Question

Find the partial fraction decomposition for the following rational expression
11x-3/(x-2)(x+3)

Answer 1

To find the partial fraction decomposition of the expression \(\frac{11x - 3}{(x - 2)(x + 3)}\), we can express it in the form:

\[
\frac{11x - 3}{(x - 2)(x + 3)} = \frac{A}{x - 2} + \frac{B}{x + 3}
\]

where \(A\) and \(B\) are constants that we need to determine.

First, we can combine the right-hand side over a common denominator:

\[
\frac{A}{x - 2} + \frac{B}{x + 3} = \frac{A(x + 3) + B(x - 2)}{(x - 2)(x + 3)}
\]

Setting the numerators equal gives us:

\[
11x - 3 = A(x + 3) + B(x - 2)
\]

Expanding the right side:

\[
A(x + 3) + B(x - 2) = Ax + 3A + Bx - 2B = (A + B)x + (3A - 2B)
\]

Now we can set the coefficients for \(x\) and the constant terms equal:

1. \(A + B = 11\) (coefficient of \(x\))
2. \(3A - 2B = -3\) (constant term)

We now have a system of equations:

1. \(A + B = 11\) (Equation 1)
2. \(3A - 2B = -3\) (Equation 2)

We can solve this system step-by-step. From Equation 1, we can express \(B\) in terms of \(A\):

\[
B = 11 - A
\]

Substituting this expression for \(B\) into Equation 2:

\[
3A - 2(11 - A) = -3
\]

Simplifying this:

\[
3A - 22 + 2A = -3
\]
\[
5A - 22 = -3
\]
\[
5A = 19
\]
\[
A = \frac{19}{5}
\]

Now, substituting \(A\) back into Equation 1 to find \(B\):

\[
\frac{19}{5} + B = 11
\]
\[
B = 11 - \frac{19}{5}
\]
\[
B = \frac{55}{5} - \frac{19}{5} = \frac{36}{5}
\]

Now we have both constants:

\[
A = \frac{19}{5}, \quad B = \frac{36}{5}
\]

Finally, we can write the partial fraction decomposition:

\[
\frac{11x - 3}{(x - 2)(x + 3)} = \frac{\frac{19}{5}}{x - 2} + \frac{\frac{36}{5}}{x + 3}
\]

Or, rewritten more cleanly:

\[
\frac{11x - 3}{(x - 2)(x + 3)} = \frac{19}{5(x - 2)} + \frac{36}{5(x + 3)}
\]

This is the partial fraction decomposition of the given rational expression.