SAT Math: Polynomials (Drill 2)

Question 1. If p(x) = x³ − 6x² + 11x − 6, which of the following is a factor of p(x)?

• A) (x − 4)
• B) (x + 1)
• C) (x − 2) ✓
• D) (x + 6)

Explanation: By the Factor Theorem, (x − a) is a factor if p(a) = 0. Test x = 2: p(2) = 8 − 24 + 22 − 6 = 0. So (x − 2) is a factor. The full factorization is (x − 1)(x − 2)(x − 3).

Question 2. What is the remainder when p(x) = x⁴ − 3x² + 5 is divided by (x + 1)?

• A) 3 ✓
• B) 5
• C) 7
• D) −1

Explanation: By the Remainder Theorem, the remainder is p(−1) = (−1)⁴ − 3(−1)² + 5 = 1 − 3 + 5 = 3.

Question 3. Which of the following is equivalent to (2x² + 3x − 1) − (x² − 4x + 5)?

• A) x² − x + 4
• B) x² + 7x − 6 ✓
• C) 3x² − x − 6
• D) x² + 7x + 4

Explanation: Distribute the negative sign: 2x² + 3x − 1 − x² + 4x − 5. Combine like terms: (2x² − x²) + (3x + 4x) + (−1 − 5) = x² + 7x − 6.

Question 4. The polynomial g(x) = 2x³ − x² + 4x − 7 has a leading term of 2x³. Which of the following describes the end behavior of g(x)?

• A) As x → −∞, g(x) → +∞ and as x → +∞, g(x) → +∞
• B) As x → −∞, g(x) → −∞ and as x → +∞, g(x) → −∞
• C) As x → −∞, g(x) → −∞ and as x → +∞, g(x) → +∞ ✓
• D) As x → −∞, g(x) → +∞ and as x → +∞, g(x) → −∞

Explanation: The leading term is 2x³. Odd degree with a positive leading coefficient means: as x → −∞, g(x) → −∞ and as x → +∞, g(x) → +∞. Choices A and B describe even-degree behavior.

Question 5. The polynomial p(x) has zeros at x = −4, x = 1, and x = 3. If the leading coefficient is 1, which of the following is p(x)?

• A) x³ − 13x + 12 ✓
• B) x³ − 7x + 12
• C) x³ + 13x − 12
• D) x³ − 12

Explanation: p(x) = (x + 4)(x − 1)(x − 3). Multiply (x − 1)(x − 3) = x² − 4x + 3. Then (x + 4)(x² − 4x + 3) = x³ − 4x² + 3x + 4x² − 16x + 12 = x³ − 13x + 12.