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

1 Answer
Mar 11, 2018

Answer:

#color(magenta)(=(x-1)(x+1)(x-2)#

Explanation:

#x^3-2x^2-x+2#

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

Grouping the #1^(st# #2# terms together and the #2^(nd# #2# together:

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

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

Using the identity: #a^2-b^2=(a+b)(a-b)#

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

#color(magenta)(=(x-1)(x+1)(x-2)#

#color(red)("Alternate Method:-"#

The above method is an easy on to solve this question. For factorizing other cubic polynomials, the following method can be used:

First, by trial and error method, you can find one factor as follows:

#x^3-2x^2-x+2#

When replacing 1,

#=>1^3-2xx1^2-1+2#

#=>cancel1-cancel2-cancel1+cancel2#

#=>0#

So, we get #(x-1)# as factor.

Then by long division, divide #(x-1)# by #(x^3-2x^2-x+2)#

You get#=>(x-1)(x^2-x-2)#

Then you have to factorize it by splitting the middle term method.

#=>(x-1)[x^2+x-2x-2]#

#=>(x-1)[x(x+1)-2(x+1)]#

#color(magenta)(=>(x-1)(x+1)(x-2)#

~Hope this helps! :)