How do you factor $4 y = 2 x^{3} + 4 x^{2} + 8 x$ $4y=2x^3 + 4x^2 + 8x$ ?

Question

1495 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-13T10:05:53

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

See explanation. Taken it to complex numbers as well.

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

Divide both sides by 2

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

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

$y = \frac{x}{2} (x^{2} + 2 x + 4)$ $y=x/2(x^2+2x+4)$ We could stop hear
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Set $y = 0 \Rightarrow x = 0 and x = \frac{- 2 \pm \sqrt{2^{2} - 4 (1) (4)}}{2 (1)}$ $y=0 => x=0 and x=(-2+-sqrt(2^2-4(1)(4)))/(2(1))$

$x = - 1 \pm \frac{\sqrt{- 12}}{2}$ $x=-1+-sqrt(-12)/2$

$x = - 1 \pm \sqrt{3} . i$ $x=-1+-sqrt(3) color(white)(.)i$

Giving: $y = x (x + 1 + \sqrt{3} i) (x + 1 - \sqrt{3} i)$ $y=x(x+1+sqrt3 i)( x+1-sqrt3 i)$

Tony B

How do you factor 4y=2x^3 + 4x^2 + 8x 4y=2x3+4x2+8x4y=2x^3 + 4x^2 + 8x ?