For the polynomial, one zero is given. Find all others.
-$$P ( x ) = x ^ { 4 } - 32 x ^ { 2 } - 144 ; - 2 i$$
A) $2 i , 6 , - 6$
B) $2 \mathrm { i } , 12 , - 12$
C) $2 i , 6 i , - 6 i$
D) $2 \mathrm { i } , 12 \mathrm { i } , - 12 \mathrm { i }$

Question

Quizplus · Accepted Answer

Given a polynomial $P(x) = x^4 - 32x^2 - 144$ with one zero as $-2i$, we know that its complex conjugate $2i$ is also a zero due to the Complex Conjugate Root Theorem. We can factor the polynomial as a quadratic in terms of $x^2$, $x^2 - 32x - 144 = 0$, and solve for $x^2$ to find the other roots. Factoring or using the quadratic formula, we find $x^2 = 36$ or $x^2 = -4$, giving us the roots $x = \pm 6$ (from $x^2 = 36$) and $x = \pm 2i$ (from $x^2 = -4$), confirming the roots are $2i, 6, -6$.

For the Polynomial, One Zero Is Given P(x)=x4−32x2−144;−2iP ( x ) = x ^ { 4 } - 32 x ^ { 2 } - 144 ; - 2 iP(x)=x4−32x2−144;−2i

For the Polynomial, One Zero Is Given $P ( x ) = x ^ { 4 } - 32 x ^ { 2 } - 144 ; - 2 i$