Rational Root Theorem Worksheet Answer Key – Precalculus

Problem 1 List Possible Rational Zeros

The constant term is −10, with factors p = ±1, ±2, ±5, ±10. The leading coefficient is 3, with factors q = ±1, ±3.

±1, ±2, ±5, ±10, ±1/3, ±2/3, ±5/3, ±10/3.

Problem 2 End Behavior from the Leading Term

(a) Odd degree, negative coefficient: rises on the left, falls on the right.

(b) Even degree, positive coefficient: rises on both ends.

(c) Even degree, negative coefficient: falls on both ends.

Problem 3 Conjugate Zeros

Non-real zeros come in conjugate pairs, so the conjugates must also be zeros: 2 + 5i and −1 − i. That gives four zeros, matching the degree.

Problem 4 Classify the Asymptote Type

(a) The numerator’s degree is 3 and the denominator’s is 2, so there is a slant asymptote. Dividing gives x + 4x + 1x² − 4, so y = x.

(b) Equal degrees: horizontal asymptote y = 3.

(c) Denominator degree is larger: horizontal asymptote y = 0.

Problem 5 Read a Sign Chart

Positive regions are between −5 and 1, and to the right of 4. The inequality is strict, so the critical numbers are excluded.

(−5, 1) ∪ (4, ∞).

Problem 6 Count the Complex Zeros

Degree = 3 + 2 + 1 = 6.

Zeros: 4 (multiplicity 3), 3i and −3i (each multiplicity 1), −1 (multiplicity 1).

Total 3 + 1 + 1 + 1 = 6 zeros, equal to the degree.

Problem 7 Find All Zeros with Candidates

Candidates: ±1, ±2, ±4, ±1/2. Try 1: 2 + 7 − 5 − 4 = 0, so 1 is a zero.

Synthetic division by (x − 1) gives 2x² + 9x + 4 = (2x + 1)(x + 4).

Zeros: 1, −1/2, −4.

Problem 8 Real and Complex Zeros Together

Candidates are the divisors of 30. Since P(−3) = −27 + 9 − 12 + 30 = 0, divide by (x + 3) to get x² − 2x + 10.

Quadratic formula: x = 2 ± √(4 − 40)2 = 1 ± 3i.

Zeros: −3, 1 + 3i, 1 − 3i.

Problem 9 Build a Polynomial from Zeros

The conjugate 3 + 2i is also a zero. Its quadratic factor is (x − 3 + 2i)(x − 3 − 2i) = (x − 3)² + 4 = x² − 6x + 13.

Multiply by (x − 5): x³ − 11x² + 43x − 65.

Problem 10 Slant Asymptote by Division

Divide: 3x² − x + 4 = (3x + 5)(x − 2) + 14.

So f(x) = 3x + 5 + 14x − 2. Slant asymptote y = 3x + 5; vertical asymptote x = 2.

Problem 11 Identify Asymptotes on a Graph

The vertical dashed line is at x = 3. The slant line passes through (−4, −5) and (10, 9), so its slope is 14/14 = 1 and its equation is y = x − 1.

Division: x² − 4x + 1 = (x − 1)(x − 3) − 2, so the quotient is x − 1, which matches.

Problem 12 Quadratic Inequality

Factor: (2x + 3)(x − 4) ≤ 0. Critical numbers: −3/2 and 4. The parabola opens upward, so it is at most 0 between them.

[−3/2, 4].

Problem 13 Rational Inequality with a Constant

Subtract 2: x + 1 − 2(x − 5)x − 5 = 11 − xx − 5 ≤ 0.

Critical numbers 5 (excluded) and 11. Signs: − for x < 5, + on (5, 11), − for x > 11. Solution: (−∞, 5) ∪ [11, ∞).

Problem 14 Zeros and Multiplicity from a Graph

Zeros −2 and 3 are crossings (multiplicity 1); 1 is a touch (multiplicity 2). The multiplicities add to 4, the degree.

Leading coefficient negative and even degree: falls on both ends.

Equation: P(x) = −(x + 2)(x − 1)²(x − 3).

Problem 15 Solve a Quartic over the Complex Numbers

Factor like a quadratic in x²: (x² − 4)(x² + 3) = 0.

x² = 4 gives x = ±2; x² = −3 gives x = ±i√3.

x = 2, −2, i√3, −i√3.

Problem 16 Dimensions of a Crate

x(x + 2)(x − 1) = 72 gives x³ + x² − 2x − 72 = 0.

Testing the divisors of 72, x = 4 works: 64 + 16 − 8 − 72 = 0. The quotient is x² + 5x + 18.

Discriminant: 25 − 72 = −47 < 0, so the other zeros are non-real. The crate is 4 ft by 6 ft by 3 ft.

Problem 17 Average Cost and Its Asymptote

Divide each term by n: A(n) = 2n + 30 + 450n, so the slant asymptote is y = 2n + 30.

Solve 2n² + 30n + 450 ≤ 110n (n > 0): 2n² − 80n + 450 ≤ 0, so n² − 40n + 225 ≤ 0. Roots: 20 ± 5√7 ≈ 6.77 and 33.23.

Check: A(7) ≈ $108.29 and A(33) ≈ $109.64, but A(6) = $117 and A(34) ≈ $111.24. So 7 to 33 kits.

Problem 18 Ball Above a Height

Solve −16t² + 48t + 64 > 32, so −16t² + 48t + 32 > 0 and t² − 3t − 2 < 0.

Roots: 3 ± √172 ≈ −0.56 and 3.56. With t ≥ 0, the ball is above 32 feet for 0 ≤ t < 3 + √172 ≈ 3.56 seconds. (It lands at t = 4.)

Problem 19 Medicine Concentration Window

Solve 40tt² + 16 ≥ 2.5. The denominator is always positive, so 40t ≥ 2.5t² + 40, which gives t² − 16t + 16 ≤ 0.

Roots: 8 ± 4√3 ≈ 1.07 and 14.93. The concentration is at least 2.5 mg/L for 8 − 4√3 ≤ t ≤ 8 + 4√3 (about 1.07 to 14.93 hours).

Problem 20 Use a Given Zero to Solve a Quartic

The conjugate −2i is a zero too, giving the factor (x − 2i)(x + 2i) = x² + 4.

Divide: P(x) = (x² + 4)(x² − 2x + 2).

The quadratic formula gives x = 2 ± √(4 − 8)2 = 1 ± i. Zeros: 2i, −2i, 1 + i, 1 − i.