To decompose the rational expression [tex]\(\frac{x}{(x-1)(x+4)}\)[/tex] into partial fractions, we assume it can be expressed in the form:
[tex]\[
\frac{x}{(x-1)(x+4)} = \frac{A}{x-1} + \frac{B}{x+4}
\][/tex]
To find the values of [tex]\(A\)[/tex] and [tex]\(B\)[/tex], we start by combining these fractions on the right-hand side over a common denominator:
[tex]\[
\frac{A}{x-1} + \frac{B}{x+4} = \frac{A(x+4) + B(x-1)}{(x-1)(x+4)}
\][/tex]
Since the denominators are the same, we can equate the numerators:
[tex]\[
x = A(x+4) + B(x-1)
\][/tex]
Next, we expand and collect like terms:
[tex]\[
x = (A + B)x + (4A - B)
\][/tex]
For the equation to be true for all [tex]\(x\)[/tex], the coefficients of [tex]\(x\)[/tex] and the constant terms on both sides must be equal. This gives us the system of equations:
[tex]\[
A + B = 1
\][/tex]
[tex]\[
4A - B = 0
\][/tex]
We now solve this system of equations:
1. From [tex]\(4A - B = 0\)[/tex], we get:
[tex]\[
B = 4A
\][/tex]
2. Substitute this into [tex]\(A + B = 1\)[/tex]:
[tex]\[
A + 4A = 1
\][/tex]
[tex]\[
5A = 1
\][/tex]
[tex]\[
A = \frac{1}{5}
\][/tex]
3. Substitute [tex]\(A = \frac{1}{5}\)[/tex] back into [tex]\(B = 4A\)[/tex]:
[tex]\[
B = 4 \left(\frac{1}{5}\right)
\][/tex]
[tex]\[
B = \frac{4}{5}
\][/tex]
Thus, the partial fraction decomposition of the given rational expression is:
[tex]\[
\frac{x}{(x-1)(x+4)} = \frac{\frac{1}{5}}{x-1} + \frac{\frac{4}{5}}{x+4}
\][/tex]
Simplifying the fractions:
[tex]\[
\frac{x}{(x-1)(x+4)} = \frac{1}{5} \cdot \frac{1}{x-1} + \frac{4}{5} \cdot \frac{1}{x+4}
\][/tex]
Therefore, the final answer is:
[tex]\[
\frac{x}{(x-1)(x+4)} = \frac{1/5}{x-1} + \frac{4/5}{x+4}
\][/tex]
