How do you factor completely #P(x)=x^3+2x^2-x-2#?

1 Answer
Roy E.
Dec 31, 2016

#(x-1)(x+2)(x+1)#

Explanation:

Spotting that #1+2-1-2=0# means that #x-1# is a factor.
Then write #x^3+2x^2-x-2-=(x-1)(...x^2 ... x ...)# and try to fill in the gaps to make the identity true by matching powers. Obviously the coefficient of #x^2# has to be #1#, and similarly the constant at the end has to be #+2#:

#(x-1)(x^2....x+2)#

Since we have got #+2x# and need #-x# we need a #+3x# added in:
#(x-1)(x^2+3x+2)# which checks out upon multiplying out.
The second bracket then factorizes easily to #(x+2)(x+1)#.

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