Answer
$P(x)=(x+3)(x-3)(2x+1)(2x-1)$
Zeros: $\displaystyle \pm 3,\pm\frac{1}{2}$
Work Step by Step
We can factor this trinomial, letting $t=x^{2}$
$4t^{2}-37t+9$=$\quad $ we need factors of 4(9)=36 whose sum is -37
... -36 and -1 ...
$4t^{2}-37t+9=4t^{2}-36t-t+9$
$=4t(t-9)-(t-9)=(t-9)(4t-1)$
$4x^{4}-37x^{2}+9=(x^{2}-9)(4x^{2}-1)$
... both parentheses are differences of squares...
$P(x)=(x+3)(x-3)(2x+1)(2x-1)$
Zeros: $\displaystyle \pm 3,\pm\frac{1}{2}$
