Alg5.1.8

From Exampleproblems

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

Splitting the denominator into factors $\frac{x^2-10x+1}{(x-1)(x^2-5x+6)}=\frac{x^2-10x+1}{(x-1)(x-3)(x-2)}\,$

Let $\frac{x^2-10x+1}{(x-1)(x-3)(x-2)}=\frac{A}{x-1}+\frac{B}{x-3}+\frac{C}{x-2}\,$

Multiplying both sides by $(x-1)(x-2)(x-3)\,$

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

Let $x=1\,$

Substituting the value in the equation above, we get $-8=2A\,$

Hence $A=-4\,$

Let $x=2\,$

Substituting the value in the equation above, we get $-15=-C\,$

Hence $C=15\,$

Let $x=3\,$

Substituting the value in the equation above, we get $-20=2B\,$

Hence $B=-10\,$

Substituting the values of A,B and C in the main equation we get $\frac{x^2-10x+1}{(x-1)(x^2-5x+6)}=\frac{-4}{x-1}+\frac{-10}{x-3}+\frac{15}{x-2}\,$