Question

Factorise x^3 - 2x^2 - x +2

Answer 1

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)#

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! :)

Answer 2

Answer:

${\bf{\green{=>x^3-2x^2-x+2=0}}}$

${\bf{\orange{=>x^2(x-2)-1(x-2)=0}}}$

${\bf{\blue{=>(x^2-1)(x-2)=0}}}$

${\bf{\red{=>(x^2-1^2)(x-2)=0}}}$

${\bf{\green{=>(x+1)(x-1)(x-2)=0}}}$

${\bf{\pink{=>x=-1}}}$ (or) ${\bf{\blue{=>x=1}}}$ (or) ${\bf{\red{x=2}}}$