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

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

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Answer 1 · 2016-01-22T15:59:56

(x + 3 )(x + 1 )(x - 1 )

Explanation:

Factorise by grouping.

ie $[x^{3} + 3 x^{2}] + [- x - 3]$ $[ x^3 + 3x^2 ] + [ - x - 3 ]$

$= x^{2} (x + 3) - 1 (x + 3)$ $= x^2 (x + 3 ) - 1(x + 3 )$

[common factor of ( x + 3 ) ]

$= (x + 3) (x^{2} - 1)$ $= (x + 3 ) (x^2 - 1 )$

$(x^{2} - 1) = (x + 1) (x - 1) difference of 2 squares$ $(x^2 - 1 ) =(x + 1 )(x - 1 ) color(black)(" difference of 2 squares ")$

$\Rightarrow x^{3} + 3 x^{2} - x - 3 = (x + 3) (x + 1) (x - 1)$ $rArr x^3 + 3x^2 - x - 3 = (x + 3 )(x + 1 )(x - 1 )$

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

1 Answer

Explanation:

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