How do you factor by grouping: $x^{3} - 2 x^{2} - 4 x + 8$ $x^3 - 2x^2 - 4x + 8$ ?

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Answer 1 · 2015-04-21T09:00:29

To factor this expression, we can make groups of two like this:

$(x^{3} - 2 x^{2}) - (4 x - 8)$ $(x^3 - 2x^2) - (4x - 8)$

$= x^{2} (x - 2) - 4 (x - 2)$ $= x^2(x-2) - 4(x - 2)$

The factor $x - 2$ $x - 2$ is common to both the terms

$= (x - 2) (x^{2} - 4)$ $= (x - 2)(x^2 - 4)$

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

We also know that $a^{2} - b^{2} = (a + b) (a - b)$ $color(blue)(a^2 - b^2 = (a+b)(a-b)$

$= (x - 2) (x + 2) (x - 2)$ $= (x - 2)(x + 2)(x - 2)$

$= {(x - 2)}^{2} \cdot (x + 2)$ $color(green)( = (x - 2)^2*(x+2)$ is the factorised form of $x^{3} - 2 x^{2} - 4 x + 8$ $x^3 - 2x^2 - 4x + 8$

How do you factor by grouping: x3−2x2−4x+8x^3 - 2x^2 - 4x + 8?