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AP Precalculus Quiz

AP Precalculus Quiz: Polynomial Functions And Complex Zeros

Practice Polynomial Functions And Complex Zeros in AP Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

Based on the passage, the degree-2 polynomial v(x)=8x2+16x+13v(x)=8x^2+16x+13v(x)=8x2+16x+13 has Δ=256−416=−160\Delta=256-416=-160Δ=256−416=−160; calculate the complex zeros using the quadratic formula and simplify −160=410 i\sqrt{-160}=4\sqrt{10}\,i−160​=410​i.​

Select an answer to continue

What this quiz covers

This quiz focuses on Polynomial Functions And Complex Zeros, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Precalculus.

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.

All questions

Question 1

Based on the passage, the degree-2 polynomial v(x)=8x2+16x+13v(x)=8x^2+16x+13v(x)=8x2+16x+13 has Δ=256−416=−160\Delta=256-416=-160Δ=256−416=−160; calculate the complex zeros using the quadratic formula and simplify −160=410 i\sqrt{-160}=4\sqrt{10}\,i−160​=410​i.​

  1. x=−1±104ix=-1\pm\dfrac{\sqrt{10}}{4}ix=−1±410​​i (correct answer)
  2. x=−1±102ix=-1\pm\dfrac{\sqrt{10}}{2}ix=−1±210​​i
  3. x=1±104ix=1\pm\dfrac{\sqrt{10}}{4}ix=1±410​​i
  4. x=−2±104ix=-2\pm\dfrac{\sqrt{10}}{4}ix=−2±410​​i
  5. x=−1+104ix=-1+\dfrac{\sqrt{10}}{4}ix=−1+410​​i only

Explanation: 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.

Question 2

Using the polynomial provided, a quadratic u(x)=3x2−18x+31u(x)=3x^2-18x+31u(x)=3x2−18x+31 has Δ=324−372=−48\Delta=324-372=-48Δ=324−372=−48; calculate the complex zeros with the quadratic formula, rewriting −48=43 i\sqrt{-48}=4\sqrt{3}\,i−48​=43​i and simplifying.​

  1. x=3±233ix=3\pm\dfrac{2\sqrt{3}}{3}ix=3±323​​i (correct answer)
  2. x=3±433ix=3\pm\dfrac{4\sqrt{3}}{3}ix=3±343​​i
  3. x=−3±233ix=-3\pm\dfrac{2\sqrt{3}}{3}ix=−3±323​​i
  4. x=18±436x=\dfrac{18\pm4\sqrt{3}}{6}x=618±43​​
  5. x=3+233ix=3+\dfrac{2\sqrt{3}}{3}ix=3+323​​i only

Explanation: 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.

Question 3

Based on the passage, let f(x)=x3+2x2+5x+10f(x)=x^3+2x^2+5x+10f(x)=x3+2x2+5x+10, a degree 333 polynomial. Factor by grouping: x3+2x2+5x+10=x2(x+2)+5(x+2)=(x+2)(x2+5)x^3+2x^2+5x+10=x^2(x+2)+5(x+2)=(x+2)(x^2+5)x3+2x2+5x+10=x2(x+2)+5(x+2)=(x+2)(x2+5). The real zero is x=−2x=-2x=−2, and the remaining zeros satisfy x2=−5x^2=-5x2=−5, so x=±5 ix=\pm\sqrt{5}\,ix=±5​i. In the complex plane, ±5i\pm\sqrt{5}i±5​i lie on the imaginary axis, symmetric about the real axis. What are the complex zeros of the polynomial function f(x)=x3+2x2+5x+10f(x)=x^3+2x^2+5x+10f(x)=x3+2x2+5x+10?

  1. x=−2, ±5 ix=-2,\ \pm\sqrt{5}\,ix=−2, ±5​i (correct answer)
  2. x=2, ±5 ix=2,\ \pm\sqrt{5}\,ix=2, ±5​i
  3. x=−2, ±5ix=-2,\ \pm 5ix=−2, ±5i
  4. x=−2, 5 ix=-2,\ \sqrt{5}\,ix=−2, 5​i
  5. x=−2, ±5x=-2,\ \pm\sqrt{5}x=−2, ±5​

Explanation: 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.

Question 4

Based on the passage, the degree-2 polynomial p(x)=4x2+4x+5p(x)=4x^2+4x+5p(x)=4x2+4x+5 has Δ=16−80=−64\Delta=16-80=-64Δ=16−80=−64; calculate the complex zeros using the quadratic formula and simplify −64=8i\sqrt{-64}=8i−64​=8i to obtain exact a+bia+bia+bi values.​

  1. x=−12±ix=-\dfrac{1}{2}\pm ix=−21​±i (correct answer)
  2. x=−12±2ix=-\dfrac{1}{2}\pm2ix=−21​±2i
  3. x=12±ix=\dfrac{1}{2}\pm ix=21​±i
  4. x=−1±ix=-1\pm ix=−1±i
  5. x=−12+ix=-\dfrac{1}{2}+ix=−21​+i only

Explanation: 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.

Question 5

Using the polynomial provided, a quadratic model h(x)=x2−6x+11h(x)=x^2-6x+11h(x)=x2−6x+11 has Δ=36−44=−8\Delta=36-44=-8Δ=36−44=−8; calculate the complex zeros using the quadratic formula, rewriting −8=22 i\sqrt{-8}=2\sqrt{2}\,i−8​=22​i and expressing answers as a±bia\pm bia±bi.​

  1. x=3±2 ix=3\pm\sqrt{2}\,ix=3±2​i (correct answer)
  2. x=3±22 ix=3\pm2\sqrt{2}\,ix=3±22​i
  3. x=−3±2 ix=-3\pm\sqrt{2}\,ix=−3±2​i
  4. x=6±22x=\dfrac{6\pm\sqrt{2}}{2}x=26±2​​
  5. x=3+2 ix=3+\sqrt{2}\,ix=3+2​i only

Explanation: 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.

Question 6

Using the polynomial provided, consider f(x)=x2+4x+13f(x)=x^2+4x+13f(x)=x2+4x+13, a degree 222 polynomial. Completing the square gives x2+4x+13=(x+2)2+9x^2+4x+13=(x+2)^2+9x2+4x+13=(x+2)2+9, so f(x)=0f(x)=0f(x)=0 implies (x+2)2=−9(x+2)^2=-9(x+2)2=−9. Taking square roots, x+2=±3ix+2=\pm 3ix+2=±3i, hence x=−2±3ix=-2\pm 3ix=−2±3i. These nonreal zeros do not create xxx-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,±3)(-2,\pm 3)(−2,±3) using real and imaginary axes. What are the complex zeros of the polynomial function f(x)=x2+4x+13f(x)=x^2+4x+13f(x)=x2+4x+13?

  1. x=2±3ix=2\pm 3ix=2±3i
  2. x=−2±3ix=-2\pm 3ix=−2±3i (correct answer)
  3. x=−2±3x=-2\pm 3x=−2±3
  4. x=−2±9ix=-2\pm 9ix=−2±9i
  5. x=−2+3ix=-2+3ix=−2+3i

Explanation: 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.

Question 7

Based on the passage, the degree-2 polynomial t(x)=6x2−12x+11t(x)=6x^2-12x+11t(x)=6x2−12x+11 has Δ=144−264=−120\Delta=144-264=-120Δ=144−264=−120; calculate the complex zeros using the quadratic formula and simplify −120=230 i\sqrt{-120}=2\sqrt{30}\,i−120​=230​i.​

  1. x=1±306ix=1\pm\dfrac{\sqrt{30}}{6}ix=1±630​​i (correct answer)
  2. x=1±303ix=1\pm\dfrac{\sqrt{30}}{3}ix=1±330​​i
  3. x=−1±306ix=-1\pm\dfrac{\sqrt{30}}{6}ix=−1±630​​i
  4. x=12±2306x=\dfrac{12\pm2\sqrt{30}}{6}x=612±230​​
  5. x=1+306ix=1+\dfrac{\sqrt{30}}{6}ix=1+630​​i only

Explanation: 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.

Question 8

Based on the passage, let f(x)=x4−2x3+5x2−8x+4f(x)=x^4-2x^3+5x^2-8x+4f(x)=x4−2x3+5x2−8x+4, a degree 444 polynomial with four zeros counting multiplicity. Suppose x=1x=1x=1 is a zero. Synthetic division by (x−1)(x-1)(x−1) gives coefficients 1,−2,5,−8,41,-2,5,-8,41,−2,5,−8,4 and produces the quotient x3−x2+4x−4x^3-x^2+4x-4x3−x2+4x−4 with remainder 000. Factor the cubic by grouping: x3−x2+4x−4=x2(x−1)+4(x−1)=(x−1)(x2+4)x^3-x^2+4x-4=x^2(x-1)+4(x-1)=(x-1)(x^2+4)x3−x2+4x−4=x2(x−1)+4(x−1)=(x−1)(x2+4). Therefore f(x)=(x−1)2(x2+4)f(x)=(x-1)^2(x^2+4)f(x)=(x−1)2(x2+4). Solving x2+4=0x^2+4=0x2+4=0 yields x=±2ix=\pm 2ix=±2i, which appear as conjugates. What are the complex zeros of the polynomial function f(x)=x4−2x3+5x2−8x+4f(x)=x^4-2x^3+5x^2-8x+4f(x)=x4−2x3+5x2−8x+4?

  1. x=1,1, ±2ix=1,1,\ \pm 2ix=1,1, ±2i (correct answer)
  2. x=1, ±2ix=1,\ \pm 2ix=1, ±2i
  3. x=−1,−1, ±2ix=-1,-1,\ \pm 2ix=−1,−1, ±2i
  4. x=1,1, 2ix=1,1,\ 2ix=1,1, 2i
  5. x=1,1, ±4ix=1,1,\ \pm 4ix=1,1, ±4i

Explanation: 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.

Question 9

Based on the passage, consider the quartic f(x)=x4+2x3+5x2+10x+13f(x)=x^4+2x^3+5x^2+10x+13f(x)=x4+2x3+5x2+10x+13, degree 444. Suppose the polynomial factors as (x2+ax+b)(x2+cx+d)(x^2+ax+b)(x^2+cx+d)(x2+ax+b)(x2+cx+d) with real coefficients, so nonreal zeros occur in conjugate pairs. Matching constant terms requires bd=13bd=13bd=13, and matching the x3x^3x3 coefficient requires a+c=2a+c=2a+c=2. The passage provides the completed-square style factorization f(x)=(x2+2x+5)(x2+1)f(x)=(x^2+2x+5)(x^2+1)f(x)=(x2+2x+5)(x2+1). Solve x2+1=0x^2+1=0x2+1=0 to get x=±ix=\pm ix=±i, and solve x2+2x+5=0x^2+2x+5=0x2+2x+5=0 to get x=−2±4−202=−1±2ix=\dfrac{-2\pm\sqrt{4-20}}{2}=-1\pm 2ix=2−2±4−20​​=−1±2i. What are the complex zeros of the polynomial function f(x)=x4+2x3+5x2+10x+13f(x)=x^4+2x^3+5x^2+10x+13f(x)=x4+2x3+5x2+10x+13?

  1. x=±i, −1±2ix=\pm i,\ -1\pm 2ix=±i, −1±2i (correct answer)
  2. x=±1, −1±2ix=\pm 1,\ -1\pm 2ix=±1, −1±2i
  3. x=i, −1±2ix=i,\ -1\pm 2ix=i, −1±2i
  4. x=±i, 1±2ix=\pm i,\ 1\pm 2ix=±i, 1±2i
  5. x=±2i, −1±ix=\pm 2i,\ -1\pm ix=±2i, −1±i

Explanation: 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.

Question 10

Based on the passage, a degree-2 polynomial g(x)=3x2+12x+17g(x)=3x^2+12x+17g(x)=3x2+12x+17 has Δ=144−204=−60\Delta=144-204=-60Δ=144−204=−60; calculate the complex zeros of g(x)g(x)g(x) using the quadratic formula and simplify −60=215 i\sqrt{-60}=2\sqrt{15}\,i−60​=215​i.​

  1. x=−2±153ix=-2\pm\dfrac{\sqrt{15}}{3}ix=−2±315​​i (correct answer)
  2. x=−2±156ix=-2\pm\dfrac{\sqrt{15}}{6}ix=−2±615​​i
  3. x=2±153ix=2\pm\dfrac{\sqrt{15}}{3}ix=2±315​​i
  4. x=−2±15 ix=-2\pm\sqrt{15}\,ix=−2±15​i
  5. x=−2+153ix=-2+\dfrac{\sqrt{15}}{3}ix=−2+315​​i only

Explanation: 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.

Question 11

Using the polynomial provided, a degree-2 function q(x)=5x2−10x+9q(x)=5x^2-10x+9q(x)=5x2−10x+9 yields Δ=100−180=−80\Delta=100-180=-80Δ=100−180=−80; calculate the complex zeros with the quadratic formula, converting −80=45 i\sqrt{-80}=4\sqrt{5}\,i−80​=45​i and simplifying fully.​

  1. x=1±255ix=1\pm\dfrac{2\sqrt{5}}{5}ix=1±525​​i (correct answer)
  2. x=1±455ix=1\pm\dfrac{4\sqrt{5}}{5}ix=1±545​​i
  3. x=−1±255ix=-1\pm\dfrac{2\sqrt{5}}{5}ix=−1±525​​i
  4. x=10±4510x=\dfrac{10\pm4\sqrt{5}}{10}x=1010±45​​
  5. x=1+255ix=1+\dfrac{2\sqrt{5}}{5}ix=1+525​​i only

Explanation: 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.

Question 12

Using the polynomial provided, consider the quadratic v(x)=4x2−4x+7v(x)=4x^2-4x+7v(x)=4x2−4x+7, a degree 222 polynomial. Compute b2−4ac=(−4)2−4(4)(7)=16−112=−96b^2-4ac=(-4)^2-4(4)(7)=16-112=-96b2−4ac=(−4)2−4(4)(7)=16−112=−96, so x=4±−968=4±96 i8=4±46 i8=12±62i.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.x=84±−96​​=84±96​i​=84±46​i​=21​±26​​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)v(x)v(x)?

  1. 12±62i\frac{1}{2}\pm\frac{\sqrt{6}}{2}i21​±26​​i (correct answer)
  2. −12±62i-\frac{1}{2}\pm\frac{\sqrt{6}}{2}i−21​±26​​i
  3. 12±6 i\frac{1}{2}\pm\sqrt{6}\,i21​±6​i
  4. 12+62i\frac{1}{2}+\frac{\sqrt{6}}{2}i21​+26​​i only
  5. 12±62\frac{1}{2}\pm\frac{\sqrt{6}}{2}21​±26​​

Explanation: 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.

Question 13

Using the polynomial provided, let f(x)=3x2+6x+10f(x)=3x^2+6x+10f(x)=3x2+6x+10, a degree 222 polynomial. Apply the quadratic formula: x=−6±62−4(3)(10)2⋅3=−6±36−1206=−6±−846x=\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}x=2⋅3−6±62−4(3)(10)​​=6−6±36−120​​=6−6±−84​​. Since −84=84 i=221 i\sqrt{-84}=\sqrt{84}\,i=2\sqrt{21}\,i−84​=84​i=221​i, we get x=−6±221 i6=−1±213ix=\dfrac{-6\pm 2\sqrt{21}\,i}{6}=-1\pm\dfrac{\sqrt{21}}{3}ix=6−6±221​i​=−1±321​​i. These zeros plot at (−1,±213)(-1,\pm\tfrac{\sqrt{21}}{3})(−1,±321​​) on the complex plane. Calculate the complex zeros of the polynomial using the quadratic formula.

  1. x=−1±213ix=-1\pm\dfrac{\sqrt{21}}{3}ix=−1±321​​i (correct answer)
  2. x=1±213ix=1\pm\dfrac{\sqrt{21}}{3}ix=1±321​​i
  3. x=−1±21 ix=-1\pm\sqrt{21}\,ix=−1±21​i
  4. x=−1±213x=-1\pm\dfrac{\sqrt{21}}{3}x=−1±321​​
  5. x=−1±213ix=\dfrac{-1\pm\sqrt{21}}{3}ix=3−1±21​​i

Explanation: 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.

Question 14

Using the polynomial provided, consider f(x)=2x2+8x+20f(x)=2x^2+8x+20f(x)=2x2+8x+20, degree 222. Divide by 222 to simplify: x2+4x+10=0x^2+4x+10=0x2+4x+10=0. Apply the quadratic formula: x=−4±42−4(1)(10)2=−4±16−402=−4±−242x=\dfrac{-4\pm\sqrt{4^2-4(1)(10)}}{2}=\dfrac{-4\pm\sqrt{16-40}}{2}=\dfrac{-4\pm\sqrt{-24}}{2}x=2−4±42−4(1)(10)​​=2−4±16−40​​=2−4±−24​​. Since −24=24 i=26 i\sqrt{-24}=\sqrt{24}\,i=2\sqrt{6}\,i−24​=24​i=26​i, we get x=−4±26 i2=−2±6 ix=\dfrac{-4\pm 2\sqrt{6}\,i}{2}=-2\pm\sqrt{6}\,ix=2−4±26​i​=−2±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.

  1. x=−2±6 ix=-2\pm\sqrt{6}\,ix=−2±6​i (correct answer)
  2. x=2±6 ix=2\pm\sqrt{6}\,ix=2±6​i
  3. x=−2±62ix=-2\pm\dfrac{\sqrt{6}}{2}ix=−2±26​​i
  4. x=−2±6x=-2\pm\sqrt{6}x=−2±6​
  5. x=−2+6 ix=-2+\sqrt{6}\,ix=−2+6​i

Explanation: 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.

Question 15

Using the polynomial provided, analyze the quadratic g(x)=3x2+6x+17g(x)=3x^2+6x+17g(x)=3x2+6x+17, a degree 222 polynomial. Compute the discriminant: b2−4ac=62−4(3)(17)=36−204=−168b^2-4ac=6^2-4(3)(17)=36-204=-168b2−4ac=62−4(3)(17)=36−204=−168, then x=−6±−1686=−6±168 i6=−6±242 i6=−1±423i.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.x=6−6±−168​​=6−6±168​i​=6−6±242​i​=−1±342​​i. Because coefficients are real, nonreal zeros must be conjugates, and the real graph has no xxx-intercepts. What are the complex zeros of the polynomial function g(x)g(x)g(x)?

  1. −1±423i-1\pm\frac{\sqrt{42}}{3}i−1±342​​i (correct answer)
  2. 1±423i1\pm\frac{\sqrt{42}}{3}i1±342​​i
  3. −1±42 i-1\pm\sqrt{42}\,i−1±42​i
  4. −1+423i-1+\frac{\sqrt{42}}{3}i−1+342​​i only
  5. −1±423-1\pm\frac{\sqrt{42}}{3}−1±342​​

Explanation: 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.

Question 16

Using the polynomial provided, consider the quadratic f(x)=2x2−4x+13f(x)=2x^2-4x+13f(x)=2x2−4x+13, a degree 222 polynomial with 222 complex solutions. Apply the quadratic formula: x=−(−4)±(−4)2−4(2)(13)2⋅2=4±16−1044=4±−884=4±222 i4=1±222i.x=\frac{-(-4)\pm\sqrt{(-4)^2-4(2)(13)}}{2\cdot 2}=\frac{4\pm\sqrt{16-104}}{4}=\frac{4\pm\sqrt{-88}}{4}=\frac{4\pm 2\sqrt{22}\,i}{4}=1\pm\frac{\sqrt{22}}{2}i.x=2⋅2−(−4)±(−4)2−4(2)(13)​​=44±16−104​​=44±−88​​=44±222​i​=1±222​​i. Complex zeros occur in conjugate pairs for real-coefficient polynomials, and they do not create xxx-intercepts on the real coordinate plane. Calculate the complex zeros of the polynomial using the quadratic formula.

  1. 1±22 i1\pm\sqrt{22}\,i1±22​i
  2. −1±222i-1\pm\frac{\sqrt{22}}{2}i−1±222​​i
  3. 1±222i1\pm\frac{\sqrt{22}}{2}i1±222​​i (correct answer)
  4. 1+222i1+\frac{\sqrt{22}}{2}i1+222​​i only
  5. 1±8841\pm\frac{\sqrt{88}}{4}1±488​​

Explanation: 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 f(x) = 2x² - 4x + 13, we apply the quadratic formula to find zeros: x = (4 ± √(-88))/4 = (4 ± √(88)i)/4 = (4 ± 2√22 i)/4 = 1 ± (√22/2)i. Choice C is correct because it accurately reflects the complex zeros calculated using the quadratic formula, showing both conjugate pairs 1 + (√22/2)i and 1 - (√22/2)i. Choice A is incorrect because it omits the division by 2 in the imaginary part, a common error when students forget to simplify the entire fraction. To help students: Practice simplifying complex fractions step-by-step and emphasize that both numerator terms must be divided by the denominator. Watch for: Students forgetting to divide both the real and imaginary parts by the denominator when simplifying the quadratic formula result.

Question 17

Using the polynomial provided, consider the quadratic r(x)=x2+8x+20r(x)=x^2+8x+20r(x)=x2+8x+20, a degree 222 polynomial. Compute b2−4ac=82−4(1)(20)=64−80=−16b^2-4ac=8^2-4(1)(20)=64-80=-16b2−4ac=82−4(1)(20)=64−80=−16, so x=−8±−162=−8±4i2=−4±2i.x=\frac{-8\pm\sqrt{-16}}{2}=\frac{-8\pm 4i}{2}=-4\pm 2i.x=2−8±−16​​=2−8±4i​=−4±2i. These nonreal zeros do not correspond to real xxx-intercepts, yet they satisfy r(x)=0r(x)=0r(x)=0 in the complex system. What are the complex zeros of the polynomial function r(x)r(x)r(x)?

  1. 4±2i4\pm 2i4±2i
  2. −4±2i-4\pm 2i−4±2i (correct answer)
  3. −4±4i-4\pm 4i−4±4i
  4. −4+2i-4+2i−4+2i only
  5. −4±2-4\pm 2−4±2

Explanation: 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.

Question 18

Using the polynomial provided, take t(x)=2x2+12x+25t(x)=2x^2+12x+25t(x)=2x2+12x+25, a degree 222 polynomial. Compute b2−4ac=122−4(2)(25)=144−200=−56b^2-4ac=12^2-4(2)(25)=144-200=-56b2−4ac=122−4(2)(25)=144−200=−56, then x=−12±−564=−12±56 i4=−12±214 i4=−3±142i.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.x=4−12±−56​​=4−12±56​i​=4−12±214​i​=−3±214​​i. These complex zeros satisfy t(x)=0t(x)=0t(x)=0 but create no real intercepts on the coordinate plane. What are the complex zeros of the polynomial function t(x)t(x)t(x)?

  1. −3±142i-3\pm\frac{\sqrt{14}}{2}i−3±214​​i (correct answer)
  2. 3±142i3\pm\frac{\sqrt{14}}{2}i3±214​​i
  3. −3±14 i-3\pm\sqrt{14}\,i−3±14​i
  4. −3, 142i-3,\ \frac{\sqrt{14}}{2}i−3, 214​​i
  5. −3±142-3\pm\frac{\sqrt{14}}{2}−3±214​​

Explanation: 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.

Question 19

Using the polynomial provided, the quadratic f(x)=x2−6x+25f(x)=x^2-6x+25f(x)=x2−6x+25 has degree 222, so it has two zeros. The discriminant is b2−4ac=(−6)2−4(1)(25)=36−100=−64b^2-4ac=(-6)^2-4(1)(25)=36-100=-64b2−4ac=(−6)2−4(1)(25)=36−100=−64, which is negative, so the zeros are nonreal. Using the quadratic formula, x=6±−642=6±8i2=3±4ix=\dfrac{6\pm\sqrt{-64}}{2}=\dfrac{6\pm 8i}{2}=3\pm 4ix=26±−64​​=26±8i​=3±4i. These zeros plot at (3,±4)(3,\pm 4)(3,±4) on the complex plane and correspond to no real xxx-intercepts. What are the complex zeros of the polynomial function f(x)=x2−6x+25f(x)=x^2-6x+25f(x)=x2−6x+25?

  1. x=3±4ix=3\pm 4ix=3±4i (correct answer)
  2. x=−3±4ix=-3\pm 4ix=−3±4i
  3. x=3±8ix=3\pm 8ix=3±8i
  4. x=3±4x=3\pm 4x=3±4
  5. x=±3±4ix=\pm 3\pm 4ix=±3±4i

Explanation: 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.

Question 20

Using the polynomial provided, consider the cubic f(x)=x3−x2+5x−5f(x)=x^3-x^2+5x-5f(x)=x3−x2+5x−5. A polynomial of degree 333 has three zeros counting multiplicity. Factor by grouping: x3−x2+5x−5=x2(x−1)+5(x−1)=(x−1)(x2+5)x^3-x^2+5x-5=x^2(x-1)+5(x-1)=(x-1)(x^2+5)x3−x2+5x−5=x2(x−1)+5(x−1)=(x−1)(x2+5). The real zero is x=1x=1x=1, and the remaining factor gives x2=−5x^2=-5x2=−5. Writing −5=5⋅(−1)-5=5\cdot(-1)−5=5⋅(−1) and using −1=i\sqrt{-1}=i−1​=i, we obtain x=±5 ix=\pm\sqrt{5}\,ix=±5​i. Because the coefficients are real, any nonreal zeros must occur as conjugate pairs, which matches 5i\sqrt{5}i5​i and −5i-\sqrt{5}i−5​i. What are the complex zeros of the polynomial function f(x)=x3−x2+5x−5f(x)=x^3-x^2+5x-5f(x)=x3−x2+5x−5?

  1. x=1, ±5 ix=1,\ \pm\sqrt{5}\,ix=1, ±5​i (correct answer)
  2. x=1, ±5ix=1,\ \pm 5ix=1, ±5i
  3. x=1, 5 ix=1,\ \sqrt{5}\,ix=1, 5​i
  4. x=−1, ±5 ix=-1,\ \pm\sqrt{5}\,ix=−1, ±5​i
  5. x=1, ±5x=1,\ \pm\sqrt{5}x=1, ±5​

Explanation: 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.