How do you express (x^5-2x^4+x^3+x+5)/(x^3-2x^2+x-2)x5−2x4+x3+x+5x3−2x2+x−2 as partial fractions?
1 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)x5−2x4+x3+x+5x3−2x2+x−2=x2+3x−2−x+1x2+1
Explanation:
First note that:
x^3-2x^2+x-2 = (x-2)(x^2+1)x3−2x2+x−2=(x−2)(x2+1)
Is the numerator divisible by either of these factors?
It is not divisible by
How about
x^5-2x^4+x^3+x+5 = (x^2+1)(x^3-2x^2+2)+(x+3)x5−2x4+x3+x+5=(x2+1)(x3−2x2+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)x5−2x4+x3+x+5x3−2x2+x−2=x5−2x4+x3−2x2+2x2+x+5x3−2x2+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)x5−2x4+x3+x+5x3−2x2+x−2=x2+2x2+x+5x3−2x2+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)x5−2x4+x3+x+5x3−2x2+x−2=x2+Ax−2+Bx+Cx2+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)x5−2x4+x3+x+5x3−2x2+x−2=x2+A(x2+1)+(Bx+C)(x−2)x2−2x2+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)x5−2x4+x3+x+5x3−2x2+x−2=x2+(A+B)x2+(−2B+C)x+(A−2C)x2−2x2+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
-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)
