Question

A[x(x-4)(x-3)]+B[x(x-4)(x-3)]+C[x(x-4)(x-3)]

Can someone please explain how to multiply this out?

Answer 1

I know the answer from the book is Ax^2-7Ax+12A but it doesnt make sense to me

Answer 2

\[x(x-4)(x-3)=x(x^2-3x-4x+12)=x(x^2-7x+12)\]

Answer 3

Does that helps?

Answer 4

yes but why doesnt the 12 at the end have an x in it? for the final answer i mean

Answer 5

The final answer would be:
x^3-7x^2+12x

Answer 6

according to my book that is incorrect, we are dealing with partial fractions here so its a little different

Answer 7

\[(x^3-7x^2+12x)(A+B+C)\]

Answer 8

What about that?

Answer 9

this is only a part of a larger problem, here is the original problem

Answer 10

write the partial faction decomposition of the rational expression : (5x^2-13x+12) / x(x-4)(x-3)

Answer 11

\[\frac{5x^2-13x+12}{x(x-4)(x-3)}\]
\[=\frac{A}{x}+\frac{B}{x-4}+\frac{C}{x-3}\]
\[=\frac{A(x-4)(x-3)+B(x)(x-3)+C(x)(x-4)}{x(x-4)(x-3)}\]

Answer 12

So then we have to simplify on the top and set it (only the numerator) equal to 5x^2-13x+12

Answer 13

Remember it?

Answer 14

okay show me the simplifying

Answer 15

Come on, give it a try, tell me what you get and I will tell you if it's ok or not, OK?

Answer 16

k 1 sec

Answer 17

x^2-7x+12+Bx^2-3x+Cx^2-4x ?

Answer 18

Ax^2*

Answer 19

hello?

Answer 20

let me a second, I'm having dinner.

Answer 21

:D

Answer 22

haha no worries, i just checked and it is correct. thanks for helping me see how to set it up

Answer 23

I'm back, Ok awsome!
Glad to help!
Thanks for the medal.
Have a wonderful day/night.

Answer 24

you too