equating coefficients ...
solving the system of equations ...
I agree with this solution.Answer: 5x^2+7x-9/4-8x+x^2-2x^3 = -1/(1-2x) - 3x+5/(x^2+4)
However, the answer from me assessment book is -(1/(2x-1)-(3x+5)/(x^2+4)
Hi guys,
as stated above, express 5x^2+7x-9/4-8x + x^2-2x^3 in partial fraction
My workings:
5x^2+7x-9/4-8x+x^2-2x^3 = 5x^2+7x-9/4(1-2x)+x^2(1-2x)
= 5x^2+7x-9/(4+x^2)(1-2x)
= A/(1-2x) + (Bx+C)/(4+x^2)
= A(4+x^2)+(Bx+C)(1-2x)/4-8x+x^2-2x^3
5x^2+7x-9 = A(4+x^2)+(Bx+C)(1-2x)
Sub x=1/2
5(1/2)^2+7(1/2)-9=A(4+(1/2)^2)
-17/4=(17/4)A
A=-1
Sub x=0
-9=-1(4)+(C)(1)
C=-5
Sub x=1
5(1)^2+7(1)-9=-1(5)+(B-5)(-1)
3=-5-B+5
B=-3
Answer: 5x^2+7x-9/4-8x+x^2-2x^3 = -1/(1-2x) - 3x+5/(x^2+4)
However, the answer from me assessment book is -(1/(2x-1)-(3x+5)/(x^2+4)
What had i done wrongly?
Thanks a lot
equating coefficients ...
solving the system of equations ...
I agree with this solution.Answer: 5x^2+7x-9/4-8x+x^2-2x^3 = -1/(1-2x) - 3x+5/(x^2+4)
However, the answer from me assessment book is -(1/(2x-1)-(3x+5)/(x^2+4)