Alg5.1.7

From Exampleproblems

Write the sum of partial fractions for $\frac{x-4}{x^2-5x+6}\,$

Splitting the denominator into factors $\frac{x-4}{x^2-5x+6}=\frac{x-4}{(x-3)(x-2)}\,$

Let $\frac{x-4}{x^2-5x+6}=\frac{A}{x-3}+\frac{B}{x-2}\,$

Multiplying bothsides by $(x-3)(x-2)\,$

then $x-4=A(x-2)+B(x-3)\,$

Let $x=3\,$

Substituting the value in $x-4=A(x-2)+B(x-3)\,$

then $-1=A\,$

Let $x=2\,$

Substituting the value in $x-4=A(x-2)+B(x-3)\,$

then $2=B\,$

Now the final partial fractions are $\frac{x-4}{x^2-5x+6}=\frac{-1}{x-3}+\frac{2}{x-2}\,$