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

$\displaystyle \frac{5x^2+7x-9}{4-8x + x^2-2x^3}$

$\displaystyle \frac{5x^2+7x-9}{4(1-2x) + x^2(1-2x)}$

$\displaystyle \frac{5x^2+7x-9}{(4+x^2)(1-2x)}$

$\displaystyle \frac{5x^2+7x-9}{(4+x^2)(1-2x)} = \frac{A}{1-2x} + \frac{Bx+C}{x^2+4}$

$\displaystyle 5x^2+7x-9 = A(x^2+4) + (Bx+C)(1-2x)$

$\displaystyle 5x^2+7x-9 = Ax^2+4A + Bx - 2Bx^2 +C - 2Cx$

$\displaystyle 5x^2+7x-9 = (A-2B)x^2 + (B-2C)x + (4A+C)$

equating coefficients ...

$\displaystyle A-2B = 5$

$\displaystyle B-2C = 7$

$\displaystyle 4A+C = -9$

solving the system of equations ...

$\displaystyle A = -1 \, , \, B = -3 \, , \, C = -5$

$\displaystyle \frac{5x^2+7x-9}{(4+x^2)(1-2x)} = \frac{1}{2x-1} - \frac{3x+5}{x^2+4}$

Quote:

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)

I agree with this solution.

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

Quote:

Originally Posted by

**skeeter** $\displaystyle \frac{5x^2+7x-9}{4-8x + x^2-2x^3}$

$\displaystyle \frac{5x^2+7x-9}{4(1-2x) + x^2(1-2x)}$

$\displaystyle \frac{5x^2+7x-9}{(4+x^2)(1-2x)}$

$\displaystyle \frac{5x^2+7x-9}{(4+x^2)(1-2x)} = \frac{A}{1-2x} + \frac{Bx+C}{x^2+4}$

$\displaystyle 5x^2+7x-9 = A(x^2+4) + (Bx+C)(1-2x)$

$\displaystyle 5x^2+7x-9 = Ax^2+4A + Bx - 2Bx^2 +C - 2Cx$

$\displaystyle 5x^2+7x-9 = (A-2B)x^2 + (B-2C)x + (4A+C)$

equating coefficients ...

$\displaystyle A-2B = 5$

$\displaystyle B-2C = 7$

$\displaystyle 4A+C = -9$

solving the system of equations ...

$\displaystyle A = -1 \, , \, B = -3 \, , \, C = -5$

$\displaystyle \frac{5x^2+7x-9}{(4+x^2)(1-2x)} = \frac{1}{2x-1} - \frac{3x+5}{x^2+4}$

I agree with this solution.

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