Express 5x^2+7x-9/4-8x+x^2-2x^3 in partial fraction

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

Re: Express 5x^2+7x-9/4-8x+x^2-2x^3 in partial fraction

Re: Express 5x^2+7x-9/4-8x+x^2-2x^3 in partial fraction

Quote:

Originally Posted by

**skeeter**

Thank you so much.

I'm assuming that the answer from my assessment book is wrong, am I right? I can't seems to find any way to achieve that answer