# How do you factor x^3 - 2x^2 - x + 2 by grouping?

Mar 11, 2018

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#### Explanation:

${x}^{3} - 2 {x}^{2} - x + 2$

$= {x}^{3} - x - 2 {x}^{2} + 2$

Grouping the 1^(st $2$ terms together and the 2^(nd $2$ together:

$= x \left({x}^{2} - 1\right) - 2 \left({x}^{2} - 1\right)$

$= \left({x}^{2} - 1\right) \left(x - 2\right)$

Using the identity: ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

$= \left({x}^{2} - {1}^{2}\right) \left(x - 2\right)$

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color(red)("Alternate Method:-"

The above method is an easy on to solve this question. For factorizing other cubic polynomials, the following method can be used:

First, by trial and error method, you can find one factor as follows:

${x}^{3} - 2 {x}^{2} - x + 2$

When replacing 1,

$\implies {1}^{3} - 2 \times {1}^{2} - 1 + 2$

$\implies \cancel{1} - \cancel{2} - \cancel{1} + \cancel{2}$

$\implies 0$

So, we get $\left(x - 1\right)$ as factor.

Then by long division, divide $\left(x - 1\right)$ by $\left({x}^{3} - 2 {x}^{2} - x + 2\right)$

You get$\implies \left(x - 1\right) \left({x}^{2} - x - 2\right)$

Then you have to factorize it by splitting the middle term method.

$\implies \left(x - 1\right) \left[{x}^{2} + x - 2 x - 2\right]$

$\implies \left(x - 1\right) \left[x \left(x + 1\right) - 2 \left(x + 1\right)\right]$

color(magenta)(=>(x-1)(x+1)(x-2)

~Hope this helps! :)