How do you factor #x^3+3x^2-2x-6#?

Question

13958 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-09T18:32:55

#(x + 3)(x + sqrt(2))(x - sqrt(2))#

We have: #x^(3) + 3 x^(2) - 2 x - 6#

#= (x^(3) + 3 x^(2)) + (- 2 x - 6)#

Let's take #x^(2)# in common in the first set of parentheses and #- 2# in common in the second set of parentheses, respectively:

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

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

Then, let's take #x + 3# in common:

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

If you allow surds to be used, then the second set of parentheses can be factorised further:

#= (x + 3)(x + sqrt(2))(x - sqrt(2))#