Algebra 2 Dividing Polynomials and the Remainder Theorem Quiz: 10 Questions with Explanations

A -10B 0C 2D 4

By the Remainder Theorem the remainder is p(2) = 4 + 6 – 10 = 0, so x – 2 is a factor. Using p(0) = -10 is a common mistake.

A \(x^{2} + x + 2\)B \(x^{2} – 5x + 6\)C \(x^{2} + x – 2 + \frac{6}{x – 3}\)D \(x^{2} + x – 2\)

The bottom row is 1, 1, -2, 0: quotient x² + x – 2 and remainder 0.

A No, because p(2) = 16B No, because the polynomial has no x-termC Yes, because p(-2) = 0D Yes, because p(2) = 0

x + 2 = x – (-2), so test p(-2) = -8 + 8 = 0. Testing p(2) uses the wrong sign.

A \(k = 2\)B \(k = -2\)C \(k = 3\)D \(k = -3\)

p(3) = 9 + 3k – 3 = 6 + 3k = 0, so k = -2.

A \(x – 2\)B \(x + 2\)C \(x – 1\)D \(x + 3\)

A zero at x = 2 means p(2) = 0, so x – 2 is a factor. The sign in the factor is opposite to the sign of the zero.

A \(x + 3\)B \(x + 2\)C \(x – 3\)D \(x + 3 + \frac{6}{x + 2}\)

x² + 5x + 6 = (x + 2)(x + 3), so the quotient is x + 3 with remainder 0.

A \(x^{2} – 2x + 4\)B \(x^{2} + 4\)C \(x^{2} + 2x – 4\)D \(x^{2} + 2x + 4\)

The bottom row is 1, 2, 4, 0: x² + 2x + 4 with remainder 0.

A \(x + 1 – \frac{3}{x + 2}\)B \(x + 3 + \frac{5}{x + 2}\)C \(x + 1 + \frac{3}{x + 2}\)D \(x + 1\)

(x + 2)(x + 1) = x² + 3x + 2, which leaves a remainder of 3.

A -4B 4C 1D 0

Write x + 4 as x – (-4), so k = -4.

A \(x + 6\)B \(x + 2\)C \(x + 4\)D \(x^{2} + 2x\)

(x + 3)(x + 1)(x + 2) = x³ + 6x² + 11x + 6, so the height is x + 2.