# How do you factor 4y=2x^3 + 4x^2 + 8x ?

May 13, 2018

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

See explanation. Taken it to complex numbers as well.

#### Explanation:

Given: $4 y = 2 {x}^{3} + 4 {x}^{2} + 8 x$

Divide both sides by 2

$2 y = {x}^{3} + 2 {x}^{2} + 4 x$

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

$y = \frac{x}{2} \left({x}^{2} + 2 x + 4\right)$ We could stop hear
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Set $y = 0 \implies x = 0 \mathmr{and} x = \frac{- 2 \pm \sqrt{{2}^{2} - 4 \left(1\right) \left(4\right)}}{2 \left(1\right)}$

$x = - 1 \pm \frac{\sqrt{- 12}}{2}$

$x = - 1 \pm \sqrt{3} \textcolor{w h i t e}{.} i$

Giving: $y = x \left(x + 1 + \sqrt{3} i\right) \left(x + 1 - \sqrt{3} i\right)$