Question

Find the partial fraction decomposition.
(-22)/(6x^2+5x-4)

Answer 1

do you know the method to decompose ?

Answer 2

@km313 ?

Answer 3

no

Answer 4

Okay. il walk you through.. :) you ready ?

Answer 5

yes

Answer 6

\(\huge \frac{-22}{6x^2+5x-4} \)

Answer 7

lets factor out the denominator

Answer 8

\(\huge = \frac{-22}{6x^2+8x - 3x-4}\)

\(\huge = \frac{-22}{2x(3x+4) - 1(3x+4)}\)

\(\huge = \frac{-22}{(3x+4)(2x-1)}\)

so far ok ?

Answer 9

yeah

Answer 10

our goal is to write the given expression in the form as below :

\(\huge = \frac{A}{(3x+4)} + \frac{B}{(2x-1)}\)

Answer 11

lets equate our expression to above

Answer 12

\(\huge \frac{-22}{(3x+4)(2x-1)} = \frac{A}{(3x+4)}+\frac{B}{(2x-1)}\)

Answer 13

@km313 you still wid me... ?

Answer 14

yes

Answer 15

\(\huge = \frac{-22}{(3x+4)(2x-1) } = \frac{A(2x-1) + B(3x+4)}{(3x+4)(2x-1)}\)

Answer 16

above simplification makes sense to u ?

Answer 17

yeah

Answer 18

\(\huge = \frac{-22}{(3x+4)(2x-1) } = \frac{2Ax-1 + 3Bx+4B}{(3x+4)(2x-1)}\)

Answer 19

\(\huge = \frac{-22}{(3x+4)(2x-1) } = \frac{x(2A+3B) + (4B-1)}{(3x+4)(2x-1)}\)

Answer 20

we just isolated the "x" term.. did u see it ?

Answer 21

yep

Answer 22

lets compare "x" terms & constants

0 = 2A + 3B ---- (1)
-22 = 4B-1 -----(2)

Answer 23

so far ok.. ?

Answer 24

yes

Answer 25

good :)

solve (1) and (2) for A & B

Answer 26

can you do that and tell me the values of A & B ?

Answer 27

actually we did one mistake

Answer 28

we should get below equations
0 = 2A + 3B ---- (1)
-22 = 4B-A -----(2)

Answer 29

solving above two equations,
you should get : A = 6, B = -4

just plugin these values in our first equation, to get the final expression :
\(\huge \frac{6}{(3x+4)} + \frac{-5}{(2x-1)}\)

Answer 30

this is the final decomposed partial expression.
hope it helped you partially... .:))

Answer 31

thank you!

Answer 32

welcome... (: