How do you express #(x^3-x^2-5x) /( x^2-3x+2)# in partial fractions?

2 Answers
Jul 19, 2018

The answer is #=x+2+5/(x-1)-6/(x-2)#

Explanation:

As the degree of the numerator is greater than the degree of the denominator, perform a long division first

#color(white)(aaaa)##x^3-x^2-5x+0##color(white)(aaaa)##|##x^2-3x+2#

#color(white)(aaaa)##x^3-3x^2+2x##color(white)(aaaaaaa)##|##x+2#

#color(white)(aaaa)##0x^3+2x^2-7x+0#

#color(white)(aaaaaaaa)##+2x^2-6x+4#

#color(white)(aaaaaaaa)##+0x^2-x-4#

Therefore,

#(x^3-x^2-5x)/(x^2-3x+2)=x+2-(x+4)/(x^2-3x+2)#

Perform the decomposition into partial fractions

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

#=A/(x-1)+B/(x-2)#

#=(A(x-2)+B(x-1))/((x-1)(x-2))#

Compare the numerators

#x+4=A(x-2)+B(x-1)#

Let #x=1#, #=>#, #5=-A#

Let #x=2#, #=>#, #6=B#

Finally,

#(x^3-x^2-5x)/(x^2-3x+2)=x+2+5/(x-1)-6/(x-2)#

#\frac{x^3-x^2-5x}{x^2-3x+2}=x+2+5/{x-1}-6/{x-2}#

Explanation:

Given rational function:

#\frac{x^3-x^2-5x}{x^2-3x+2}#

#=x+2-\frac{x+4}{x^2-3x+2}#

#=x+2-\frac{x+4}{(x-1)(x-2)}#

#=x+2-(\frac{A}{x-1}+B/{x-2})#

Comparing corresponding coefficients to find #A=-5# & #B=6#

#=x+2-(\frac{-5}{x-1}+6/{x-2})#

#=x+2+5/{x-1}-6/{x-2}#