Question

Find the values of A and B to complete the partial fraction decomposition. (x-12)/(x^2-4x)=A/x + B/(x-4)

Answer 1

have a look

Answer 2

So what step are you stuck on? :o

Answer 3

Finding the value of A and B. I dont understand at all.

Answer 4

Factoring out an x from each term in the denominator gives us 2 `linear` factors.
\[\Large\bf\sf \frac{x-12}{x^2-4x}\quad=\quad \frac{x-12}{x(x-4)}\]
We can break this down according to the factors in the denominator.
Our new numerators will be one degree less than the degree of the denominator.
Since each of our factors is `linear`, our numerators will be `constant`.
So they gave us the constants A and B.\[\Large\bf\sf \frac{x-12}{x(x-4)}\quad=\quad \frac{A}{x}+\frac{B}{x-4}\]We'll start out by multiplying both sides by the denominator on the left side of the equation.

Answer 5

\[\Large\bf\sf x(x-4)\cdot\frac{x-12}{x(x-4)}\quad=\quad \left(\frac{A}{x}+\frac{B}{x-4}\right)\cdot x(x-4)\]That will cancel the denominator on the left side.

Answer 6

And it will cancel some denominators on the right as well.

Answer 7

\[\Large\bf\sf x-12\quad=\quad A(x-4)+Bx\]Do you see how I got to this point?

Answer 8

Looking at this and the pdf the other person posted, I figured it out!

Answer 9

ok cool c: