To find the zeros of the polynomial function [tex]\( F(x)=x^3+x^2-6x \)[/tex], we need to determine the values of [tex]\( x \)[/tex] for which [tex]\( F(x) = 0 \)[/tex].
Let's go through the process step-by-step:
1. Set the polynomial equal to zero:
[tex]\[ F(x) = x^3 + x^2 - 6x = 0 \][/tex]
2. Factor the polynomial:
We can factor out an [tex]\( x \)[/tex] from all terms in the polynomial:
[tex]\[ x(x^2 + x - 6) = 0 \][/tex]
3. Factor the quadratic expression inside the parentheses:
Let's factor [tex]\( x^2 + x - 6 \)[/tex]. We need two numbers that multiply to [tex]\(-6\)[/tex] and add to [tex]\(1\)[/tex]. These numbers are [tex]\(3\)[/tex] and [tex]\(-2\)[/tex]:
[tex]\[ x^2 + x - 6 = (x + 3)(x - 2) \][/tex]
4. Write the fully factored form of the polynomial:
[tex]\[ x(x + 3)(x - 2) = 0 \][/tex]
5. Set each factor equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[
\begin{cases}
x = 0 \\
x + 3 = 0 \Rightarrow x = -3 \\
x - 2 = 0 \Rightarrow x = 2
\end{cases}
\][/tex]
Therefore, the zeros of the polynomial function [tex]\( F(x) = x^3 + x^2 - 6x \)[/tex] are [tex]\( x = -3 \)[/tex], [tex]\( x = 0 \)[/tex], and [tex]\( x = 2 \)[/tex].
Given the options provided:
A. [tex]\( x = -2, x = 0, \)[/tex] and [tex]\( x = 3 \)[/tex]
B. [tex]\( x = -6, x = 0, \)[/tex] and [tex]\( x = 1 \)[/tex]
C. [tex]\( x = -3, x = 0, \)[/tex] and [tex]\( x = 2 \)[/tex]
D. [tex]\( x = -1, x = 0, \)[/tex] and [tex]\( x = 6 \)[/tex]
The correct answer is:
C. [tex]\( x = -3, x = 0, \)[/tex] and [tex]\( x = 2 \)[/tex]
