partial fractions - linear factors

Topics

• partial fractions

If the denominator $Q(x)$ can be factored into distinct linear factors $(x-a)$, $(x-b)$, etc., then the partial fraction decomposition takes the form:

$$\frac{P(x)}{Q(x)}=\frac{A}{(x-a)}+\frac{B}{(x-b)}+...$$

where $A$, $B$, … are constants to be determined.

Method to find A, B, … :

1. Multiply both sides by $Q(x)$
2. Substitute values of $x$ that make each linear factor zero (e.g., $x=a$, $x=b$). This eliminates all but one unknown constant, allowing you to solve for it directly
3. Alternatively, equate coefficients of like terms on both sides to create a system of linear equations. Solve this system to find the constants

Example:
Decompose: $\frac{5x-4}{(x^2-x-2)}$

Factoring the denominator, the expression is equal to $\frac{5x-4}{(x-2)(x+1)}$.
Using the method above we get:

$$\frac{5x-4}{(x-2)(x+1)}=\frac{A}{(x-2)}+\frac{B}{(x+1)}$$

Multiply both sides by the denominator:

$$5x-4=A(x+1)+B(x-2)$$

Substitute $x=2$ into the equation to get $6=3A$, thus $A=2$.
Substitute $x=-1$ into the equation to get $-9=-3B$, thus $B=3$.

Final result:

$$\frac{5x-4}{(x^2-x-2)}=\frac{2}{(x-2)}+\frac{3}{(x+1)}$$