How do you factor completely $x^{3} - 2 x^{2} - 9 x + 18$ $x^3-2x^2-9x+18$ ?

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Answer 1 · 2017-05-08T02:14:38

$(x - 3) (x + 3) (x - 2)$ $(x-3)(x+3)(x-2)$

Factor by grouping:

$(x^{3} - 2 x^{2})$ $(color(blue)(x^3-2x^2))$ $+$ $+$ $(- 9 x + 18)$ $(color(red)(-9x+18))$

Starting on the left we can factor out an $x^{2}$ $x^2$

$x^{2} (x - 2)$ $color(blue)(x^2(x-2))$

On the right we can then factor out a $- 9$ $-9$

$- 9 (x - 2)$ $color(red)(-9(x-2))$

Observe:

$x^{2} (x - 2)$ $color(blue)(x^2(x-2))$ $+$ $+$ $- 9 (x - 2)$ $color(red)(-9(x-2))$

*Notice how we have two $x - 2$ $x-2$ . We can then simply rewrite the expression as follows.

$(x^{2} - 9) (x - 2)$ $(x^2-9)(x-2)$

*Note: all we did was combine $x^{2}$ $color(blue)(x^2)$ and $- 9$ $color(red)(-9)$ and wrote $(x - 2)$ $(x-2)$ as one term instead of two.

We're not done just yet. We can still factor $(x^{2} - 9)$ $(x^2-9)$ into $(x - 3) (x + 3)$ $(x-3)(x+3)$

So a completely factored expression is then:

$(x - 3) (x + 3) (x - 2)$ $(x-3)(x+3)(x-2)$

How do you factor completely x3−2x2−9x+18x^3-2x^2-9x+18?