Question

I need to find the sum of this equation

Answer 1

Answer:

$\large\boxed{\dfrac{3s^2+19s-4}{(s-1)^2(s+5)}=\dfrac{3s^2+19s-4}{s^3+3s^2-9s+5}}$

Step-by-step explanation:

$s^2-2s+1=s^2-s-s+1=s(s-1)-1(s-1)=(s-1)(s-1)=(s-1)^2\\\\s^2+4s-5=s^2+5s-s-5=s(s+5)-1(s+5)=(s+5)(s-1)\\\\\text{Therefore}\ LCD\ \text{is}\ (s-1)^2(s+5).$

$\dfrac{3s}{s^2-2s+1}+\dfrac{4}{s^2+4s-5}=\dfrac{3s(s+5)}{(s-1)^2(s+5)}+\dfrac{4(s-1)}{(s-1)^2(s+5)}\\\\\text{use the distributive property}\\\\=\dfrac{(3s)(s)+(3s)(5)+(4)(s)+(4)(-1)}{(s-1)^2+(s+5)}=\dfrac{3s^2+15s+4s-4}{(s-1)^2(s+5)}\\\\=\dfrac{3s^2+19s-4}{(s-1)^2(s+5)}$