How to long divide #(x^3-2x^2-4x-4)/(x^2+x-2)#?

1 Answer
Apr 16, 2017

#(x^3-2x^2-4x-4)/(x^2+x-2)=x-3+4/(x+2)+3/(x-1)#

Explanation:

#(x^3-2x^2-4x-4)/(x^2+x-2)#

By long division,

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Hence,

#(x^3-2x^2-4x-4)/(x^2+x-2)=x-3+color(green)((x-10)/(x^2+x-2)#

Then, let #a# and #b# be unknowns,

#color(green)((x-10)/(x^2+x-2))=(x-10)/((x+2)(x-1))#
#color(white)(xxxxxx//x)=a/(x+2)+b/(x-1)#

Multiply throughout by #x^2+x-2#,

#x-10=a(x-1)+b(x+2)#

When #color(red)(x=1#,

#color(red)(1)-10=a(color(red)(1)-1)+b(color(red)(1)+2)#
#color(white)(xxx)3b=-9#
#color(white)(xxx3)b=-3#

When #color(blue)(x=-2#,

#color(blue)(-2)-10=a(color(blue)(-2)-1)+b(color(blue)(-2)+2)#
#color(white)(....)-3a=-12#
#color(white)(....-3)a=4#

Hence, substitute #a=4# and #b=-3#,

#(x^3-2x^2-4x-4)/(x^2+x-2)=x-3+4/(x+2)+3/(x-1)#