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:


$\frac{(-16a^2+80a)-(-16+80)}{a-1}$ # 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.


We have that




indeed by quadratic equation for $a^2-5a+4=0$

$$a_{1,2}=\frac{5\pm \sqrt{25-16}}{2}=4,1$$

that is


therefore providing that $a\neq 1$



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").

$$\begin{align} \frac{16(a^2-5a+4)}{a-1} &= \frac{16(a^2-a-4a+4)}{a-1}\\ &= \frac{16(a(a-1)-4(a-1))}{a-1}\\ &= \frac{16(a-4)(a-1)}{a-1}\\ &= \frac{16(a-4)}{1}\cdot\frac{a-1}{a-1}\\ &= \frac{16(a-4)}{1}\cdot1\\ &= \frac{16(a-4)}{1}\\ &= 16(a-4)\\ \end{align}$$

  • $\begingroup$ Thanks, this was also very helpful for understanding. $\endgroup$ – Doug Fir Nov 13 '19 at 15:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.