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

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

ryanmk54 Group TitleBest ResponseYou've already chosen the best response.0
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?
 8 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
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?
 8 months ago

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

myininaya Group TitleBest 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).  \[1=A(x+2)(x^2+4)+B(x2)(x^2+4)+(Cx+D)(x2)(x+2)\]
 8 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Did you get that far?
 8 months ago

ryanmk54 Group TitleBest ResponseYou've already chosen the best response.0
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 months ago

myininaya Group TitleBest 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: \[A+B+C=0\] \[2A2B+D=0\] \[4A+4B4C=0\] \[8A8B4D=1\]
 8 months ago

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

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

myininaya Group TitleBest ResponseYou've already chosen the best response.1
if 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. \[A+B+C=0 => \frac{1}{32}+\frac{1}{32}+C=0 \] \[2(\frac{1}{32})2\frac{1}{32}+D=0\]
 8 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
multiplication and combining like terms
 8 months ago
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