## ryanmk54 Group Title If I am doing partial fraction decomposition and I have (Ax+B)(x^2+4) +... = . Is A= 0 4 months ago 4 months ago

1. myininaya Group Title

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

2. ryanmk54 Group Title

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?

3. myininaya Group Title

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?

4. ryanmk54 Group Title

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

5. myininaya Group Title

$x^4-16=(x^2-4)(x^2+4)=(x-2)(x+2)(x^2+4)$ --- $\frac{1}{x^4-16}=\frac{A}{x-2}+\frac{B}{x+2}+\frac{Cx+D}{x^2+4}$ --- Combine fractions to find A,B,C, and D. --- $\frac{1}{x^4-16}=\frac{A(x+2)(x^2+4)+B(x-2)(x^2+4)+(Cx+D)(x-2)(x+2)}{(x-2)(x+2)(x^2+4)}$ --- 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). --- $1=A(x+2)(x^2+4)+B(x-2)(x^2+4)+(Cx+D)(x-2)(x+2)$

6. myininaya Group Title

Did you get that far?

7. ryanmk54 Group Title

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.

8. myininaya Group Title

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

9. ryanmk54 Group Title

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

10. ryanmk54 Group Title

How did you split up the A and the B?

11. myininaya Group Title

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. $A+B+C=0 => \frac{1}{32}+\frac{-1}{32}+C=0$ $2(\frac{1}{32})-2\frac{-1}{32}+D=0$

12. myininaya Group Title

multiplication and combining like terms