Question #fe6ca

May 14, 2017

Because all the attempts to factor this by breaking it up into the form $\left(A x + B\right) \left(C x + D\right)$ fail, a good method to try is using the quadratic equation.

The given equation $y = 4 {x}^{2} + 16 x + 9$ is in standard form
$y = a {x}^{2} + b x + c$
with
$a = 4$
$b = 16$
$c = 9$

The quadratic equation finds $x$ for $4 {x}^{2} + 16 x + 9 = 0$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{- 16 \pm \sqrt{{16}^{2} - 4 \left(4\right) \left(9\right)}}{2 \left(4\right)}$
$\frac{- 16 \pm \sqrt{{16}^{2} - 144}}{8} = \frac{- 16 \pm \sqrt{112}}{8} = \frac{- 16 \pm 4 \sqrt{7}}{8}$

The "$\pm$" symbol means "both + and $-$", so this gives us that the two values of $x$ that make $4 {x}^{2} + 16 x + 9 = 0$ are

$x = \frac{- 16 + 4 \sqrt{7}}{8}$ and $x = \frac{- 16 - 4 \sqrt{7}}{8}$

To create the final factorization of the original question, solve the two above equations for zero and multiply the two expressions:

$4 {x}^{2} + 16 x + 9 = \left(\frac{8}{4 \sqrt{7}} x + \frac{16}{4 \sqrt{7}}\right) \left(\frac{8}{- 4 \sqrt{7}} x + \frac{16}{- 4 \sqrt{7}}\right)$

The final step is technically optional, but teachers like it enough to grade you on it. Simplify:

$4 {x}^{2} + 16 x + 9 = \left(\frac{2 \sqrt{7}}{7} x + \frac{4 \sqrt{7}}{7}\right) \left(\frac{- 2 \sqrt{7}}{7} x - \frac{4 \sqrt{7}}{7}\right)$