If $h(t)=-16t^2+80t$ then simplify $\frac{h(a)-h(1)}{a-1}$ I am to simplify $\frac{h(a)-h(1)}{a-1}$ given $h(t)=-16t^2+80t$. The solution provided is 
$\frac{-64+80a-16a^2}{-1+a}$ = $-16a+64$
I cannot see how this was arrived at. Here's as far as I got:
$\frac{h(a)-h(1)}{a-1}$
$\frac{(-16a^2+80a)-(-16+80)}{a-1}$ # substitute in the function h(t)
$\frac{-16a^2+80a-64}{a-1}$ # simplify numerator
$\frac{16(-a^2+5a-4)}{1-a}$ # 16 is a common factor in the numerator, attempted to simplify
...
I was not able to factor $-a^2+5a-4$
How can I arrive at the provided solution? More granular baby steps appreciated.
 A: We have that
$$\frac{-16a^2+80a-64}{a-1}=\frac{-16(a^2-5a+4)}{a-1}$$
and
$$a^2-5a+4=(a-4)(a-1)$$
indeed by quadratic equation for $a^2-5a+4=0$
$$a_{1,2}=\frac{5\pm \sqrt{25-16}}{2}=4,1$$
that is 
$$a^2-5a+4=(a-a_1)(a-a_2)=(a-4)(a-1)$$
therefore providing that $a\neq 1$
$$\frac{-16(a^2-5a+4)}{a-1}=\frac{-16(a-4)(a-1)}{a-1}=-16(a-4)=-16a+64$$
A: Sometimes it's helpful (but not really necessary) to have positive leading coefficients on the terms of highest power.
$$\begin{align}
\frac{16(-a^2+5a-4)}{1-a} &= \frac{-16(a^2-5a+4)}{-(a-1)}\\
&= \frac{-1}{-1}\cdot\frac{16(a^2-5a+4)}{a-1}\\
&= \frac{16(a^2-5a+4)}{a-1}\\
\end{align}$$
In order to factor the polynomial in the numerator, we need to find two numbers that when multiplied together give $4$ and when added give $-5$. But, we can also make the guess that the denominator is a factor of the numerator (because the question asks us to "simplify").
$$\begin{align}
\frac{16(a^2-5a+4)}{a-1} &= \frac{16(a^2-a-4a+4)}{a-1}\\
&= \frac{16(a(a-1)-4(a-1))}{a-1}\\
&= \frac{16(a-4)(a-1)}{a-1}\\
&= \frac{16(a-4)}{1}\cdot\frac{a-1}{a-1}\\
&= \frac{16(a-4)}{1}\cdot1\\
&= \frac{16(a-4)}{1}\\
&= 16(a-4)\\
\end{align}$$
