PARTIAL FRACTION EXPANSION

[tawny jungle]

Could you please help me figure out where my working went wrong in partial fraction expansion? Here is my working for $ \frac{3x-2}{x^2-5x-24}$. First factor $x^2-5x-24$ that is (x+3)(x-8). Then make $\frac{A}{factor} + \frac{B}{factor}$, $\frac{A}{x+3} + \frac{B}{x-8}$ that is $\frac{Ax-8A+Bx+3B}{(x+3)(x-8)}$ then $\frac{Ax-8A+Bx+3B}{(x+3)(x-8)}=\frac{A}{x+3} + \frac{B}{x-8}$ then multiply both sides by (x+3)(x-8), Ax-8A+Bx+3B=A(x-8)+B(x+3). To solve for A you make x=-3 so plug that into the whole equation, 8A+-3B+3B=A(-11)+B(0). Then simplify both sides. -3A-8A+-3B+3B=A(-11)+B(0), -11A=-11A. Finally, divide both sides by negative eleven, A+A? Then for solving for B make x=8, plug that into the equation, 8A-8A+8B+3B=A(8-8)+B(8+3), simplify 11B=11B so then divide both sides by 11, B=B? Plug that in, $ \frac{3x-2}{x^2-5x-24}=\frac{A}{x+3} + \frac{B}{x-8}$ I went horribly wrong somewhere here could you please tell me where I went wrong and If you want do the problem with me.

[ocean gull]

kangaroo rat

[limber badger]

@tawny jungle

$\frac{Ax-8A+Bx+3B}{(x+3)(x-8)}=\frac{A}{x+3} + \frac{B}{x-8}$

[ocean gull]

k

[limber badger]

why did you do this step?

it is known that both sides are equal

you need to do $\frac{Ax-8A+Bx+3B}{(x+3)(x-8)}=\frac{3x-2}{(x+3)(x-8)}$

[ocean gull]

k

[tawny jungle]

Oh

[tawny jungle]

Is that the only mistake?

[limber badger]

well, everything after that is wrong....

[tawny jungle]

Yes

[limber badger]

because you were simplifying the wrong thing

[tawny jungle]

Yea

[tawny jungle]

Thanks

[tawny jungle]

@limber badger Here is my working for $ \frac{-x+3}{x^2-9x+20} $. First factor $x^2-9x+20$ which is (x-4)(x-5). $\frac{A}{x-5}+\frac{B}{x-4}=\frac{ax-5a+bx-4b}{(x-4)(x-5)}$ then $\frac{ax-5a+bx-4b}{(x-4)(x-5)}=$ $\frac{-x+3}{(x-4)(x-5)}$ to get rid of the fraction multiply both sides by (x-4)(x-5) that is $ax-5a+bx-4b=-x+3$ solve for variables $b=\frac{-x+3-ax+5a}{x-4}$, $a=\frac{-x+3-bx+4b}{x-5}$. Where did I go wrong (if it was in solving for variables please solve for the variable with me). Thanks...

[ocean gull]

kangaroo rat

[limber badger]

that is wrong

[limber badger]

that part is wrong i mean

[tawny jungle]

How!!!

[violet cloud]

$\frac{a}{x - 4} + \frac{b}{x - 5} = \frac{a(x - 5) + b(x - 4)}{(x - 4)(x - 5)} = \frac{ax - 5a + bx - 4b}{(x - 4)(x - 5)}$

[ocean gull]

Mizere

[violet cloud]

@tawny jungle

[limber badger]

a/(x-4)+b/(x-5) is not equal to a/(x-5)+b/(x-4)

take note of the denominator

[tawny jungle]

ye that is what i did

@violet cloud @limber badger was i not supposed to simplify

[limber badger]

$\frac{A}{x-5}+\frac{B}{x-4}=\frac{-x+3}{(x-5)(x-4)}\newline$ Multiply both sides by $(x-5)(x-4)$ and you get $\newline A(x-4)+B(x-5)=-x+3\newline x(A+B)+(-4A-5B)=-x+3\newline$ Now compare coefficients $\newline A+B=-1\newline -4A-5B=3\newline$ which you can easily solve

[ocean gull]

k

[tawny jungle]

So when you are doing the solving for A and B you are not using $\frac{ax-5a+bx-4b}{(x-4)(x-5)}$ so why add $\frac{A}{x-5}+\frac{B}{x-4}$.

[ocean gull]

kangaroo rat

[tawny jungle]

@limber badger

[limber badger]

first of all, the two things you have written in this message are not equal

[tough sonnet]

Innit

[limber badger]

..... not according to me

A=b and B=a is not a natural thing to do

[tough sonnet]

So you contradict yourself eh

[limber badger]

nope

[tough sonnet]

I said innit to you means I'm agreeing with you

[limber badger]

I assumed the opposite

[tough sonnet]

Then I hope you get contradiction to your assumption someday

[limber badger]

I just did

[tough sonnet]

NZ lost yesterday

How disappointing

[limber badger]

.....

from pakistan out of all....

the team that lost from zimbabwe

[tough sonnet]

Yes 🤡

Now we defeat them in finals

[limber badger]

inshallah

[tawny jungle]

r u from pakistan

I go for NZ I am part New Zealander

[limber badger]

nah¿the guy's indian, I'm in NZ

[tawny jungle]

nice

where? (in nz)

[tawny jungle]

So anyway, how do you explain this???

[limber badger]

this is correct

[tawny jungle]

Is symbolab scamming or something?

[tawny jungle]

So y u say it isn't

[tawny jungle]

.

[limber badger]

I'm saying what you were writing was incorrect

[limber badger]

yes

[tawny jungle]

🧐

[tawny jungle]

wdym?

[limber badger]

a/(x-5) and a/(x-4) are different things

[tawny jungle]

oh shoot

t

lol

I didn't see that

[tawny jungle]

So what should have I done (From the beginning)?

If your not to busy

[limber badger]

^

I've already written it

[tawny jungle]

oh okk

thanks

I am just realizing how tiered I must have been earlier lol

[tawny jungle]

wdym by compare coefficients?

@limber badger

[limber badger]

well, x(A+B)=-x, so A+B=-1

like, the coefficient in front of the x, and the constant terms must equal each other

[tawny jungle]

what u do here?

[limber badger]

do you agree that x(A+B)+(-4A-5B)=-x+3?

[tawny jungle]

no?

[limber badger]

..... shit

hm, then we need to get you to agree

till which part do you agree

[tawny jungle]

I distributed x, Ax+Bx-4A-5b

@limber badger

I might need to go eat soon

[limber badger]

that is correct

[tawny jungle]

i GTg

thanks

[limber badger]

now, AREn"t THeSE The SaMe THIng

[tawny jungle]

c ya

soon