For the polynomial, one zero is given. Find all others.
-$$P ( x ) = x ^ { 3 } - 8 x ^ { 2 } + 17 x - 30 ; 6$$
A) $1 + 2 \mathrm { i } , 1 - 2 \mathrm { i }$
B) $1 + \sqrt { 5 } \mathrm { i } , 1 - \sqrt { 5 } \mathrm { i }$
C) $- 1 + 2 \mathrm { i } , - 1 - 2 \mathrm { i }$
D) $1 + \sqrt { 5 } , 1 - \sqrt { 5 }$

Question

Quizplus · Accepted Answer

Given that 6 is a zero of the polynomial, we can use synthetic division or polynomial division to divide the polynomial by $x - 6$ to find the quotient. The quotient will be a quadratic polynomial, and the roots of this quadratic polynomial will be the other zeros of the original polynomial. Since the coefficients of the original polynomial are real numbers and one of the given zeros is real, the other zeros must either be real or come in complex conjugate pairs due to the Complex Conjugate Root Theorem. The only option that provides a complex conjugate pair is option A, $1 + 2i$ and $1 - 2i$.

For the Polynomial, One Zero Is Given P(x)=x3−8x2+17x−30;6P ( x ) = x ^ { 3 } - 8 x ^ { 2 } + 17 x - 30 ; 6P(x)=x3−8x2+17x−30;6

For the Polynomial, One Zero Is Given $P ( x ) = x ^ { 3 } - 8 x ^ { 2 } + 17 x - 30 ; 6$