anonymous
  • anonymous
If I am doing partial fraction decomposition and I have (Ax+B)(x^2+4) +... = . Is A= 0
Calculus1
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
myininaya
  • myininaya
Seems like we are missing things in that equation above? You have some dots and also and equal sign with nothing to follow.
anonymous
  • anonymous
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?
myininaya
  • myininaya
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?

Looking for something else?

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

More answers

anonymous
  • anonymous
use partial fractions: integral of 1/(x^4-16)
myininaya
  • myininaya
\[x^4-16=(x^2-4)(x^2+4)=(x-2)(x+2)(x^2+4) \] --- \[\frac{1}{x^4-16}=\frac{A}{x-2}+\frac{B}{x+2}+\frac{Cx+D}{x^2+4}\] --- 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). --- \[1=A(x+2)(x^2+4)+B(x-2)(x^2+4)+(Cx+D)(x-2)(x+2)\]
myininaya
  • myininaya
Did you get that far?
anonymous
  • anonymous
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.
myininaya
  • myininaya
\[1=x^3[A+B+C]+x^2[2A-2B+D]+x[4A+4B-4C]+[8A-8B-4D]\] We should have \[1=x^3(0)+x^2(0)+x(0)+1\] This implies the following equations: \[A+B+C=0\] \[2A-2B+D=0\] \[4A+4B-4C=0\] \[8A-8B-4D=1\]
anonymous
  • anonymous
Oops, sorry I guess I didn't get that far. I got:\[(Ax+B)(x+2)(x-2)+C(x ^{2}+4)(x-2)+D(x ^{2}+4)(x+2)\]
anonymous
  • anonymous
How did you split up the A and the B?
myininaya
  • myininaya
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. \[A+B+C=0 => \frac{1}{32}+\frac{-1}{32}+C=0 \] \[2(\frac{1}{32})-2\frac{-1}{32}+D=0\]
myininaya
  • myininaya
multiplication and combining like terms

Looking for something else?

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