# Irreducible Quadratic Factors (Partial Fractions)

• Feb 22nd 2009, 10:06 PM
wyhwang7
Hello, I've been reading the examples in my text over and over and I'm having trouble solving problems with irreducible quadratic factors.

The problem: Find the constants in the partial fraction decomposition

(2x-3)/(x-2)((x^2)+4) = A/(x-2) + (Bx+C)/((x^2)+4)

I solved for A

2x+4 = A((x^2)+4) + (Bx+C)(x-2)

Setting x=2, 8 = 8A + 0

A=1

Then, like in the example in my text, I distributed the equation and grouped by x's

2x + 4 = (x^2) + 4 + B(x^2) - 2B + Cx - 2C

0 = (x^2)(1+B) + x(C-2) - 2(B-C)

This is where I am lost. I do not know how to solve for B or C.

Thanks!
• Feb 22nd 2009, 11:23 PM
Thomas154321
Quote:

Originally Posted by wyhwang7
Hello, I've been reading the examples in my text over and over and I'm having trouble solving problems with irreducible quadratic factors.

The problem: Find the constants in the partial fraction decomposition

(2x-3)/(x-2)((x^2)+4) = A/(x-2) + (Bx+C)/((x^2)+4)

I solved for A

2x+4 = A((x^2)+4) + (Bx+C)(x-2)

Setting x=2, 8 = 8A + 0

A=1

Then, like in the example in my text, I distributed the equation and grouped by x's

2x + 4 = (x^2) + 4 + B(x^2) - 2Bx + Cx - 2C

0 = (x^2)(1+B) + x(C-2) - 2(B-C)

This is where I am lost. I do not know how to solve for B or C.

Thanks!

Missed the x there.

Gives

2x + 4 = (x^2)(1+B) + x(C-2B) - 2(C-2)

Comparing coeffts gives B = -1 and C = 0 which works for all 3 powers of x (x^2, x, const) so is correct.
• Feb 22nd 2009, 11:37 PM
Thomas154321
I want the number of x^2 on the right to be the same as on the left.

So (1+B) = 0 and B = -1

Then I want 4 = -2(C-2) which gives C = 0.

Then I check for x: 2 = C - 2B = 0 - (-1)*2 = 2 so it works.
• Feb 22nd 2009, 11:37 PM
wyhwang7
Quote:

Originally Posted by Thomas154321
Missed the x there.

Gives

2x + 4 = (x^2)(1+B) + x(C-2B) - 2(C-2)

Comparing coeffts gives B = -1 and C = 0 which works for all 3 powers of x (x^2, x, const) so is correct.

Thanks for replying! But I'm confused. How exactly did you find out that B = -1? That would get rid of the x^2, but what about x(C-2B), and how did you derive C = 0? The examples in my text do something like this too, but it doesn't say exactly how these values were derived. Is there a method?
• Feb 22nd 2009, 11:43 PM
wyhwang7
Quote:

Originally Posted by Thomas154321
I want the number of x^2 on the right to be the same as on the left.

So (1+B) = 0 and B = -1

Then I want 4 = -2(C-2) which gives C = 0.

Then I check for x: 2 = C - 2B = 0 - (-1)*2 = 2 so it works.

Oh, I see! I thought you had to move everything to the right to set the equation to zero.

Thanks so much!