Is it possible to factor y=2x^2 + 8x+4? If so, what are the factors?

Dec 8, 2015

First you can take out the factor $2$

Explanation:

$\to y = 2 \left({x}^{2} + 4 x + 2\right)$

This cannot be factored any further without using radical expressions.

Dec 8, 2015

$2 {x}^{2} + 8 x + 4 = 2 \left({x}^{2} + 4 x + 2\right)$

$= 2 \left(x + 2 + \sqrt{2}\right) \left(x + 2 - \sqrt{2}\right)$

Explanation:

Given $y = 2 {x}^{2} + 8 x + 4$ first separate out the common scalar factor $2$ to get:

$y = 2 \left({x}^{2} + 4 x + 2\right)$

The quadratic factor is in the form $a {x}^{2} + b x + c$ with $a = 1$, $b = 4$, $c = 2$. This has zeros given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{- 4 \pm \sqrt{{\left(- 4\right)}^{2} - \left(4 \times 1 \times 2\right)}}{2 \times 1}$

$= \frac{- 4 \pm \sqrt{8}}{2} = \frac{- 4 \pm \sqrt{{2}^{2} \cdot 2}}{2} = \frac{- 4 \pm 2 \sqrt{2}}{2} = - 2 \pm \sqrt{2}$

These zeros correspond to factors $\left(x + 2 + \sqrt{2}\right)$ and $\left(x + 2 - \sqrt{2}\right)$

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Another way of finding this is by completing the square, then using the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = \left(x + 2\right)$ and $b = \sqrt{2}$ as follows:

$y = 2 {x}^{2} + 8 x + 4$

$= 2 \left({x}^{2} + 4 x + 2\right)$

$= 2 \left({x}^{2} + 4 x + 4 - 2\right)$

$= 2 \left({\left(x + 2\right)}^{2} - 2\right)$

$= 2 \left({\left(x + 2\right)}^{2} - {\left(\sqrt{2}\right)}^{2}\right)$

$= 2 \left(\left(x + 2\right) - \sqrt{2}\right) \left(\left(x + 2\right) + \sqrt{2}\right)$

$= 2 \left(x + 2 - \sqrt{2}\right) \left(x + 2 + \sqrt{2}\right)$