Question

Someone refresh my memory on partial fractions?

Answer 1

\(\Huge \dfrac{1}{(x-a)(x-b)}=\dfrac{A}{x-a}+\dfrac{B}{x-b}\)

Answer 2

\(\Huge \dfrac{1}{x^2(x-a)}=\dfrac{A}{x}+\dfrac{B}{x^2}+\dfrac{C}{x-a}\)

Answer 3

any particular problem ?

Answer 4

Let's try this one:
\[\LARGE \frac{3x+5}{x^2+3x+2}\]

Answer 5

factor the denom

Answer 6

\[\LARGE \frac{3x+5}{(x+2)(x+1)}\]right?

Answer 7

yes, ofcourse
\(\huge \dfrac{A}{x+1}+\dfrac{B}{x+2} \\ \Large 3x+5 =...? \)

Answer 8

\(\Large 3x+5 = A(x+2)+B(x+1)\)
makes sense ?

Answer 9

So set them equal, get common denominators, then cancel. So yea

Answer 10

correct.
now that expression is true for any value of x
so, to find A we can out x=-1
what do u get then ?

Answer 11

can put**

Answer 12

Oh, this is that one method I never learned:
\[3(-1)+5=A(-1+2)+B(-1+1)\]
\[-3+5=A(1)+B(0)\]
So \[2=A\]
?

Answer 13

correct! :)
what will you put x =... ?
to find B ?
then B=...?

Answer 14

-2?

Answer 15

yes, put x=-2, find B

Answer 16

\[3(-2)+5=A(-2+2)+B(-2+1)\]
\[-6+5=A(0)+B(-1)\]
\[-1=-B\]
\[1=B\]

Answer 17

correct! :)
so, \(\Huge \dfrac{3x+5}{x^2+3x+2}=\dfrac{2}{x+1}+\dfrac{1}{x+2}\)

Answer 18

Just like I thought, thanks hartnn :)

Answer 19

welcome ^_^