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

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myininaya
 10 months ago
Best ResponseYou've already chosen the best response.1Seems like we are missing things in that equation above? You have some dots and also and equal sign with nothing to follow.

ryanmk54
 10 months ago
Best ResponseYou've already chosen the best response.0Its 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?

myininaya
 10 months ago
Best ResponseYou've already chosen the best response.1You 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?

ryanmk54
 10 months ago
Best ResponseYou've already chosen the best response.0use partial fractions: integral of 1/(x^416)

myininaya
 10 months ago
Best ResponseYou've already chosen the best response.1\[x^416=(x^24)(x^2+4)=(x2)(x+2)(x^2+4) \]  \[\frac{1}{x^416}=\frac{A}{x2}+\frac{B}{x+2}+\frac{Cx+D}{x^2+4}\]  Combine fractions to find A,B,C, and D.  \[\frac{1}{x^416}=\frac{A(x+2)(x^2+4)+B(x2)(x^2+4)+(Cx+D)(x2)(x+2)}{(x2)(x+2)(x^2+4)}\]  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).  \[1=A(x+2)(x^2+4)+B(x2)(x^2+4)+(Cx+D)(x2)(x+2)\]

myininaya
 10 months ago
Best ResponseYou've already chosen the best response.1Did you get that far?

ryanmk54
 10 months ago
Best ResponseYou've already chosen the best response.0Yes. 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.

myininaya
 10 months ago
Best ResponseYou've already chosen the best response.1\[1=x^3[A+B+C]+x^2[2A2B+D]+x[4A+4B4C]+[8A8B4D]\] We should have \[1=x^3(0)+x^2(0)+x(0)+1\] This implies the following equations: \[A+B+C=0\] \[2A2B+D=0\] \[4A+4B4C=0\] \[8A8B4D=1\]

ryanmk54
 10 months ago
Best ResponseYou've already chosen the best response.0Oops, sorry I guess I didn't get that far. I got:\[(Ax+B)(x+2)(x2)+C(x ^{2}+4)(x2)+D(x ^{2}+4)(x+2)\]

ryanmk54
 10 months ago
Best ResponseYou've already chosen the best response.0How did you split up the A and the B?

myininaya
 10 months ago
Best ResponseYou've already chosen the best response.1if x=2, we have \[1=A(2+2)(2^2+4)+B(22)(2^2+4)+(C(2)+D)(22)(2+2) => 1=A(4)(8)\] if x=2, we have \[1=A(2+2)((2)^2+4)+B(22)((2)^2+4)+(C(2)+D)(22)(2+2) \] \[=> 1=B(4)(8)\] so we have \[A=\frac{1}{32} \text{ and } B=\frac{1}{32}\] By my equation 1 you can find C. \[A+B+C=0 => \frac{1}{32}+\frac{1}{32}+C=0 \] \[2(\frac{1}{32})2\frac{1}{32}+D=0\]

myininaya
 10 months ago
Best ResponseYou've already chosen the best response.1multiplication and combining like terms
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