Find the zeros of the polynomial function.
-$$f ( x ) = x ^ { 3 } + 3 x ^ { 2 } - x - 3$$
A) $x = - 1 , x = 1 , x = - 3$
B) $x = 1 , x = - 3 , x = 3$
C) $x = - 3 , x = 3$
D) $x = 9$

Question

Quizplus · Accepted Answer

To find the zeros of the polynomial $f(x) = x^3 + 3x^2 - x - 3$, we can factor it. Factoring by grouping, we get:$$x^2(x + 3) - 1(x + 3) = (x^2 - 1)(x + 3) = (x + 1)(x - 1)(x + 3)$$Setting each factor equal to zero gives the solutions:$$x + 1 = 0 \Rightarrow x = -1$$$$x - 1 = 0 \Rightarrow x = 1$$$$x + 3 = 0 \Rightarrow x = -3$$Therefore, the zeros of the polynomial are $x = -1$, $x = 1$, and $x = -3$.

Find the Zeros of the Polynomial Function f(x)=x3+3x2−x−3f ( x ) = x ^ { 3 } + 3 x ^ { 2 } - x - 3f(x)=x3+3x2−x−3

Find the Zeros of the Polynomial Function $f ( x ) = x ^ { 3 } + 3 x ^ { 2 } - x - 3$