Question

Decompose $\frac{px+q}{(x-a)(x-b)(x-c)}$ into partial fractions.

• A. $\frac{ap+q}{(x-a)(a-b)(a-c)}+\frac{pb+q}{(x-b)(b-a)(b-c)}+\frac{pc+q}{(x-c)(c-a)(c-b)}$
• B. $\frac{C}{(x-a)(x-b)}$
• C. $\frac{C}{(x-a{)}^2}+\frac{D}{(x-b{)}^2}$
• D. None of these

Answer 1

Answer: (A)

$\frac{ap+q}{(x-a)(a-b)(a-c)}+\frac{pb+q}{(x-b)(b-a)(b-c)}+\frac{pc+q}{(x-c)(c-a)(c-b)}$

The first thing is to factor the denominator and get the form of the partial fraction decomposition

$\Rightarrow$ $\frac{px+q}{(x-a)(x-b)(x-c)}=\frac{P}{x-a}+\frac{Q}{x-b}+\frac{R}{x-c}$

The next step is to set numerators equal

$\Rightarrow$$px+q=P(x-b)(x-c)+Q(x-a)(x-c)+R(x-b)(x-c)$

we can just pick a few values of $x$ and find the constraints

$\Rightarrow$ $x=a$ $pa+q=P(a-b)(a-c)$ $P=\frac{pa+q}{(a-b)(a-c)}$

$\Rightarrow$ $x=b$ $pb+q=Q(b-a)(b-c)$ $Q=\frac{pb+q}{(b-a)(b-c)}$

$\Rightarrow$ $x=c$ $pc+q=P(c-b)(c-a)$ $R=\frac{pc+q}{(c-b)(c-a)}$

$\Rightarrow$ $\frac{px+q}{(x-a)(x-b)(x-c)}=\frac{ap+q}{(x-a)(a-b)(a-c)}+\frac{bp+q}{(x-b)(b-a)(b-c)}+\frac{cp+q}{(x-c)(c-a)(c-b)}$

Hence, option A is correct.