Answer
$P(x)=(2x-1)(2x+1)(x+1)$
zeros: $\displaystyle \pm\frac{1}{2}, -1$
Work Step by Step
Descart's rule of signs:
$P(x)=4x^{3}+4x^{2}-x-1$
... 1 sign changes: 1 positive zeros
$P(-x)=-4x^{3}+4x^{2}+x-1$
... $2$ sign changes: 2 or 0 negative zeros
Rational Zeros Theorem:
Possible rational zeros $\displaystyle \frac{p}{q}$: $\displaystyle \pm\frac{1}{1,2,4}$
Testing with synthetic division, try 1
$\left.\begin{array}{l}
1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrrrrr}
4 &4 &-1 & -1 & & & \\\hline
& 4 & 8 & 7 & & & \\\hline
4 &8 &7 & |\ \ 6 & & & \end{array}$
$1$ fails (not a zero), but we now know that
1 is an upper bound (all entries in the bottom row are positive)
Try $1/2$
$\left.\begin{array}{l}
1/2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrrrrr}
4 &4 &-1 & -1 & & & \\\hline
& 2 & 3 & 1 & & & \\\hline
4 &6 &2 & |\ \ 0 & & & \end{array}$
$P(x)=(x-\displaystyle \frac{1}{2})(4x^{2}+6x+2)$
$=(x-\displaystyle \frac{1}{2})\cdot 2(2x^{2}+3x+1)$
... for the trinomial, look for factors of $2$($1$)=$2$ whose sum is $3$
... $2$ and $1$:
$2x^{2}+3x+1=2x^{2}+2x+x+1=$
$=2x(x+1)+(x+1)=(x+1)(2x+1)$
$P(x)=(2x-1)(2x+1)(x+1)$
zeros: $\displaystyle \pm\frac{1}{2}, -1$
