Question

Why do we need to put linear numerators of the form $B_{i}x+C_i$in every term of a partial-fraction decomposition which involves irreducible quadratics? Think about the example $\frac{p\left(x\right)}{q\left(x\right)}=\frac{1}{{\left(x^2+1\right)}^2}$and figure out what goes wrong if we attempt to make a decomposition of the form$\frac{C_1}{x^2+1}+\frac{C_2}{{\left(x^2+1\right)}^2}$.

Short answer

The $q\left(x\right)$$=\frac{1}{x^2+1}$, whichis a reducible quadratic then $q\left(x\right)$is a product of $l_1\left(x\right)$and $l_2\left(x\right)$of two linear functions. So,$l_1\left(x\right)=x-1,l_2\left(x\right)=x-2$.

Step-by-step solution

Step 1. Given information

$\frac{p\left(x\right)}{q\left(x\right)}=\frac{1}{{\left(x^2+1\right)}^2}$.

Step 2. Let us see some concepts about qx.

If $q\left(x\right)$is an irreducible quadratic then $\frac{p\left(x\right)}{q\left(x\right)}$is its own partial fractions decomposition, there is nothing further we can decompose.

If $q\left(x\right)$is a reducible quadratic then $q\left(x\right)$is a product of $l_1\left(x\right).l_2\left(x\right)$of two linear functions and the obtain a partial function decomposition of the form $\frac{A_1}{l_1\left(x\right)}+\frac{A_2}{l_2\left(x\right)}$.

The given function is, $q\left(x\right)=\frac{1}{x^2+1}$.

So,$l_1\left(x\right)=x-1,l_2\left(x\right)=x-2$.