Precalculus Polynomial Division and the Factor Theorem Quiz: 10 Questions with Explanations

A -4B 0C 1D 4

f(1) = 1 + 3 – 4 = 0, so x – 1 is a factor. Using -4 confuses the remainder with the constant term.

A \(5\)B \(\frac15\)C \(-\frac15\)D \(-5\)

x + 5 = x – (-5), so c = -5.

A \(x-3\)B \(x+3\)C \(3x\)D \(x-9\)

By the Factor Theorem, f(c) = 0 means x – c is a factor.

A \(x^{2}+2x-3\)B \(x^{2}-2x-3\)C \(x^{2}-2x+3\)D \(x^{3}-2x^{2}-3x\)

The bottom row 1, -2, -3, 0 gives x^2 – 2x – 3 with remainder 0.

A \(\frac32\)B \(\frac{3}{4}\cdot 3\)C \(\frac52\)D \(\frac43\)

p divides -4 and q divides 3, so 4/3 is allowed. Reversing p and q is a common error.

A \(3\)B \(-3\)C \(5\)D \(-5\)

f(-2) = -8 + 4 + 1 = -3. Dividing by x + 2 uses c = -2, not 2.

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

A zero at -2 gives the factor x – (-2) = x + 2.

A 3B 2C 4D 5

A polynomial of degree n has exactly n zeros counting multiplicity and complex zeros.

A \(0\)B \(1\)C Nothing, skip itD \(x^{2}\)

Each power needs a coefficient, so the missing x^2 term has coefficient 0.

A \((x-3)(x-1)(x+2)\)B \((x-3)(x+1)(x-2)\)C \((x+3)(x-1)(x+2)\)D \((x-3)(x^{2}+x-2)(x+2)\)

The quotient x^2 + x – 2 factors as (x – 1)(x + 2).