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?
I know the answer from the book is Ax^2-7Ax+12A but it doesnt make sense to me
\[x(x-4)(x-3)=x(x^2-3x-4x+12)=x(x^2-7x+12)\]
Does that helps?
yes but why doesnt the 12 at the end have an x in it? for the final answer i mean
The final answer would be: x^3-7x^2+12x
according to my book that is incorrect, we are dealing with partial fractions here so its a little different
\[(x^3-7x^2+12x)(A+B+C)\]
What about that?
this is only a part of a larger problem, here is the original problem
write the partial faction decomposition of the rational expression : (5x^2-13x+12) / x(x-4)(x-3)
\[\frac{5x^2-13x+12}{x(x-4)(x-3)}\] \[=\frac{A}{x}+\frac{B}{x-4}+\frac{C}{x-3}\] \[=\frac{A(x-4)(x-3)+B(x)(x-3)+C(x)(x-4)}{x(x-4)(x-3)}\]
So then we have to simplify on the top and set it (only the numerator) equal to 5x^2-13x+12
Remember it?
okay show me the simplifying
Come on, give it a try, tell me what you get and I will tell you if it's ok or not, OK?
k 1 sec
x^2-7x+12+Bx^2-3x+Cx^2-4x ?
Ax^2*
hello?
let me a second, I'm having dinner.
:D
haha no worries, i just checked and it is correct. thanks for helping me see how to set it up
I'm back, Ok awsome! Glad to help! Thanks for the medal. Have a wonderful day/night.
you too
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