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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.

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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 $a\neq 1$

$$\frac{-16(a^2-5a+4)}{a-1}=\frac{-16(a-4)(a-1)}{a-1}=-16(a-4)=-16a+64$$

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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").

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

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

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