Answer
$x\in \{-2, \frac{1}{2}\}$
Work Step by Step
See The Rational Zero Theorem:
... If $\displaystyle \frac{p}{q}$ is a zero of the polynomial $f(x) $with integer coefficients,
then $p$ is a factor of the constant term, $a_{0}$, and
$q$ is a factor of the leading coefficient, $a_{n}$.
------------------------
$f(x)=4x^3+4x^2-7x+2$
a. candidates for zeros, $\displaystyle \frac{p}{q}:$
$p:\qquad \pm 1, \pm 2, $
$q:\qquad \pm 1, \pm2,\pm4$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2,\pm\frac{1}{2}, \pm \frac{1}{4}$
b. Try for $x=-2:$
$\begin{array}{lllll}
\underline{-2}| & 4 & 4 & -7 & 2\\
& & -8 & 8 & -2\\
& -- & -- & -- & --\\
& 4 & -4 & 1 & |\underline{0}
\end{array}$
$-2$ is a zero,
$f(x)=(x+2)(4x^2-4x+1)$
c. Factorize the trinomial factor $(4x^2-4x+1)$
(find two factors of $1(4)=4$ whose sum is $-4):$
$(-2$ and $-2$)
$4x^{2} -4x+1=4x^{2} -2x-2x+1 \quad$...factor in pairs ...
$2x(2x-1)-1(2x-1)=(2x-1)^2$
$f(x)=(x+2)(2x-1)^2$
The zeros of f satisfy $f(x)=0$
$x\in \{-2, \frac{1}{2}\}$
