Find the average rate of change of $g(t)=t^2+3t+1$ on the interval [0,a] I am working on a textbook question "Find the average rate of change of $g(t)=t^2+3t+1$ on the interval $[0,a]$". The solution provided, along with the steps in between is:
Avg. rate of change: 
= $\frac{g(a)-g(0)}{a-0}$
= $\frac{(a^2+3a+1)-(0^2+3(0)+1)}{a-0}$
= $\frac{a^2+3a+1-1}{a}$ # Here - how did they get rid of -0 in denominator?
= $\frac{a(a+3)}{a}$
= $a+3$
Where I'm stuck is between the second and third step. How does one go from denominator of $a-0$ to just $a$?

For any number $a, a-0=a$.  Subtracting $0$ makes no change.
