# Question #9c3e2

Feb 14, 2017

$\left\{- 3 , - \frac{1}{2}\right\}$

#### Explanation:

If $\frac{1}{2}$ and $3$ are zeroes then $y = 4 {x}^{4} - 37 {x}^{2} + 9 = {p}_{2} \left(x\right) \left(x - \frac{1}{2}\right) \left(x - 3\right)$ where ${p}_{2} \left(x\right) = a {x}^{2} + b x + c$ is a generic order $2$ polynomial.

Substituting and equating coefficients we have

$4 {x}^{4} - 37 {x}^{2} + 9 = \left(a {x}^{2} + b x + c\right) \left(x - \frac{1}{2}\right) \left(x - 3\right)$

so

$\left\{\begin{matrix}\frac{3 c}{2} = 9 \\ - \frac{3 b}{2} + \frac{7 c}{2} = 0 \\ - \frac{3 a}{2} + \frac{7 b}{2} - c = 37 \\ \frac{7 a}{2} - b = 0 \\ a = 4\end{matrix}\right.$

now solving for $a , b , c$ we obtain

$a = 4 , b = 14 , c = 6$

so ${p}_{2} \left(x\right) = 4 {x}^{2} + 14 x + 6$ with roots

$x = \left\{- 3 , - \frac{1}{2}\right\}$ so finally

$y = 4 {x}^{4} - 37 {x}^{2} + 9 = 4 \left({x}^{2} - {3}^{2}\right) \left({x}^{2} - {\left(\frac{1}{2}\right)}^{2}\right)$