How do you factor $x^{3} + x^{2} - x - 1$ $x^3+x^2-x-1$ ?

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Answer 1 · 2015-05-06T20:07:50

Factor by grouping

$x^{3} + x^{2} - x - 1 = [x^{3} + x^{2}] + [- x - 1]$ $x^3+x^2-x-1 =[x^3+x^2]+[-x-1]$

The first bracket has a common factor of $x^{2}$ $x^2$ and the second bracket has a common factor of $- 1$ $-1$ . take those out to get:

$x^{3} + x^{2} - x - 1 = x^{2} [x + 1] + (- 1) [x + 1]$ $x^3+x^2-x-1 = x^2[x+1] color(red)(+) (-1)[x+1]$

Now we have two terms, one on each side of the red $+$ $color(red)(+)$ .

Each term has a factor (in brackets) of $[x + 1]$ $[x+1]$ . Tlhat is a common factor, so we can factor it out:

$x^{3} + x^{2} - x - 1 = x^{2} [x + 1] + (- 1) [x + 1]$ $x^3+x^2-x-1 = x^2[x+1] color(red)(+) (-1)[x+1]$

$ssssssssssssssssss$ $color(white)"ssssssssssssssssss"$ $= (x^{2} + (- 1)) [x + 1]$ $=( x^2 color(red)(+) (-1))[x+1]$

$ssssssssssssssssss$ $color(white)"ssssssssssssssssss"$ $= (x^{2} - 1) (x + 1)$ $=( x^2 - 1)(x+1)$

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

Are we finished or can anything be factored more?

$x^{2} - 1$ $x^2-1$ is a difference of twp squares, so we can factor it.

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

Now we are finished.

How do you factor x^3+x^2-x-1x3+x2−x−1x^3+x^2-x-1?