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ryanmk54

  • 2 years ago

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

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  1. myininaya
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    use partial fractions: integral of 1/(x^4-16)

  5. myininaya
    • 2 years ago
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    \[x^4-16=(x^2-4)(x^2+4)=(x-2)(x+2)(x^2+4) \] --- \[\frac{1}{x^4-16}=\frac{A}{x-2}+\frac{B}{x+2}+\frac{Cx+D}{x^2+4}\] --- 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
    • 2 years ago
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    Did you get that far?

  7. ryanmk54
    • 2 years ago
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    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
    • 2 years ago
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    \[1=x^3[A+B+C]+x^2[2A-2B+D]+x[4A+4B-4C]+[8A-8B-4D]\] 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
    • 2 years ago
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    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
    • 2 years ago
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    How did you split up the A and the B?

  11. myininaya
    • 2 years ago
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    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
    • 2 years ago
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    multiplication and combining like terms

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