Distinct roots: general case + shortcut

(Friday, September 3, 2021
Revised after class)

$\displaystyle\frac{7x-5}{(x-1)(x+2)(x+3)}$	$\displaystyle=$	$\displaystyle\frac{A}{x-1}+\cdots?$
	$\displaystyle=$	$\displaystyle\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{x+3}$
	$\displaystyle=$	$\displaystyle\frac{A(x+2)(x+3)+B(x-1)(x+3)+C(x-1)(x+2)}{(x-1)(x+2)(x+3)}$
	$\displaystyle=$	$\displaystyle\frac{(A+B+C)x^{2}+(5A+2B+C)x+(6A-3B-2C)}{(x-1)(x+2)(x+3)}$

We then solve

by substitution.

A clever shortcut.

Solving linear equations is a lot of work! Here’s a faster way to find $A$ : observe

$\displaystyle\frac{7x-5}{(x-1)(x+2)(x+3)}$	$\displaystyle=$	$\displaystyle\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{x+3}$
$\displaystyle\frac{7x-5}{(x-1)(x+2)(x+3)}\cdot(x-1)$	$\displaystyle=$	$\displaystyle\left(\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{x+3}\right)(x-1)$
$\displaystyle\frac{7x-5}{(x+2)(x+3)}$	$\displaystyle=$	$\displaystyle A+B\frac{x-1}{x+2}+C\frac{x_{1}}{x+3}.$

If we now set $x=1$ , then

\frac{7-5}{(1+2)(1+3)}=A+B\cdot 0+C\cdot 0=A.