Alg5.1.6

From Exampleproblems

Find the partial decomposition of $\frac{3x+7}{x^2-3x+2}\,$

Splitting the denominator into factors $\frac{3x+7}{(x-2)(x-1)}\,$

Now $\frac{3x+7}{x^2-3x+2}=\frac{3x+7}{(x-2)(x-1)}\,$

Decomposing the expression Let $\frac{3x+7}{(x-2)(x-1)}=\frac{A}{x-2}+\frac{B}{x-1}\,$

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

Then $3x+7=A(x-1)+B(x-2)\,$

Let $x=2\,$

Then $13=A\,$

Now, let $x=1\,$

then $10=-B\,$

Now $\frac{3x+7}{x^2-3x+2}=\frac{1}{x-2}-\frac{10}{x-1}\,$