Question

In Exercises \(1-4,\) find the values of \(A\) and \(B\) that complete the partial fraction decomposition. $$\frac{x-12}{x^{2}-4 x}=\frac{A}{x}+\frac{B}{x-4}$$

Short answer

The values of \(A\) and \(B\) that complete the partial fraction decomposition are \(A=-3\) and \(B=-1\).

Step-by-step solution

Setup the equation for manipulating

We'll start by setting up the equation like this: \(x-12 = A(x-4) + Bx\). We can get this equation by multiplying both sides of the initial equation by \(x^{2}-4x\), the common denominator, which will result in this equation.

Separate the equation into two parts

Next, we'll set up two equations, setting \(x\) to make either \(A\) or \(B\) drop out. We get \(x=0\) and \(x=4\) which are the roots of the denominator \(x^{2}-4x\). When \(x=0\), the equation becomes \(-12=4A\), and solving for \(A\) gives \(A=-3\). When \(x=4\), the equation becomes \(-4=4B\), and solving for \(B\) gives \(B=-1\).

Summary

We found that the values of \(A\) and \(B\) that complete the partial fraction decomposition of the given equation are \(A=-3\) and \(B=-1\) respectively. Therefore, the original equation becomes \(\frac{x-12}{x^{2}-4 x}=\frac{-3}{x}+\frac{-1}{x-4}\).