Question

find the partial fraction decomp of (5x+3)/(X^2-3x-10)

Answer 1

Woo partial fractionss

Answer 2

Alright so first you want to find the (x-a)(x-b) form of the bottom)

Answer 3

After which you can assume that (5x+3)/(x^2-3x-10) = A/(x-5) + B/(x+2)

Answer 4

Next you want to multiply everything by the x^2 denominator term to get to
5x+3=A(x+2) + B(x-5)

Answer 5

Plug in x=-2, x=5 to get your terms for A and B respectively and put them in to the A/(x-5) + B/(x+2) and there's your answer:)

Answer 6

im lost sry I was out for a month with menijitis

Answer 7

Oh okay

Answer 8

so its 7/(x-5) + 2/(x+2) ?

Answer 9

Going off of where you have things now. When x = -2, you have:
5(-2) + 3 = A(-2+2) + B(-2-5)
-7 = -7B
B = 1
When x is 5 you have:
5(5) + 3 = A(5+2) + B(5-5)
28 = 7A
A = 4

So using those values for A and B, they are plugged into the first fractions that were invented involving A and B, namely the A/(x-5) and B/(x+2), so this means the decomp is

4/(x-5) + 1/(x+2)

Answer 10

you want a quick way to do this?

Answer 11

yes plz

Answer 12

\[\frac{5x+3}{(x-5)(x+2)}=\frac{A}{x-5}+\frac{B}{x+2}\]
if \(x-5=0\) then \(x=5\) put your hand over the factor of \(x-5\) and replace \(x\) by \(5\) in the left hand side \[\frac{5\times 5+3}{\cancel{(x-5)}(5+2)}\]
\[=\frac{28}{7}=4\]

Answer 13

you can do it in your head with a minimum amount of practice
repeat with \(x=-2\)
\[\frac{5\times (-2)+3}{(-2-5)\cancel{(x+2)}}=\frac{-7}{-7}=1\]

Answer 14

Cover-up method, hehe.

Answer 15

thanks a million guys

Answer 16

just make sure to remember which is A and which is B, it doesn't matter, but you have to keep track
also this method only works for simple linear denominator