Zeros, Factors, and the Factor Theorem - College Algebra

1. A cubic polynomial $f(x)$ has leading coefficient 1 and zeros at $x = -1$, $x = 2$, and $x = 5$. If $g(x) = f(x - 3)$, what are the zeros of $g(x)$?
   Answer: $x = 2, x = 5, x = 8$ 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.
2. Consider the polynomial $q(x) = x^3 - 7x + 6$. If you know that $x = 1$ is a zero, and you want to find the remaining zeros by factoring, which of the following correctly represents $q(x)$ in factored form?
   Answer: $(x - 1)(x^2 + x - 6)$ 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.
3. The polynomial $f(x) = x^4 - 10x^2 + 9$ can be factored by recognizing it as a quadratic in $x^2$. How many real zeros does $f(x)$ have?
   Answer: Four real zeros, all with multiplicity 1 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.
4. The polynomial $h(x) = x^4 + 2x^3 - 7x^2 - 8x + 12$ has $(x - 1)$ and $(x + 3)$ as factors. What is the product of all four zeros of $h(x)$?
   Answer: 12 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.
5. If $f(x) = 2x^3 - 7x^2 + 2x + 3$, and the Factor Theorem confirms that $(x - 3)$ is a factor of $f(x)$, which of the following represents the complete factorization of $f(x)$?
   Answer: $(x - 3)(2x^2 - x - 1)$ 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.

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