Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

ryanmk54

  • 10 months ago

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

  • This Question is Closed
  1. myininaya
    • 10 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 10 months ago
    Best Response
    You've already chosen the best response.
    Medals 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?

  3. myininaya
    • 10 months ago
    Best Response
    You've already chosen the best response.
    Medals 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?

  4. ryanmk54
    • 10 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  5. myininaya
    • 10 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[x^4-16=(x^2-4)(x^2+4)=(x-2)(x+2)(x^2+4) \] --- \[\frac{1}{x^4-16}=\frac{A}{x-2}+\frac{B}{x+2}+\frac{Cx+D}{x^2+4}\] --- 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
    • 10 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Did you get that far?

  7. ryanmk54
    • 10 months ago
    Best Response
    You've already chosen the best response.
    Medals 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. myininaya
    • 10 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[1=x^3[A+B+C]+x^2[2A-2B+D]+x[4A+4B-4C]+[8A-8B-4D]\] 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
    • 10 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 10 months ago
    Best Response
    You've already chosen the best response.
    Medals 0

    How did you split up the A and the B?

  11. myininaya
    • 10 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 10 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    multiplication and combining like terms

  13. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.