Question

Resolve into partial fractions:
$\frac{x^2-10x+13}{(x-1)(x^2-5x+6)}$

Answer 1

$\frac{x^2-10x+3}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}$ where A ,B ,C are to be determined .

$=>x^2-10x+3=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)$

For $x=1,=>1^2-10\left(1\right)+3=A(1-2)(1-3)+B\left(0\right)+C\left(0\right)$

$=>-6=2A$

$=>A=-3$ -(1)

For $x=2,=>2^2-10\left(2\right)+3=A\left(0\right)+B(2-1)(2-3)+C\left(0\right)$

$=>-13=-B$

$=>B=13$ -(2)

For $x=3,=>3^2-10\left(3\right)+3=A\left(0\right)+B\left(0\right)+C(3-1)(3-2)$

$=>-18=2C$

$=>C=-9$ -(3)

Therefore from (1),(2),(3) , we get $\frac{x^2-10x+3}{(x-1)(x-2)(x-3)}=\frac{-3}{x-1}+\frac{13}{x-2}-\frac{9}{x-3}$