Polynomial Functions and Complex Zeros - AP Precalculus

1. Using the polynomial provided, a quadratic $s(x)=x^2+2x+10$ has $\Delta=4-40=-36$; calculate the complex zeros via the quadratic formula and simplify $\sqrt{-36}=6i$ to exact $a\pm bi$ form.​
   Answer: $x=-1\pm3i$ This question tests AP Precalculus skills: understanding polynomial functions and complex zeros. When a quadratic polynomial has complex zeros, they are conjugates of the form a ± bi. For s(x) = x² + 2x + 10 with Δ = -36, we calculate: x = (-2 ± √(-36))/(2·1) = (-2 ± 6i)/2 = -1 ± 3i. Choice A is correct because it shows the simplified complex zeros with real part -1 and imaginary parts ±3i. Choice C is incorrect because it shows -1 ± 6i, failing to divide the imaginary part by 2. To help students: Remember that √(-36) = 6i, and always divide both parts of the numerator by 2a. Watch for: Forgetting to divide the imaginary part by the denominator in the quadratic formula.
2. Using the polynomial provided, consider the quadratic $v(x)=4x^2-4x+7$, a degree $2$ polynomial. Compute $b^2-4ac=(-4)^2-4(4)(7)=16-112=-96$, so $x=\frac{4\pm\sqrt{-96}}{8}=\frac{4\pm\sqrt{96}\,i}{8}=\frac{4\pm 4\sqrt{6}\,i}{8}=\frac{1}{2}\pm\frac{\sqrt{6}}{2}i.$ These conjugate zeros indicate the parabola has no real zeros on the coordinate plane. What are the complex zeros of the polynomial function $v(x)$?
   Answer: $\frac{1}{2}\pm\frac{\sqrt{6}}{2}i$ This question tests AP Precalculus skills: understanding polynomial functions and complex zeros. Polynomial functions can have complex zeros, which are solutions to the polynomial equation in the form a + bi, where a and b are real numbers. For v(x) = 4x² - 4x + 7, the discriminant is -96, yielding: x = (4 ± √(-96))/8 = (4 ± 4√6 i)/8 = 1/2 ± (√6/2)i. Choice A is correct because it shows the properly simplified complex zeros with real part 1/2 and imaginary parts ±(√6/2)i. Choice E is incorrect because it treats the result as entirely real (1/2 ± √6/2), missing the imaginary unit i that must be included when taking the square root of -96. To help students: Practice simplifying complex fractions step-by-step and emphasize that √(-96) = √96 · i = 4√6 i. Watch for: Students omitting the imaginary unit i when working with negative discriminants.
3. Using the polynomial provided, consider the cubic $f(x)=x^3-x^2+5x-5$. A polynomial of degree $3$ has three zeros counting multiplicity. Factor by grouping: $x^3-x^2+5x-5=x^2(x-1)+5(x-1)=(x-1)(x^2+5)$. The real zero is $x=1$, and the remaining factor gives $x^2=-5$. Writing $-5=5\cdot(-1)$ and using $\sqrt{-1}=i$, we obtain $x=\pm\sqrt{5}\,i$. Because the coefficients are real, any nonreal zeros must occur as conjugate pairs, which matches $\sqrt{5}i$ and $-\sqrt{5}i$. What are the complex zeros of the polynomial function $f(x)=x^3-x^2+5x-5$?
   Answer: $x=1,\ \pm\sqrt{5}\,i$ This question tests AP Precalculus skills: understanding polynomial functions and complex zeros. A cubic polynomial must have exactly three zeros (counting multiplicity), and when coefficients are real, complex zeros occur in conjugate pairs. The polynomial f(x) = x³ - x² + 5x - 5 can be factored by grouping: x²(x-1) + 5(x-1) = (x-1)(x²+5). Setting x² + 5 = 0 gives x² = -5, so x = ±√(5)i. Choice A is correct because it identifies all three zeros: the real zero x = 1 and the conjugate pair ±√(5)i. Choice E is incorrect because it shows ±√5 as real numbers rather than imaginary. To help students: Practice factoring by grouping and recognizing when a quadratic factor yields complex zeros. Emphasize that complex zeros of polynomials with real coefficients always come in conjugate pairs.
4. Using the polynomial provided, take $t(x)=2x^2+12x+25$, a degree $2$ polynomial. Compute $b^2-4ac=12^2-4(2)(25)=144-200=-56$, then $x=\frac{-12\pm\sqrt{-56}}{4}=\frac{-12\pm\sqrt{56}\,i}{4}=\frac{-12\pm 2\sqrt{14}\,i}{4}=-3\pm\frac{\sqrt{14}}{2}i.$ These complex zeros satisfy $t(x)=0$ but create no real intercepts on the coordinate plane. What are the complex zeros of the polynomial function $t(x)$?
   Answer: $-3\pm\frac{\sqrt{14}}{2}i$ This question tests AP Precalculus skills: understanding polynomial functions and complex zeros. Polynomial functions can have complex zeros, which are solutions to the polynomial equation in the form a + bi, where a and b are real numbers. For t(x) = 2x² + 12x + 25, the discriminant is -56, giving: x = (-12 ± √(-56))/4 = (-12 ± 2√14 i)/4 = -3 ± (√14/2)i. Choice A is correct because it properly simplifies to show the real part -3 and imaginary parts ±(√14/2)i. Choice E is incorrect because it presents the result as entirely real (-3 ± √14/2), omitting the imaginary unit i that must accompany the square root of -56. To help students: Emphasize that negative discriminants always produce complex zeros and practice simplifying √(-56) = √56 · i = 2√14 i. Watch for: Students treating square roots of negative numbers as if they were real.
5. Using the polynomial provided, a degree-2 function $q(x)=5x^2-10x+9$ yields $\Delta=100-180=-80$; calculate the complex zeros with the quadratic formula, converting $\sqrt{-80}=4\sqrt{5}\,i$ and simplifying fully.​
   Answer: $x=1\pm\dfrac{2\sqrt{5}}{5}i$ This question tests AP Precalculus skills: understanding polynomial functions and complex zeros. For polynomials with negative discriminants, the quadratic formula yields complex conjugate zeros. Given q(x) = 5x² - 10x + 9 with Δ = -80, we calculate: x = (10 ± √(-80))/(2·5) = (10 ± √80 i)/10. Since √80 = √(16·5) = 4√5, we have √(-80) = 4√5 i, giving x = (10 ± 4√5 i)/10 = 1 ± (2√5/5)i. Choice A is correct because it shows the fully simplified form with proper rationalization. Choice B is incorrect because it shows 1 ± (4√5/5)i, failing to simplify the fraction 4/10 to 2/5. To help students: Practice simplifying radicals and reducing fractions with irrational denominators. Watch for: Errors in simplifying √80 and reducing the final fraction.
6. Using the polynomial provided, consider the quadratic $r(x)=x^2+8x+20$, a degree $2$ polynomial. Compute $b^2-4ac=8^2-4(1)(20)=64-80=-16$, so $x=\frac{-8\pm\sqrt{-16}}{2}=\frac{-8\pm 4i}{2}=-4\pm 2i.$ These nonreal zeros do not correspond to real $x$-intercepts, yet they satisfy $r(x)=0$ in the complex system. What are the complex zeros of the polynomial function $r(x)$?
   Answer: $-4\pm 2i$ This question tests AP Precalculus skills: understanding polynomial functions and complex zeros. Polynomial functions can have complex zeros, which are solutions to the polynomial equation in the form a + bi, where a and b are real numbers. For the polynomial r(x) = x² + 8x + 20, the negative discriminant (64 - 80 = -16) indicates complex zeros: x = (-8 ± √(-16))/2 = (-8 ± 4i)/2 = -4 ± 2i. Choice B is correct because it shows the properly calculated complex zeros with real part -4 and imaginary parts ±2i. Choice E is incorrect because it treats the result as if it were entirely real (-4 ± 2), missing the imaginary unit i that must accompany the square root of a negative number. To help students: Practice the quadratic formula with negative discriminants and emphasize that √(-k) = √k · i. Watch for: Students dropping the imaginary unit i when simplifying square roots of negative numbers.
7. Based on the passage, a quadratic polynomial $f(x)=2x^2-4x+13$ has degree $2$, so it has two zeros counting multiplicity. Complex zeros satisfy $f(x)=0$ even when the graph on the real coordinate plane never crosses the $x$-axis. Using the quadratic formula, set $2x^2-4x+13=0$ so $x=\dfrac{4\pm\sqrt{(-4)^2-4(2)(13)}}{2\cdot 2}=\dfrac{4\pm\sqrt{16-104}}{4}=\dfrac{4\pm\sqrt{-88}}{4}$. Rewrite $\sqrt{-88}=\sqrt{88}\,i=2\sqrt{22}\,i$, giving $x=\dfrac{4\pm 2\sqrt{22}\,i}{4}=1\pm\dfrac{\sqrt{22}}{2}i$. On the complex plane, these zeros plot at $(1,\pm\tfrac{\sqrt{22}}{2})$ using real and imaginary axes. Calculate the complex zeros of the polynomial using the quadratic formula.
   Answer: $x=1\pm\dfrac{\sqrt{22}}{2}i$ This question tests AP Precalculus skills: understanding polynomial functions and complex zeros. When a quadratic polynomial has a negative discriminant, its zeros are complex conjugates of the form a ± bi. The polynomial f(x) = 2x² - 4x + 13 requires the quadratic formula: x = (4 ± √(16-104))/4 = (4 ± √(-88))/4. Since √(-88) = √(88)i = 2√(22)i, we get x = (4 ± 2√(22)i)/4 = 1 ± (√(22)/2)i. Choice A is correct because it shows the proper simplification with the imaginary part √(22)/2. Choice B is incorrect because it fails to divide the imaginary part by 2, showing 1 ± √(22)i instead. To help students: Practice simplifying complex expressions step-by-step, especially when dividing both real and imaginary parts by a common factor. Watch for: Students forgetting to divide both the real and imaginary parts when simplifying fractions containing complex numbers.
8. Using the polynomial provided, the quadratic $f(x)=x^2-6x+25$ has degree $2$, so it has two zeros. The discriminant is $b^2-4ac=(-6)^2-4(1)(25)=36-100=-64$, which is negative, so the zeros are nonreal. Using the quadratic formula, $x=\dfrac{6\pm\sqrt{-64}}{2}=\dfrac{6\pm 8i}{2}=3\pm 4i$. These zeros plot at $(3,\pm 4)$ on the complex plane and correspond to no real $x$-intercepts. What are the complex zeros of the polynomial function $f(x)=x^2-6x+25$?
   Answer: $x=3\pm 4i$ This question tests AP Precalculus skills: understanding polynomial functions and complex zeros. A negative discriminant in a quadratic indicates complex conjugate zeros, and the quadratic formula directly yields these zeros. For f(x) = x² - 6x + 25, the discriminant is 36 - 100 = -64 < 0. Using the quadratic formula: x = (6 ± √(-64))/2 = (6 ± 8i)/2 = 3 ± 4i. Choice A is correct because it shows the proper simplification with real part 3 and imaginary parts ±4i. Choice C is incorrect because it doubles the imaginary part to ±8i, failing to divide by 2. To help students: Practice recognizing when to expect complex zeros (negative discriminant) and carefully simplify both parts of the result. Visualize these zeros as points (3, ±4) on the complex plane.
9. Based on the passage, let $s(x)=5x^2-10x+29$, a degree $2$ polynomial. The discriminant is $(-10)^2-4(5)(29)=100-580=-480$, and $x=\frac{10\pm\sqrt{-480}}{10}=\frac{10\pm\sqrt{480}\,i}{10}=\frac{10\pm 4\sqrt{30}\,i}{10}=1\pm\frac{2\sqrt{30}}{5}i.$ Conjugate zeros reflect across the real axis in the complex plane for real coefficients. What are the complex zeros of the polynomial function $s(x)$?
   Answer: $1\pm\frac{2\sqrt{30}}{5}i$ This question tests AP Precalculus skills: understanding polynomial functions and complex zeros. Polynomial functions can have complex zeros, which are solutions to the polynomial equation in the form a + bi, where a and b are real numbers. For s(x) = 5x² - 10x + 29, the discriminant is -480, yielding: x = (10 ± √(-480))/10 = (10 ± 4√30 i)/10 = 1 ± (2√30/5)i. Choice A is correct because it shows the fully simplified form with real part 1 and imaginary parts ±(2√30/5)i. Choice C is incorrect because it shows ±(√30/5)i instead of ±(2√30/5)i, missing the factor of 2 that comes from simplifying √480 = 4√30. To help students: Practice simplifying radicals completely before dividing, and show that √480 = √(16·30) = 4√30. Watch for: Students incorrectly simplifying the radical or forgetting to divide all terms by the denominator.
10. Using the polynomial provided, a quadratic model $h(x)=x^2-6x+11$ has $\Delta=36-44=-8$; calculate the complex zeros using the quadratic formula, rewriting $\sqrt{-8}=2\sqrt{2}\,i$ and expressing answers as $a\pm bi$.​
    Answer: $x=3\pm\sqrt{2}\,i$ 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.
11. Using the polynomial provided, analyze the quadratic $g(x)=3x^2+6x+17$, a degree $2$ polynomial. Compute the discriminant: $b^2-4ac=6^2-4(3)(17)=36-204=-168$, then $x=\frac{-6\pm\sqrt{-168}}{6}=\frac{-6\pm\sqrt{168}\,i}{6}=\frac{-6\pm 2\sqrt{42}\,i}{6}=-1\pm\frac{\sqrt{42}}{3}i.$ Because coefficients are real, nonreal zeros must be conjugates, and the real graph has no $x$-intercepts. What are the complex zeros of the polynomial function $g(x)$?
    Answer: $-1\pm\frac{\sqrt{42}}{3}i$ This question tests AP Precalculus skills: understanding polynomial functions and complex zeros. Polynomial functions can have complex zeros, which are solutions to the polynomial equation in the form a + bi, where a and b are real numbers. In the polynomial g(x) = 3x² + 6x + 17, the discriminant is negative (36 - 204 = -168), indicating complex zeros: x = (-6 ± √(-168))/6 = (-6 ± 2√42 i)/6 = -1 ± (√42/3)i. Choice A is correct because it shows the properly simplified complex zeros with both the real part (-1) and the imaginary parts (±√42/3)i. Choice E is incorrect because it treats the square root of a negative number as if it were real, missing the imaginary unit i entirely. To help students: Emphasize that negative discriminants always yield complex zeros and practice simplifying radicals with negative arguments. Watch for: Students forgetting to include the imaginary unit i when taking square roots of negative numbers.
12. Based on the passage, let $h(x)=x^3-2x^2+5x-10$, a degree $3$ polynomial. Factor by grouping: $x^2(x-2)+5(x-2)=(x-2)(x^2+5)$. The real root is $x=2$, and the remaining quadratic gives $x^2+5=0\Rightarrow x=\pm\sqrt{5}\,i$. Real-coefficient polynomials force nonreal zeros to appear as conjugate pairs. Which statement about the zeros of $h(x)$ is true?
    Answer: The zeros are $2,\ \sqrt{5}i,\ -\sqrt{5}i$. This question tests AP Precalculus skills: understanding polynomial functions and complex zeros. Polynomial functions can have complex zeros, which are solutions to the polynomial equation in the form a + bi, where a and b are real numbers. The polynomial h(x) = x³ - 2x² + 5x - 10 can be factored by grouping to get (x - 2)(x² + 5), revealing one real zero at x = 2 and two complex zeros from x² + 5 = 0, which gives x = ±√5 i. Choice A is correct because it lists all three zeros: the real zero 2 and the conjugate pair of complex zeros √5 i and -√5 i. Choice D is incorrect because it treats x² + 5 = 0 as if it yields real solutions ±√5, failing to recognize that x² = -5 requires imaginary solutions. To help students: Practice factoring by grouping and emphasize that x² = -k (where k > 0) always yields imaginary solutions ±√k i. Watch for: Students confusing x² - 5 = 0 (real solutions) with x² + 5 = 0 (complex solutions).
13. Based on the passage, let $f(x)=x^4-2x^3+5x^2-8x+4$, a degree $4$ polynomial with four zeros counting multiplicity. Suppose $x=1$ is a zero. Synthetic division by $(x-1)$ gives coefficients $1,-2,5,-8,4$ and produces the quotient $x^3-x^2+4x-4$ with remainder $0$. Factor the cubic by grouping: $x^3-x^2+4x-4=x^2(x-1)+4(x-1)=(x-1)(x^2+4)$. Therefore $f(x)=(x-1)^2(x^2+4)$. Solving $x^2+4=0$ yields $x=\pm 2i$, which appear as conjugates. What are the complex zeros of the polynomial function $f(x)=x^4-2x^3+5x^2-8x+4$?
    Answer: $x=1,1,\ \pm 2i$ 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.
14. Using the polynomial provided, consider $f(x)=2x^2+8x+20$, degree $2$. Divide by $2$ to simplify: $x^2+4x+10=0$. Apply the quadratic formula: $x=\dfrac{-4\pm\sqrt{4^2-4(1)(10)}}{2}=\dfrac{-4\pm\sqrt{16-40}}{2}=\dfrac{-4\pm\sqrt{-24}}{2}$. Since $\sqrt{-24}=\sqrt{24}\,i=2\sqrt{6}\,i$, we get $x=\dfrac{-4\pm 2\sqrt{6}\,i}{2}=-2\pm\sqrt{6}\,i$. These conjugate zeros reflect across the real axis in the complex plane. Calculate the complex zeros of the polynomial using the quadratic formula.
    Answer: $x=-2\pm\sqrt{6}\,i$ This question tests AP Precalculus skills: understanding polynomial functions and complex zeros. When a quadratic has a common factor, dividing it out first simplifies the calculation while preserving the zeros. For f(x) = 2x² + 8x + 20, dividing by 2 gives x² + 4x + 10 = 0. The quadratic formula yields x = (-4 ± √(-24))/2, where √(-24) = 2√(6)i, so x = (-4 ± 2√(6)i)/2 = -2 ± √(6)i. Choice A is correct because it shows the proper form with real part -2 and imaginary parts ±√(6)i. Choice C is incorrect because it shows ±√(6)/2i, incorrectly dividing the imaginary part by an extra factor of 2. To help students: Practice simplifying before applying formulas and carefully track all division steps. Verify answers by substitution when possible.
15. Based on the passage, let $f(x)=x^3+2x^2+5x+10$, a degree $3$ polynomial. Factor by grouping: $x^3+2x^2+5x+10=x^2(x+2)+5(x+2)=(x+2)(x^2+5)$. The real zero is $x=-2$, and the remaining zeros satisfy $x^2=-5$, so $x=\pm\sqrt{5}\,i$. In the complex plane, $\pm\sqrt{5}i$ lie on the imaginary axis, symmetric about the real axis. What are the complex zeros of the polynomial function $f(x)=x^3+2x^2+5x+10$?
    Answer: $x=-2,\ \pm\sqrt{5}\,i$ 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.
16. Using the polynomial provided, consider $p(x)=x^3+3x^2+4x+12$, a degree $3$ polynomial. Factor by grouping: $x^2(x+3)+4(x+3)=(x+3)(x^2+4)$. The real root is $x=-3$, and $x^2+4=0$ yields $x=\pm 2i$. Complex conjugates correspond to symmetric points in the complex plane, while only real zeros appear as $x$-intercepts. What are the complex zeros of the polynomial function $p(x)$?
    Answer: $-3,\ 2i,\ -2i$ This question tests AP Precalculus skills: understanding polynomial functions and complex zeros. Polynomial functions can have complex zeros, which are solutions to the polynomial equation in the form a + bi, where a and b are real numbers. The polynomial p(x) = x³ + 3x² + 4x + 12 factors by grouping as (x + 3)(x² + 4), giving one real zero at x = -3 and two complex zeros from x² + 4 = 0, yielding x = ±2i. Choice B is correct because it identifies all three zeros: the real zero -3 and the conjugate pair 2i and -2i. Choice D is incorrect because it lists 2i twice instead of recognizing that complex zeros of polynomials with real coefficients must occur in conjugate pairs. To help students: Emphasize the fundamental theorem that polynomials with real coefficients have complex zeros in conjugate pairs. Watch for: Students listing the same complex zero twice instead of including both members of the conjugate pair.
17. Using the polynomial provided, let $f(x)=3x^2+6x+10$, a degree $2$ polynomial. Apply the quadratic formula: $x=\dfrac{-6\pm\sqrt{6^2-4(3)(10)}}{2\cdot 3}=\dfrac{-6\pm\sqrt{36-120}}{6}=\dfrac{-6\pm\sqrt{-84}}{6}$. Since $\sqrt{-84}=\sqrt{84}\,i=2\sqrt{21}\,i$, we get $x=\dfrac{-6\pm 2\sqrt{21}\,i}{6}=-1\pm\dfrac{\sqrt{21}}{3}i$. These zeros plot at $(-1,\pm\tfrac{\sqrt{21}}{3})$ on the complex plane. Calculate the complex zeros of the polynomial using the quadratic formula.
    Answer: $x=-1\pm\dfrac{\sqrt{21}}{3}i$ This question tests AP Precalculus skills: understanding polynomial functions and complex zeros. The quadratic formula applied to polynomials with negative discriminants yields complex conjugate zeros. For f(x) = 3x² + 6x + 10, the discriminant is 36 - 120 = -84, indicating complex zeros. Applying the formula: x = (-6 ± √(-84))/6 = (-6 ± 2√(21)i)/6 = -1 ± (√(21)/3)i. Choice A is correct because it properly simplifies both the real part (-1) and imaginary part (±√(21)/3). Choice C is incorrect because it fails to divide the imaginary part by 3, showing ±√(21)i instead. To help students: Practice careful arithmetic when simplifying complex fractions. Emphasize checking work by verifying that both parts are divided by the same denominator.
18. Using the polynomial provided, consider $f(x)=x^2+4x+13$, a degree $2$ polynomial. Completing the square gives $x^2+4x+13=(x+2)^2+9$, so $f(x)=0$ implies $(x+2)^2=-9$. Taking square roots, $x+2=\pm 3i$, hence $x=-2\pm 3i$. These nonreal zeros do not create $x$-intercepts on the real coordinate plane, yet they still satisfy the equation in the complex number system. On the complex plane, they plot at $(-2,\pm 3)$ using real and imaginary axes. What are the complex zeros of the polynomial function $f(x)=x^2+4x+13$?
    Answer: $x=-2\pm 3i$ 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.
19. Based on the passage, consider the quartic $f(x)=x^4+2x^3+5x^2+10x+13$, degree $4$. Suppose the polynomial factors as $(x^2+ax+b)(x^2+cx+d)$ with real coefficients, so nonreal zeros occur in conjugate pairs. Matching constant terms requires $bd=13$, and matching the $x^3$ coefficient requires $a+c=2$. The passage provides the completed-square style factorization $f(x)=(x^2+2x+5)(x^2+1)$. Solve $x^2+1=0$ to get $x=\pm i$, and solve $x^2+2x+5=0$ to get $x=\dfrac{-2\pm\sqrt{4-20}}{2}=-1\pm 2i$. What are the complex zeros of the polynomial function $f(x)=x^4+2x^3+5x^2+10x+13$?
    Answer: $x=\pm i,\ -1\pm 2i$ 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.
20. Based on the passage, analyze the degree $3$ polynomial $q(x)=x^3-x^2+x-1$. Factor by grouping: $x^2(x-1)+1(x-1)=(x-1)(x^2+1)$. Thus $x=1$ is real, and $x^2+1=0$ gives $x=\pm i$. For real coefficients, the complex zeros must be conjugates, so both $i$ and $-i$ occur. Which statement about the zeros of the polynomial is true?
    Answer: The zeros are $1,\ i,\ -i$. This question tests AP Precalculus skills: understanding polynomial functions and complex zeros. Polynomial functions can have complex zeros, which are solutions to the polynomial equation in the form a + bi, where a and b are real numbers. The polynomial q(x) = x³ - x² + x - 1 factors by grouping as (x - 1)(x² + 1), revealing one real zero at x = 1 and two complex zeros from x² + 1 = 0, which gives x = ±i. Choice B is correct because it correctly identifies all three zeros: the real zero 1 and the conjugate pair i and -i. Choice A is incorrect because it assumes all zeros must be real for odd-degree polynomials, but odd-degree polynomials only guarantee at least one real zero, not that all zeros are real. To help students: Clarify that odd-degree polynomials have at least one real zero but can have complex zeros in conjugate pairs. Watch for: The misconception that odd-degree polynomials can only have real zeros.

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