Find a polynomial of degree 3 with real coefficients that satisfies the given conditions.
-Zeros of $- 4$ and $1 - \mathrm { i } , \mathrm { P } ( 0 ) = 8$
A) $P ( x ) = x ^ { 3 } + \frac { 3 } { 2 } x ^ { 2 } - 2 x + 4$
B) $P ( x ) = x ^ { 3 } + 2 x ^ { 2 } + 6 x + 8$
C) $P ( x ) = x ^ { 3 } + 2 x ^ { 2 } - 6 x$
D) $P ( x ) = x ^ { 3 } + 2 x ^ { 2 } - 6 x + 8$

Question

Quizplus · Accepted Answer

Given the zeros $ -4 $ and $ 1 - i $, the third zero must be $ 1 + i $ because complex zeros of polynomials with real coefficients come in conjugate pairs. The polynomial can be constructed from its zeros as follows: $(x + 4)(x - (1 - i))(x - (1 + i))$. Simplifying this expression gives $P(x) = x^3 + 2x^2 - 6x + 8$, which matches option D. Additionally, $P(0) = 8$ confirms the correct choice, as substituting $x = 0$ into the polynomial should yield $8$.

Find a Polynomial of Degree 3 with Real Coefficients That −4- 4−4

Find a Polynomial of Degree 3 with Real Coefficients That $- 4$