1. ## fraction decomposition

(3x^2+7) / ( (x-1)(x^2+2x+5) )

2. Originally Posted by cityismine
(3x^2+7) / ( (x-1)(x^2+2x+5) )
Your set up will look like this,

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

Clearing the denominators gives us,

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

Test a couple of values of x like x=1 to solve for A , B and C

Your solution should come out as,

$A=\frac{5}{4} \\\ B= \frac{7}{4} \\\ and \\\ C=-\frac{3}{4}$

Which would give you,

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

Hope that helps!!

3. I figured it out, my answer matches yours, so it's right. thx,