## College Algebra 7th Edition

$P(x)=(x+3)(x-3)(2x+1)(2x-1)$ Zeros: $\displaystyle \pm 3,\pm\frac{1}{2}$
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: $\displaystyle \pm 3,\pm\frac{1}{2}$