Unit 4: Polynomial and Rational Functions

Answer: B — The correct answer is C because the area of a rectangle is given by the formula A = length * width. In this case, the area is 48 and the length is given by the polynomial 2x + 1. So, the width can be found by dividing the area by the length: width = 48 / (2x + 1). Options A, B, and D are incorrect because they do not accurately represent the width of the bed based on the given information.

Answer: A — To find the zeros of a polynomial in factored form, set each factor equal to zero and solve. From (x + 2) = 0, we get x = -2. From (x - 5) = 0, we get x = 5. From (x + 1) = 0, we get x = -1. Therefore, the zeros are -2, 5, and -1. Option B is incorrect because students may have mistakenly changed the signs of all terms. Option C incorrectly keeps the sign of the 5 as negative, suggesting a misunderstanding of solving (x - 5) = 0. Option D reverses all the signs, indicating confusion about the relationship between factors and zeros. The key insight is that if (x - a) is a factor, then a is a zero; if (x + a) is a factor, then -a is a zero.

Answer: A — To solve this problem, we must factor both the numerator and denominator completely. The numerator x³ - 3x² - 4x factors as x(x² - 3x - 4) = x(x - 4)(x + 1). The denominator x² - 5x + 4 factors as (x - 1)(x - 4). After factoring, we have f(x) = x(x - 4)(x + 1)/[(x - 1)(x - 4)]. The factor (x - 4) appears in both numerator and denominator, so it cancels, creating a hole at x = 4. The factor (x - 1) remains only in the denominator, creating a vertical asymptote at x = 1. Option B is incorrect because x = 1 is a vertical asymptote (not canceled) and x = 4 is a hole (canceled factor). Option C reverses these, incorrectly placing the hole at x = 4 and asymptote at x = 1. Option D is incorrect because it suggests both create holes with no asymptotes, which misidentifies the role of uncanceled denominator factors. Understanding the distinction between canceled factors (holes) and uncanceled denominator factors (vertical asymptotes) is essential to analyzing rational function behavior.

Answer: A — To find holes versus vertical asymptotes, we must factor both numerator and denominator completely. The numerator factors as x(x² - 2x - 3) = x(x - 3)(x + 1). The denominator factors as x(x - 3). After canceling the common factors x and (x - 3), we get the simplified form f(x) = (x + 1)/1 = x + 1 for x ≠ 0 and x ≠ 3. A hole occurs at x-values where factors cancel from both numerator and denominator. Since both x and (x - 3) appear in both the numerator and denominator, holes occur at x = 0 and x = 3. However, the question asks specifically which point has a hole rather than a vertical asymptote, and x = 3 is the answer that appears in the options. At x = 3, the hole occurs at the point (3, 4) since the simplified function gives y = 3 + 1 = 4. Option A is incorrect because x = 0 is also a hole (at the point (0, 1)), but if forced to choose one answer, x = 3 is the intended answer testing whether students can identify cancelled factors. Option C is incorrect because x = -1 makes the simplified numerator zero, creating an x-intercept, not a hole. Option D is incorrect because the function does have holes where common factors cancel.

Answer: A — Both $(x + 3)(x^2 - x - 4)$ and $(x + 4)(x^2 - 2x - 3)$ are correct factorizations because when expanded, they both result in the original polynomial $x^3 + 2x^2 - 7x - 12$. The other options are incorrect because they claim that only one or neither of the factorizations is correct.