Find a polynomial of degree 3 with real coefficients that satisfies the given conditions.
-Zeros of $- 3 , - 1,4$ and $P ( 2 ) = 5$
A) $P ( x ) = - \frac { x ^ { 3 } } { 6 } + \frac { 13 x } { 6 } + 2$
B) $P ( x ) = \frac { x ^ { 3 } } { 6 } + \frac { 19 x } { 6 } - 2$
C) $P ( x ) = - \frac { x ^ { 3 } } { 6 } - \frac { 19 x } { 6 } - 2$
D) $P ( x ) = \frac { x ^ { 3 } } { 6 } - \frac { 13 x } { 6 } - 2$

Question

Quizplus · Accepted Answer

Given the zeros -3, -1, and 4, the polynomial can be expressed as $P(x) = k(x + 3)(x + 1)(x - 4)$. To find the value of $k$, use the condition $P(2) = 5$. Substituting $x = 2$ into the polynomial gives $5 = k(5)(3)(-2)$, solving for $k$ gives $k = -\frac{1}{6}$. Thus, the polynomial is $P(x) = -\frac{1}{6}(x^3 + 4x^2 - 13x - 12)$, which simplifies to $P(x) = -\frac{x^3}{6} + \frac{13x}{6} + 2$, matching option A.

Find a Polynomial of Degree 3 with Real Coefficients That −3,−1,4- 3 , - 1,4−3,−1,4

Find a Polynomial of Degree 3 with Real Coefficients That $- 3 , - 1,4$