Answer
$P(x)=(2x-1)(x+2)^{2}$
zeros: $-2, \displaystyle \frac{1}{2}$
Work Step by Step
Descart's rule of signs:
$P(x)=2x^{3}+7x^{2}+4x-4$
... 1 sign changes: 1 positive zeros
$P(-x)=-2x^{3}+7x^{2}-4x-4$
... $2$ sign changes: 2 or 0 negative zeros
Rational Zeros Theorem:
Possible rational zeros $\displaystyle \frac{p}{q}$: $\displaystyle \pm\frac{1,2,4}{1,2}$
Testing with synthetic division, try 1
$\left.\begin{array}{l}
1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrrrrr}
2 &7 &4 & -4 & & & \\\hline
& 2 &18 & 44 & & & \\\hline
2 &9 &22 & |\ \ 40 & & & \end{array}$
$1$ 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
$\left.\begin{array}{l}
1/2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrrrrr}
2 &7 &4 & -4 & & & \\\hline
& 1 &4 & -4 & & & \\\hline
2 &8 &8 & |\ \ 0 & & & \end{array}$
$P(x)=(x-\displaystyle \frac{1}{2})(2x^{2}+8x+8)$
$=(x-\displaystyle \frac{1}{2})\cdot 2(x^{2}+4x+4)$
... recognize a perfect square
$=2(x-\displaystyle \frac{1}{2})(x+2)^{2}$
$P(x)=(2x-1)(x+2)^{2}$
zeros: $-2, \displaystyle \frac{1}{2}$
