Answer
$P(x)=(x-1)(2x-1)(2x+3)$
zeros: $-\displaystyle \frac{3}{2}, \frac{1}{2}, 1$
Work Step by Step
Descart's rule of signs:
$P(x)=4x^{3}-7x+3$
... $2$ sign changes: 2 or 0 positive zeros
$P(-x)=-4x^{3}+7x+3$
... 1 sign changes: 1 negative zeros
Rational Zeros Theorem:
Possible rational zeros $\displaystyle \frac{p}{q}$: $\displaystyle \pm\frac{1,3}{1,2,4}$
Testing with synthetic division, ( missing powers: 0 in top row)
try 1
$\left.\begin{array}{l}
1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrrrrr}
4 &0 &-7 & 3 & & & \\\hline
& 4 & 4 & -3 & & & \\\hline
4 &4 &-3 & |\ \ 0 & & & \end{array}$
$P(x)=(x-1)(4x^{2}+4x-3)$
... search for two factors of 4(-3)=-12 whose sum is 4
... find 6 and -2
$4x^{2}+4x-3 =4x^{2}-2x+6x-3$
$=2x(2x-1)+3(2x-1)$
$=(2x-1)(2x+3)$
$P(x)=(x-1)(2x-1)(2x+3)$
zeros: $-\displaystyle \frac{3}{2}, \frac{1}{2}, 1$
