How to write the form of partial fraction decomposition of the function. How to write the form of partial fraction decomposition of the function.
$$I= \dfrac{2\,x^3+24\,x^2+20\,x+10}{x^2+10\,x+25}$$
I used long division I get$$Quotient= 2\,x+4\\Remainder =-70\,x - 90\\$$
But don't know how to solve further steps. Thanks in advance.
 A: Dividing the function you get$\text{Quotient}= 2\,x+4\\\text{Remainder} =-70\,x - 90\\\text{Divisor}= x^2+10\,x+25\tag*{}$So,$2\,x^3+24\,x^2 +20\,x+10=(2x+4)(x^2+10x+25)+(-70x-90)\tag*{}$Rewrite the function,$I = \dfrac{(x^2+10\,x+25)(2\,x+4)+(-70\,x-90)}{x^2+10\,x+25}\\=(2\,x+4) -\dfrac{70\,x+90}{(x+5)^2}\tag*{}$Now decompose $\dfrac{70\,x+90}{(x+5)^2}$  by using  partial fraction.$\Rightarrow\dfrac{70\,x+90}{(x+5)^2} = \dfrac{A}{(x+5)} +\dfrac{B}{(x+5)^2}...(1)\\=70\,x+90 = A\,x+5\,A+B\tag*{}$comparing coefficients of both side of equation$Ax=70x,5\,A+B=90\\A=70,B=-260\tag*{}$Therefore,$ I=(2\,x+4) -\dfrac{70}{x+5}+ \dfrac{260}{(x+5)^2}\tag*{}$
here is the similar problem with complete solution:
partial fraction decomposition
A: HINT:
As $x^2+10x+25=(x+5)^2$
using Partial Fraction Decomposition, write $$\dfrac{-70x-90}{(x+5)^2}=\dfrac A{x+5}+\dfrac B{(x+5)^2}$$
A: Here, $I=(2x+4)+\frac{-70x-90}{x^2+10x+25}=(2x+4)+\frac{-70x-90}{(x+5)^2}=(2x+4)+\frac{A}{x+5}+\frac{B}{(x+5)^2}$.$\ $To determine A and B, we solve the equations $$A=-70;\\5A+B=-90$$ to obtain $I=(2x+4)+\frac{-70}{x+5}+\frac{260}{(x+5)^2}$
