How do you factor $x^{5} + 2 x^{4} - x - 2$ $x^5+2x^4-x-2$ ?

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How do you factor $x^{5} + 2 x^{4} - x - 2$ $x^5+2x^4-x-2$ ?

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Answer 1 · 2016-05-12T07:37:12

$x^{5} + 2 x^{4} - x - 2 = (x - 1) (x + 1) (x^{2} + 1) (x + 2)$ $x^5+2x^4-x-2=(x-1)(x+1)(x^2+1)(x+2)$

Explanation:

Use the difference of squares identity:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2 = (a-b)(a+b)$

after factoring by grouping:

$x^{5} + 2 x^{4} - x - 2$ $x^5+2x^4-x-2$

$= (x^{5} + 2 x^{4}) - (x + 2)$ $=(x^5+2x^4)-(x+2)$

$= x^{4} (x + 2) - 1 (x + 2)$ $=x^4(x+2)-1(x+2)$

$= (x^{4} - 1) (x + 2)$ $=(x^4-1)(x+2)$

$= ({(x^{2})}^{2} - 1^{2}) (x + 2)$ $=((x^2)^2-1^2)(x+2)$

$= (x^{2} - 1) (x^{2} + 1) (x + 2)$ $=(x^2-1)(x^2+1)(x+2)$

$= (x^{2} - 1^{2}) (x^{2} + 1) (x + 2)$ $=(x^2-1^2)(x^2+1)(x+2)$

$= (x - 1) (x + 1) (x^{2} + 1) (x + 2)$ $=(x-1)(x+1)(x^2+1)(x+2)$

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