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How do you factor $w^{3} - 3 w^{2} - 9 w + 27$ $w^3-3w^2-9w+27$ ?

1 Answer

Sep 26, 2017

$(w^{3} - 3 w^{2} - 9 w + 27) = (x - 3) (x - 3) (x + 3)$ $(w^3-3w^2-9w+27)=color(blue)((x-3)(x-3)(x+3))$

Explanation:

Writing $w^{3} - 3 w^{2} - 9 w + 27$ $w^3-3w^2-9w+27$
as
$XXX w^{3} - 3 w^{2} - 3^{3} w + 3^{3}$ $color(white)("XXX")w^3-3w^2-3^3w+3^3$
makes it fairly clear that
$XXX w = 3$ $color(white)("XXX")w=3$ is a zero of this expression.

That is $(w - 3)$ $(w-3)$ is a factor of $(w^{3} - 3 w^{2} - 9 w + 27)$ $(w^3-3w^2-9w+27)$

If we divided $(w^{3} - 3 w^{2} - 9 w + 27)$ $(w^3-3w^2-9w+27)$ by $(w - 3)$ $(w-3)$
(either by polynomial long division or synthetic division)
we find
$XXX (x^{2} - 9)$ $color(white)("XXX")(x^2-9)$ is also a factor

...but $(x^{2} - 9) = (x^{2} - 3^{2}) = (x - 3) (x + 3)$ $(x^2-9)=(x^2-3^2)=(x-3)(x+3)$

So a complete factoring is
$XXX (w^{3} - 3 w^{2} - 9 w + 27) = (x - 3) (x - 3) (x + 3)$ $color(white)("XXX")(w^3-3w^2-9w+27)=(x-3)(x-3)(x+3)$

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