Alan

f(a) = a² + 8

f(a+h) = (a+h)² + 8 = a² + 2ah + h² + 8

So: f(a+h) - f(a) =  a² + 2ah + h² + 8 - ( a² + 8) = 2ah + h²

and: (f(a+h) - f(a))/h = 2a + h

(Now if you do take the limit of this as h tends to zero, you are left with just 2a)

On the calculator you don't need to - just make sure it's in degree mode (select deg at the bottom left of the calculator).

On the Question page, select from the Insert menu (Special character) or just press Alt_0176.

Use the calculator on this site:

$${factor}{\left({{\mathtt{a}}}^{{\mathtt{6}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{b}}}^{{\mathtt{6}}}\right)} = \left({{\mathtt{b}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{a}}}^{{\mathtt{2}}}\right){\mathtt{\,\times\,}}\left({{\mathtt{b}}}^{{\mathtt{4}}}{\mathtt{\,-\,}}{{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{{\mathtt{b}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{a}}}^{{\mathtt{4}}}\right)$$

Thanks Melody, your way is much more concise. However, I deliberately wanted to construct the result step by step, in such a way that the questioner might spot the pattern, and understand where the 77 months comes from.  I guess that, since a question of this type was asked, the basic formula is already available to the questioner (though that doesn't mean they know how to identify the right one if they've been given several, I suppose!).

Looking again at what I did, I notice I skipped a few steps in summarising the formula for the amount owed at the end of each month; so perhaps my way won't be so useful in practice!

First rewrite them as 

y<-(5+x)/2

y<x

Then plot y=-(5+x)/2 and y=x and shade in the region below both curves (which is the region that satisfies both inequalities):

The red shaded area is where both inequalities are satisfied simultaneously (of course this region should cover the

range -∞<x<∞ in principle, but it is infeasible to plot all of that!)

Looks good when enlarged (as on your Profile page), but looks a little smudged in the small version on the Questions page!!  Skilfully done though!

Use the on-site calculator to get

$${\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{\mathtt{x}} = -{\mathtt{5}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{41}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{9}}\right)}{{\mathtt{4}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{41}}}}{\mathtt{\,-\,}}{\mathtt{9}}\right)}{{\mathtt{4}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{3.850\: \!781\: \!059\: \!358\: \!212\: \!2}}\\
{\mathtt{x}} = -{\mathtt{0.649\: \!218\: \!940\: \!641\: \!787\: \!8}}\\
\end{array} \right\}$$

Perhaps this will help:

So, rounding up to the nearest month, n = 77 months.

Correction:  There are 6, not 8, multiplications for the raw polynomial.

a₃*x*x*x = 3 multiplications (4 terms) and a₂*x*x = 2 multiplications (3 terms)