How do you factor #x^3+x^2-x-1#?

Answer 1

By grouping, factor

#x^3+x^2-x-1 =[x^3+x^2]+[-x-1]#

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

#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]#. Tlhat is a common factor, so we can factor it out:

#x^3+x^2-x-1 = x^2[x+1] color(red)(+) (-1)[x+1]#

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

#color(white)"ssssssssssssssssss"# #=( x^2 - 1)(x+1)#

#x^3+x^2-x-1 = ( x^2 - 1)(x+1)#

Can anything else be factored in, or are we done?

#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)#

We're done now.

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Answer 2

Eva Abramson

To factor (x^3+x^2-x-1), you can group the terms and factor by grouping or use synthetic division to find the roots.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.