Question

how do i find the partial fraction decompisition -20x/3x^2+2x-8

Answer 1

does the denominator factor?

Answer 2

oh yes, it does
\[-\frac{20x}{(3x-4)(x+2)}=\frac{a}{3x-4}+\frac{b}{x+2}\]
the minus sign out front is kind of annoying, it would be easier to compute
\[\frac{20x}{(3x-4)(x+2)}=\frac{a}{3x-4}+\frac{b}{x+2}\] then stick a minus sign outside the result

Answer 3

what to do it the snap way?

Answer 4

what do i do with the numerator?

Answer 5

not sure what you mean
your job is to find \(a\) and \(b\)

Answer 6

oh i see, okay thank you

Answer 7

do you know how to do it? there is a rather easy method for this one

Answer 8

i would love to learn the easier method because i have no clue what to do

Answer 9

ok lets do it the donkey way first, then we can do it the snap way. it amounts to the same thing

Answer 10

okay

Answer 11

\[\frac{20x}{(3x-4)(x+2)}=\frac{a}{3x-4}+\frac{b}{x+2}\] is the start. now add up on the right
pay no attention to the denominator because they will be the same
you get
\[20x=a(x+2)+b(3x-4)\]
this has to be true not matter what \(x\) is , so pick \(x=-2\) to solve for \(b\)
literally you get
\[20\times -2=a(-2+2)+b(3\times -2-4)\] but you should really start with
\[-40=b(-6-4)\] this tells you \(-40=-10b\implies b=10\)

Answer 12

oops that was wrong
\[-40=-10b\implies b=4\] sorry

Answer 13

now repeat the process, this time with the value of \(x\) that will make \(3x-4=0\) namely pick \(x=\frac{4}{3}\)

Answer 14

ok, im a little confused, how did you get rid of the (3x-4)(x+2) that was in the denomenator earlier, with the 20x in the numerator, to get 20x= a(3x-4)+b(x+2)

Answer 15

\[\begin{array}{l} \text{factor the denominator into} \\ (x+2) (3 x-4) \\ \text{Then the partial fraction expansion is of the form} \\ -\frac{20 x}{(2+x) (-4+3 x)}=\frac{A}{x+2}+\frac{B}{3 x-4} \\ \text{Multiply both sides by }(x+2) (3 x-4)\text{ and simplify} \\ -20 x=A (3 x-4)+B (x+2) \\ \text{You need to make either A or B zero, so, try let x=-2,B will be zero,} \\ -20(-2)=A(3(-2)-4) \\ A=-4 \\ \text{Now do the same for A, let x=4/3, Then A will be zero, solve for B}\\\end{array}\]

Answer 16

its the simplifying that im struggling with, it seems that everytime i simplify i end up with -20x/(3x-4)(x+2)=a(x+2)+b(3x-4)/3(x-4)(x+2)

Answer 17

You just multiply both sides by (2+x)(-4+3x)

Answer 18

okay thanks

Answer 19

\[-\frac{20 x}{(2+x) (-4+3 x)}=\frac{A}{x+2}+\frac{B}{3 x-4}\]
\[-\frac{20 x}{(2+x) (-4+3 x)}(2+x) (-4+3 x)=\frac{A}{x+2}(2+x) (-4+3 x)+\frac{B}{3 x-4}(2+x) (-4+3 x)\]
\[-\frac{20 x}{\cancel{(2+x) (-4+3 x)}}\cancel{(2+x) (-4+3 x)}=\frac{A}{\cancel{x+2}}\cancel{(2+x)} (-4+3 x)+\frac{B}{\cancel{3 x-4}}(2+x) \cancel{(-4+3 x)}\]
\[-20x=A(-4+3x)+B(2+x)\]

Answer 20

thank you soooo much , i get it now.