Advanced Math/Partial Fractions
Hey , Im struggling on a few questions and was wondering if you could help me out so that i can see how it done correctly, the question is on Partial Fractions and reads
Resolve the following into partial fractions
Any help would be appreciated!
Try to write the fraction as a sum of fractions over each of the factors of the denominator.
(6x²-24x-3)/(x+4)(x+1)(2x-1) = a/(x+4) + b/(x+1) + c(2x-1)
where a,b,c are constants we need to find .
Add the fractions on the right side, then equate numerators from both sides.
6x²-24x-3 = a(x+1)(2x-1) + b(x+4)(2x-1) + c(x+4)(x+1)
set coefficients of the same powers of x from the left and right sides equal to each other
2a+2b+c = 6
a+7b+5c = -24
-a-4b+4c = 3
Solve the system and get
a = 65/9 , b = -11/3 , c = -10/9
(6x²-24x-3)/(x+4)(x+1)(2x-1) = 65/9(x+4) - 11/3(x+1) - 10/9(2x-1)
You can always start by getting the degree of the numerator smaller than the degree of the denominator , if it isn't already.
(3x²-8x-63)/(x²-3x-10) = [3(x²-3x-10) + x-33]/(x²-3x-10) = 3 + (x-33)/(x²-3x-10)
So we want to write (x-33)/(x²-3x-10) as a sum of partial fractions.
Procede as in part a)
(x-33)/(x²-3x-10) = (x-33)/(x-5)(x+2) = a/(x-5) + b/(x+2)
a(x+2) + b(x-5) = x-33
a+b = 1
2a-5b = -33
a=-4 , b=5
(3x²-8x-63)/(x²-3x-10) = 3 - 4/(x-5) + 5/(x+2)