Question

If I am doing partial fraction decomposition and I have (Ax+B)(x^2+4) +... = . Is A= 0

Answer 1

Seems like we are missing things in that equation above?
You have some dots and also and equal sign with nothing to follow.

Answer 2

Its a long equation. It starts with Ax^3 + 4Ax + my B's C's and D's this all = 1How can I solve for A if there is no A without an X?

Answer 3

You need to get all your x^3 together, all you x^2's together, and all your x's together, all your constants together.

Can you give me your original fraction at least?

Answer 4

use partial fractions: integral of 1/(x^4-16)

Answer 5

\[x^4-16=(x^2-4)(x^2+4)=(x-2)(x+2)(x^2+4) \]
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\[\frac{1}{x^4-16}=\frac{A}{x-2}+\frac{B}{x+2}+\frac{Cx+D}{x^2+4}\]
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Combine fractions to find A,B,C, and D.
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\[\frac{1}{x^4-16}=\frac{A(x+2)(x^2+4)+B(x-2)(x^2+4)+(Cx+D)(x-2)(x+2)}{(x-2)(x+2)(x^2+4)}\]
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So we have that the numerators have to be equal since the denominators are equal and these are after all equal fractions (by the equal sign).
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\[1=A(x+2)(x^2+4)+B(x-2)(x^2+4)+(Cx+D)(x-2)(x+2)\]

Answer 6

Did you get that far?

Answer 7

Yes. I solved for D and C by setting x=2 and x=-2. But I didn't see a way to do that with this one. I started to multiply the whole thing out but realized that I wouldn't get a A without an x.

Answer 8

\[1=x^3[A+B+C]+x^2[2A-2B+D]+x[4A+4B-4C]+[8A-8B-4D]\]
We should have
\[1=x^3(0)+x^2(0)+x(0)+1\]
This implies the following equations:
\[A+B+C=0\]
\[2A-2B+D=0\]
\[4A+4B-4C=0\]
\[8A-8B-4D=1\]

Answer 9

Oops, sorry I guess I didn't get that far. I got:\[(Ax+B)(x+2)(x-2)+C(x ^{2}+4)(x-2)+D(x ^{2}+4)(x+2)\]

Answer 10

How did you split up the A and the B?

Answer 11

if x=2, we have
\[1=A(2+2)(2^2+4)+B(2-2)(2^2+4)+(C(2)+D)(2-2)(2+2) => 1=A(4)(8)\]
if x=-2, we have
\[1=A(-2+2)((-2)^2+4)+B(-2-2)((-2)^2+4)+(C(-2)+D)(-2-2)(-2+2) \]
\[=> 1=B(-4)(8)\]
so we have
\[A=\frac{1}{32} \text{ and } B=\frac{-1}{32}\]
By my equation 1 you can find C.
\[A+B+C=0 => \frac{1}{32}+\frac{-1}{32}+C=0 \]
\[2(\frac{1}{32})-2\frac{-1}{32}+D=0\]

Answer 12

multiplication and combining like terms