This question tests AP Precalculus skills: understanding polynomial functions and complex zeros. Complex zeros occur in conjugate pairs when a quadratic has a negative discriminant. For v(x) = 8x² + 16x + 13 with Δ = -160, we use: x = (-16 ± √(-160))/(2·8) = (-16 ± √160 i)/16. Since √160 = √(16·10) = 4√10, we have √(-160) = 4√10 i, giving x = (-16 ± 4√10 i)/16 = -1 ± (√10/4)i. Choice A is correct because it shows the properly simplified form with real part -1 and imaginary parts ±√10/4. Choice B is incorrect because it shows ±√10/2, which would result from dividing by 8 instead of 16. To help students: Always simplify radicals by factoring out perfect squares, then reduce fractions carefully. Watch for: Confusion about whether to divide by 2a or just a in the quadratic formula.

This question tests AP Precalculus skills: understanding polynomial functions and complex zeros. For quadratics with negative discriminants, the zeros are complex conjugates found via the quadratic formula. Given u(x) = 3x² - 18x + 31 with Δ = -48, we calculate: x = (18 ± √(-48))/(2·3) = (18 ± √48 i)/6. Since √48 = √(16·3) = 4√3, we have √(-48) = 4√3 i, giving x = (18 ± 4√3 i)/6 = 3 ± (2√3/3)i. Choice A is correct because it properly simplifies to the standard form a ± bi with rationalized denominators. Choice B is incorrect because it shows ±4√3/3, failing to simplify the fraction 4/6 to 2/3. To help students: Factor perfect squares from radicals before reducing fractions, and always simplify completely. Watch for: Missing the simplification of 4/6 to 2/3 in the imaginary part.

This question tests AP Precalculus skills: understanding polynomial functions and complex zeros. Factoring by grouping is an efficient method for certain polynomials, revealing both real and complex zeros. For f(x) = x³ + 2x² + 5x + 10, grouping gives x²(x+2) + 5(x+2) = (x+2)(x²+5). The real zero is x = -2, and from x² + 5 = 0, we get x² = -5, so x = ±√(5)i. Choice A is correct because it identifies all three zeros: the real zero -2 and the conjugate pair ±√(5)i. Choice B is incorrect because it has the wrong sign for the real zero (2 instead of -2). To help students: Practice recognizing grouping patterns and connecting factored form to zeros. Emphasize that x² + positive number = 0 always yields purely imaginary zeros.

This question tests AP Precalculus skills: understanding polynomial functions and complex zeros. Complex zeros of polynomials with real coefficients always come in conjugate pairs a ± bi. For p(x) = 4x² + 4x + 5 with Δ = -64, we apply the quadratic formula: x = (-4 ± √(-64))/(2·4) = (-4 ± 8i)/8. Simplifying by dividing both terms by 8, we get x = -1/2 ± i. Choice A is correct because it shows the properly reduced form with real part -1/2 and imaginary parts ±i. Choice B is incorrect because it shows -1/2 ± 2i, which would come from incorrectly simplifying (-4 ± 16i)/8. To help students: Remember that √(-64) = 8i (not 16i), and always reduce fractions completely. Watch for: Confusion between √64 = 8 and mistakenly doubling it when dealing with imaginary numbers.

This question tests AP Precalculus skills: understanding polynomial functions and complex zeros. When a quadratic has a negative discriminant, it yields two complex conjugate zeros that can be found using the quadratic formula. For h(x) = x² - 6x + 11 with Δ = -8, we calculate: x = (6 ± √(-8))/(2·1) = (6 ± √8 i)/2. Since √8 = √(4·2) = 2√2, we have √(-8) = 2√2 i, giving x = (6 ± 2√2 i)/2 = 3 ± √2 i. Choice A is correct because it shows the simplified form with real part 3 and imaginary parts ±√2 i. Choice B is incorrect because it shows 3 ± 2√2 i, failing to divide the coefficient of i by 2. To help students: Practice simplifying square roots by factoring out perfect squares, and always divide both the real and imaginary parts by 2a. Watch for: Errors in simplifying √8 and forgetting to divide the entire expression by 2.

This question tests AP Precalculus skills: understanding polynomial functions and complex zeros. Completing the square is an alternative method to find complex zeros when the discriminant is negative. For f(x) = x² + 4x + 13, completing the square gives (x+2)² + 9 = 0, so (x+2)² = -9. Taking square roots yields x + 2 = ±3i, therefore x = -2 ± 3i. Choice B is correct because it shows the proper form with real part -2 and imaginary parts ±3i. Choice A is incorrect because it has the wrong real part (2 instead of -2). To help students: Practice completing the square and connecting it to the quadratic formula. Visualize complex zeros on the complex plane to reinforce that they don't create x-intercepts on the real coordinate system.

This question tests AP Precalculus skills: understanding polynomial functions and complex zeros. Complex zeros of quadratics with real coefficients come in conjugate pairs when the discriminant is negative. For t(x) = 6x² - 12x + 11 with Δ = -120, we apply: x = (12 ± √(-120))/(2·6) = (12 ± √120 i)/12. Since √120 = √(4·30) = 2√30, we have √(-120) = 2√30 i, giving x = (12 ± 2√30 i)/12 = 1 ± (√30/6)i. Choice A is correct because it shows the fully reduced form with proper simplification of both parts. Choice B is incorrect because it shows ±√30/3, which would come from dividing by 6 instead of 12. To help students: Always simplify radicals first, then reduce the entire fraction. Watch for: Errors in simplifying √120 and confusion about which denominator to use.

This question tests AP Precalculus skills: understanding polynomial functions and complex zeros. A degree 4 polynomial has exactly four zeros counting multiplicity, and the problem demonstrates synthetic division and factoring techniques. Starting with f(x) = x⁴ - 2x³ + 5x² - 8x + 4 and knowing x = 1 is a zero, synthetic division yields x³ - x² + 4x - 4. Factoring by grouping gives (x-1)(x²+4), so f(x) = (x-1)²(x²+4). Choice A is correct because it lists all four zeros with proper multiplicity: x = 1 with multiplicity 2, and x = ±2i. Choice B is incorrect because it only lists three zeros, missing the multiplicity of x = 1. To help students: Practice synthetic division and recognize repeated factors. Emphasize counting zeros with their multiplicities to match the polynomial's degree.

This question tests AP Precalculus skills: understanding polynomial functions and complex zeros. A quartic polynomial has four zeros, and when factored into quadratics with real coefficients, each quadratic may yield a conjugate pair of complex zeros. The polynomial f(x) = x⁴ + 2x³ + 5x² + 10x + 13 factors as (x² + 2x + 5)(x² + 1). From x² + 1 = 0, we get x = ±i; from x² + 2x + 5 = 0, using the quadratic formula gives x = (-2 ± √(-16))/2 = -1 ± 2i. Choice A is correct because it lists all four zeros: ±i and -1 ± 2i. Choice D is incorrect because it has the wrong real part for the second pair (1 instead of -1). To help students: Practice factoring quartics into quadratics and solving each factor separately. Verify factorizations by expanding to check coefficients.

This question tests AP Precalculus skills: understanding polynomial functions and complex zeros. For a quadratic with negative discriminant, the zeros are complex conjugates a ± bi where both a and b are real numbers. Given g(x) = 3x² + 12x + 17 with Δ = -60, we use the quadratic formula: x = (-12 ± √(-60))/(2·3) = (-12 ± √60 i)/6. Since √60 = √(4·15) = 2√15, we have √(-60) = 2√15 i, giving x = (-12 ± 2√15 i)/6 = -2 ± (√15/3)i. Choice A is correct because it accurately shows both the real part (-2) and the properly simplified imaginary part (±√15/3). Choice D is incorrect because it shows -2 ± √15 i, failing to divide the imaginary coefficient by 3. To help students: Always simplify radicals by factoring out perfect squares, and ensure both parts of the complex number are divided by 2a. Watch for: Common errors in simplifying the imaginary part and forgetting to reduce fractions.