Find an nth degree polynomial function with real coefficients satisfying the given conditions.
-$\mathrm { n } = 3 ; - 4$ and $\mathrm { i }$ are zeros; $\mathrm { f } ( - 3 ) = 60$
A) $f ( x ) = 6 x ^ { 3 } + 24 x ^ { 2 } + 6 x + 24$
B) $f ( x ) = 6 x ^ { 3 } + 24 x ^ { 2 } - 6 x - 24$
C) $f ( x ) = - 6 x ^ { 3 } - 24 x ^ { 2 } - 6 x - 24$
D) $f ( x ) = - 6 x ^ { 3 } - 24 x ^ { 2 } + 6 x + 24$

Question

Quizplus · Accepted Answer

Given the zeros -4 and i, the third zero must be the complex conjugate of i, which is -i. The polynomial can be constructed as $f(x) = a(x + 4)(x - i)(x + i)$. Simplifying, we get $f(x) = a(x + 4)(x^2 + 1)$. To find the coefficient $a$, we use the condition $f(-3) = 60$, leading to $a(-3 + 4)((-3)^2 + 1) = 60$, which simplifies to $a \cdot 1 \cdot 10 = 60$, giving $a = 6$. Thus, the polynomial is $f(x) = 6(x + 4)(x^2 + 1)$, which expands to $f(x) = 6x^3 + 24x^2 + 6x + 24$.

Find an Nth Degree Polynomial Function with Real Coefficients Satisfying n=3;−4\mathrm { n } = 3 ; - 4n=3;−4

Find an Nth Degree Polynomial Function with Real Coefficients Satisfying $\mathrm { n } = 3 ; - 4$