# decompose fraction question

• May 28th 2008, 03:58 PM
cityismine
decompose fraction question
(3x^2+7) / ((x-1)(x^2+2x+5))
• May 28th 2008, 05:41 PM
Jonboy
what's stumpin' ya?
• May 28th 2008, 08:33 PM
cityismine
I can decompose fractions of the type A/(x-1)+ B/(x^2+2x+5) but I'm having trouble with fractions of the type A/(x-1)+(Bx+c)(x^2+2x+5). Can someone show me a step by step method of decomposing fractions of the (Bx+C) type?
• May 28th 2008, 09:59 PM
cityismine
Finally figured it out. Answer is 5/(4(x-1)) + (7x-3)/(4(x^2+2x+5)). Hope I'm right.
• May 28th 2008, 10:40 PM
Chris L T521
Quote:

Originally Posted by cityismine
(3x^2+7) / ((x-1)(x^2+2x+5))

Quote:

Originally Posted by cityismine
Finally figured it out. Answer is 5/(4(x-1)) + (7x-3)/(4(x^2+2x+5)). Hope I'm right.

$\frac{3x^2+7}{(x-1)(x^2+2x+5)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+2x+5}$

Multiply both sides by the denominator of the left term:

$3x^2+7=A(x^2+2x+5)+(Bx+C)(x-1)$

Now, we get:

$3x^2+7=Ax^2+2Ax+5A+Bx^2-Bx+Cx-C$

$3x^2+7=(A+B)x^2+(2A-B+C)x+5A-C$

We now compare the coefficients:

\begin{aligned}
A+B &=3 \\
2A-B+C &=0 \\
5A-C &=7
\end{aligned}

From the first equation, we see that $B=3-A$. From the third equation, we see that $C=5A-7$. Now substitute these values for B and C into the second equation:

$2A-(3-A)+(5A-7)=0 \implies 8A-10=0\implies \color{red}\boxed{A=\frac{5}{4}}$

Thus, $\color{red}\boxed{C=-\frac{3}{4}}$ and $\color{red}\boxed{B=\frac{7}{4}}$

Thus the PFD is:

$\frac{\frac{5}{4}}{x-1}+\frac{-\frac{3}{4}x+\frac{7}{4}}{x^2+2x+5}\implies \color{red}\boxed{\frac{5}{4(x-1)}+\frac{7-3x}{4(x^2+2x+5)}}$.

It's Good!! :D