Algebra 2 Zeros and Graphs of Polynomial Functions Quiz: 10 Questions with Explanations

A \(-5 and 2\)B \(5 and 2\)C \(-5 and -2\)D \(5 and -2\)

Set each factor to zero: x – 5 = 0 gives 5 and x + 2 = 0 gives -2. The sign flips from the factor to the zero.

A \(x = -2\)B \(x = 1\)C \(x = 0\)D \(x = 2\)

At x = 1 the graph turns around, so the zero has even multiplicity. It crosses at x = -2.

A 4B 1C 3D 5

The multiplicity is the exponent of the factor (x – 3), which is 4.

A Down on the left, up on the rightB Up on both endsC Down on both endsD Up on the left, down on the right

Odd degree with a negative leading coefficient rises on the left and falls on the right.

A Both ends downB Left down, right upC Both ends up (y = x⁴ panel)D Left up, right down

Even degree with a positive leading coefficient rises on both ends.

A \(\frac{3}{4}\)B \(\frac{1}{3}\)C \(\frac{4}{3}\)D \(2\)

Candidates have the form p/q with p dividing 4 and q dividing 3. The fraction 3/4 has the numerator 3, which does not divide 4.

A \(1, -1\)B \(0, 1, -1\)C \(0, 1\)D \(1 only\)

x³ – x = x(x – 1)(x + 1), so the zeros are 0, 1, and -1.

A \(x + 1\)B \(x – 1\)C \(x + 2\)D \(x – 4\)

p(-1) = 0, so x – (-1) = x + 1 is a factor. Zeros give factors with the opposite sign.

A \(2\)B \(-2i\)C \(-3i\)D \(4\)

Complex zeros of real polynomials come in conjugate pairs, so -2i must also be a zero.

A \(x(x + 2)(x – 5)\)B \((x – 2)(x + 5)\)C \(x(x – 2)(x + 5)\)D \(x(x – 2)(x – 5)\)

Each zero r gives the factor (x – r): x, x – 2, and x + 5.