Find the zeros of the polynomial function and state the multiplicity of each.
-$$\left( x ^ { 2 } + x - 12 \right) ^ { 4 } ( x - 1 + \sqrt { 7 } ) ^ { 3 }$$
A) $- 4 , - 3$ (multiplicity $4 ), - 1 - \sqrt { 7 }$ (multiplicity 3 )
B) $4 , - 3$ (multiplicity 4 ), $1 - \sqrt { 7 }$ (multiplicity 3 )
C) $- 4,3$ (multiplicity 4 ), $1 - \sqrt { 7 }$ (multiplicity 3 )
D) $- 4,3$ (multiplicity 4 ), $1 + \sqrt { 7 }$ (multiplicity 3 )

Question

Quizplus · Accepted Answer

The zeros of the polynomial can be found by setting each factor equal to zero. For $(x^2 + x - 12)^4$, we factor to find its zeros: $x^2 + x - 12 = (x + 4)(x - 3)$, so the zeros are $x = -4$ and $x = 3$, each with multiplicity 4 because of the exponent 4. For $(x - 1 + \sqrt{7})^3$, setting it equal to zero gives $x = 1 - \sqrt{7}$, with multiplicity 3, as indicated by the exponent 3.

Find the Zeros of the Polynomial Function and State the Multiplicity