I'm trying to find the partial fraction decomposition of:
10x^2+2x-2/x^2(x+1)
I already found B=-2 and C=6
Plugging in the numbers, I get:
10x^2+2x-2=Ax(x+1)+(-2)(x+1)+(6)x^2
How do I solve this for A? Maybe this is just some algebra that I'm rusty on. But I'm having trouble with this. Help please?
so far, so good. You now have
10x^2+2x-2 = Ax(x+1)+(-2)(x+1)+(6)x^2
10x^2+2x-2 = Ax^2+Ax-2x-2+6x^2
10x^2+2x-2 = (A+6)x^2 - (A-2)x - 2
for those two polynomials to be identical, all the coefficients must match. That is
A+6 = 10
A-2 = 2
So, A=4
Luckily, those two equations give the same solution for A. I'm just curious how you found B and C without seeing how to get A. B I can see, since you wound up with
(A+C)x^2 + (A+B)x + B
To get B, I plugged in A=0. To get C I plugged in B=-1
No, I plugged in X=0 and X=-1
To solve for the unknown coefficient A, let's simplify the equation you've written:
10x^2 + 2x - 2 = Ax(x + 1) - 2(x + 1) + 6x^2
Now, let's combine like terms:
10x^2 + 2x - 2 = Ax^2 + Ax - 2x - 2 - 2 + 6x^2
Simplify further by rearranging the terms and combining like terms:
10x^2 + 2x - 2 = (A + 6)x^2 + (A - 2)x - 4
Since the coefficients of the x^2 terms on both sides of the equation must be equal, we have:
A + 6 = 10
Solving this equation for A:
A = 10 - 6
A = 4
Therefore, the value of A in the partial fraction decomposition is A = 4.
To summarize, to solve for A, you set the coefficients of x^2 on both sides of the equation equal to each other and solve for A.