How do you factor $y = x^{3} + 6 x^{2} + 8 x$ $y=x^3 + 6x^2 + 8x$ ?

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Answer 1 · 2016-07-22T19:56:33

$y = x (x + 2) (x + 4)$ $y = x(x+2)(x+4)$

Take common factor of x out leaving a quadratic inside:

$y = x (x^{2} + 6 x + 8)$ $y = x(x^2+6x + 8)$

Fairly easy to solve by eye but will use quadratic formula for completeness:

$x = \frac{- 6 \pm \sqrt{6^{2} - 4 \cdot 1 \cdot 8}}{2} = \frac{- 6 \pm \sqrt{4}}{2}$ $x = (-6+-sqrt(6^2 - 4*1*8))/2 = (-6+-sqrt(4))/2$

$\Rightarrow x = - 2 or x = - 4$ $implies x = -2 or x = -4$

Hence

$y = x (x + 2) (x + 4)$ $y = x(x+2)(x+4)$

How do you factor y=x^3 + 6x^2 + 8x y=x3+6x2+8xy=x^3 + 6x^2 + 8x ?