)

Explanation: When you're given that a specific value is a zero of a polynomial, you can use polynomial division to factor out the corresponding linear factor and find the remaining zeros. Since $x = 1$ is a zero of $q(x) = x^3 - 7x + 6$, we know $(x - 1)$ is a factor. To find the complete factorization, you need to divide $q(x)$ by $(x - 1)$. Using polynomial long division or synthetic division: $x^3 - 7x + 6 = (x - 1)(x^2 + x - 6)$ You can verify this by expanding: $(x - 1)(x^2 + x - 6) = x^3 + x^2 - 6x - x^2 - x + 6 = x^3 - 7x + 6$ ✓ Choice A gives $(x - 1)(x^2 - x + 6)$. Expanding this yields $x^3 - x^2 + 6x - x^2 + x + 6 = x^3 - 2x^2 + 7x + 6$, which doesn't match our original polynomial. Choice B gives $(x - 1)(x^2 - x - 6)$. Expanding yields $x^3 - x^2 - 6x - x^2 + x + 6 = x^3 - 2x^2 - 5x + 6$, also incorrect. Choice C gives $(x - 1)(x^2 + x + 6)$. Expanding yields $x^3 + x^2 + 6x - x^2 - x - 6 = x^3 + 5x - 6$, which is wrong. Only choice D produces the correct expansion. Study tip: When factoring polynomials with a known zero, always verify your factorization by expanding it back out. This catches arithmetic errors and ensures you've correctly performed the polynomial division.

Question 4

1. Two real zeros, both with multiplicity 1
2. Four real zeros, all with multiplicity 1 (correct answer)
3. Two real zeros, both with multiplicity 2
4. Four real zeros, two with multiplicity 2 each

Explanation: Let $u = x^2$, so $f(x) = u^2 - 10u + 9 = (u - 1)(u - 9) = (x^2 - 1)(x^2 - 9)$. Further factoring: $f(x) = (x - 1)(x + 1)(x - 3)(x + 3)$. This gives four distinct real zeros: $x = -3, -1, 1, 3$, each with multiplicity 1. Choice A incorrectly counts only two zeros. Choice C incorrectly states multiplicity 2. Choice D incorrectly describes the multiplicities.

Question 5

1. 8
2. -12
3. 12 (correct answer)
4. -8

Explanation: When you encounter a polynomial with known factors, you're dealing with relationships between zeros and coefficients. The key insight here is that you don't need to find all four zeros individually—there's a more direct path using Vieta's formulas. For any polynomial $ax^n + bx^{n-1} + \cdots + k = 0$, the product of all zeros equals $(-1)^n \cdot \frac{\text{constant term}}{\text{leading coefficient}}$. Since $h(x) = x^4 + 2x^3 - 7x^2 - 8x + 12$ is degree 4 with leading coefficient 1 and constant term 12, the product of all zeros is $(-1)^4 \cdot \frac{12}{1} = 1 \cdot 12 = 12$. This confirms answer choice C. Let's examine why the other answers are incorrect. Choice A (8) might come from incorrectly multiplying the known zeros: since $(x-1)$ and $(x+3)$ are factors, two zeros are $x = 1$ and $x = -3$, giving $1 \times (-3) = -3$, not leading to 8. Choice B (-12) represents the constant term with incorrect sign—this would be the answer if you forgot the $(-1)^n$ factor in Vieta's formula for even-degree polynomials. Choice D (-8) combines both errors: wrong sign and possibly incorrect calculation. Study tip: Memorize Vieta's formulas—they're powerful shortcuts on polynomial questions. For the product of zeros, always remember: $(-1)^n \times \frac{\text{constant term}}{\text{leading coefficient}}$. This saves time compared to factoring completely and is less error-prone than synthetic division.