Answer
$x\in\{-2,-6, 1\}$
Work Step by Step
If we know that $-2$ is a zero of the equation $f(x)=x^3+7x^2+4x-12$, we can solve for $x^3+7x^2+4x-12=0$, using synthetic divison and factoring it out.
$\begin{array}{lllll}
\underline{-2}| & 1 & 7 & 4 & -12 \\
& & -2 & -10 & 12 \\
\hline & & & & \\
& 1 & 5 & -6 & 0
\end{array}$
$x^3+7x^2+4x-12=(x+2)(x^2+5x-6)=0$.
and then, we can factor out the quadratic equation by grouping.
To factor the quadratic equation by factoring,
1.With the quadratic in standard form, ax2+bx+c=0, multiply a⋅c
2.Find two numbers whose product equals ac and whose sum equals b
3.Rewrite the equation replacing the bx term with two terms using the numbers found in step 1 as coefficients of x
4.Factor the first two terms and then factor the last two terms. The expressions in parentheses must be exactly the same to use grouping.
5.Factor out the expression in parentheses. Set the expressions equal to zero and solve for the variable.
$(x+2)(x^2-x+6x-6)=(x+2)(x(x-1)+6(x-1))=(x+2)(x+6)(x-1)$,
Therefore, $x\in${$-2,-6, 1$}
