How do you express #(x^5-2x^4+x^3+x+5)/(x^3-2x^2+x-2)# as partial fractions?

1 Answer
Oct 15, 2016

Answer:

#(x^5-2x^4+x^3+x+5)/(x^3-2x^2+x-2) = x^2 + 3/(x-2) - (x+1)/(x^2+1)#

Explanation:

First note that:

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

Is the numerator divisible by either of these factors?

It is not divisible by #(x-2)# since #5# is odd.

How about #(x^2+1)# ?

#x^5-2x^4+x^3+x+5 = (x^2+1)(x^3-2x^2+2)+(x+3)#

So there are no common factors of the numerator and denominator.

Next note that:

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

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

#color(white)((x^5-2x^4+x^3+x+5)/(x^3-2x^2+x-2)) = x^2 + A/(x-2) + (Bx+C)/(x^2+1)#

#color(white)((x^5-2x^4+x^3+x+5)/(x^3-2x^2+x-2)) = x^2 + (A(x^2+1) + (Bx+C)(x-2))/(x^2-2x^2+x-2)#

#color(white)((x^5-2x^4+x^3+x+5)/(x^3-2x^2+x-2)) = x^2 + ((A+B)x^2+(-2B+C)x+(A-2C))/(x^2-2x^2+x-2)#

Equating coefficients we get the following system of linear equations:

#{(A+B=2), (-2B+C=1), (A-2C=5) :}#

Subtracting the third equation from the first, we get:

#B+2C = -3#

Adding twice this to the second equation, we find:

#5C = -5#

Hence:

#C = -1#

Substituting this value of #C# into the second equation we find:

#-2B-1=1#

Hence:

#B=-1#

Then from the first equation we find:

#A = 2-B = 3#

So:

#(x^5-2x^4+x^3+x+5)/(x^3-2x^2+x-2) = x^2 + 3/(x-2) - (x+1)/(x^2+1)#