# Basic Algebra?

• May 6th 2009, 10:18 AM
Fallen186
Basic Algebra?
Problem
$\displaystyle \int\frac{10x-2x^2}{(x-1)^2(x+3)}$
I did all the steps correctly until this part

A+C = -2
2A+ B -2C = 10
-3A +3B + C = 0

Attempt:
I said C = -A-2
B = -2A +2C + 10

So
-3A + 3(2A + 2(-A-2) + 10) + (-A-2) = 0
-3A + 3(2A -2A -4 + 10) -A-2 = 0
-3A + -12 +30 -A - 2 = 0
-4A +16 = 0
-4A = -16

A = 4

A= 1, B= 2, C= -3

*This is related to partial fractions in Calculus.
• May 6th 2009, 10:31 AM
running-gag
Hi

So
-3A + 3(-2A + 2(-A-2) + 10) + (-A-2) = 0
-3A + 3(-2A -2A -4 + 10) -A-2 = 0
-3A + -12A -12 +30 -A - 2 = 0
-16A +16 = 0
-16A = -16

A = 1
• May 6th 2009, 10:31 AM
masters
Quote:

Originally Posted by Fallen186
Problem
$\displaystyle \int\frac{10x-2x^2}{(x-1)^2(x+3)}$
I did all the steps correctly until this part

A+C = -2
2A+ B -2C = 10
-3A +3B + C = 0

Attempt:
I said C = -A-2
B = -2A +2C + 10

So
-3A + 3(-2A + 2(-A-2) + 10) + (-A-2) = 0
-3A + 3(-2A -2A -4 + 10) -A-2 = 0
-3A + -12 +30 -A - 2 = 0
-4A +16 = 0
-4A = -16

A = 4

A= 1, B= 2, C= -3

*This is related to partial fractions in Calculus.

You missed a sign at the beginning (See chage in red). See if that fixes it. I didn't bother to work it out for you.
• May 6th 2009, 10:33 AM
skeeter
Quote:

Originally Posted by Fallen186
Problem
$\displaystyle \int\frac{10x-2x^2}{(x-1)^2(x+3)}$

A= 1, B= 2, C= -3

*This is related to partial fractions in Calculus.

$\displaystyle \frac{10x-2x^2}{(x-1)^2(x+3)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+3}$

$\displaystyle 10x-2x^2 = A(x-1)(x+3) + B(x+3) + C(x-1)^2$

$\displaystyle x = 1$ ... $\displaystyle 4B = 8$ ... $\displaystyle B = 2$

$\displaystyle x = -3$ ... $\displaystyle 16C = -48$ ... $\displaystyle C = -3$

$\displaystyle x = 0$ ... $\displaystyle -3A + 3B + C = 0$ ... $\displaystyle -3A + 6 - 3 = 0$ ... $\displaystyle A = 1$