What the hell, McClane? Why would you wear a sign like that?!?

Yeah. Anyway. So, quick reminder of the most important things we’ve worked out:

Finding the roots was done for the sake of finding factors of 1-x-x². With those factors, the expression in the top red box is in a better position to be solved, by way of partial fractions, which look like this:

Those somethings are the factors of 1-x-x² (that’s just how partial fractions work). In the previous post, I mentioned the ‘factor theorem’, which says that if ‘r’ is a root of an equation, then x-r is a factor. Our roots are ‘a’ and ‘b’. Which implies that (x-a) and (x-b) are factors of 1-x-x². Well, let’s check and see what happens when we multiply (x-a) by (x-b), and then substitute in the actual values of a and b.

Obviously, now we wanna know what a+b, and a*b are:

and:

This will come in super-handy later, so take note:

Let’s throw these into what we’ve got:

Hang on. We factored 1-x-x², and when we multiplied those factors back together, we ended up with x²+x-1. But those are different expressions! Or… are they?

**We’re not so different, you and I…**

There’s some subtleties to the factor theorem, but essentially, we’ve just found a way of looking at our original equation from a different perspective, a perspective we can work with and happen to know the factors of.

Anyway, remember when I said that a partial fraction could help reduce the degree/power of an expression? Well, that’s what we can do now:

We know little ‘a’ and ‘b’. All we do now is work out big ‘A’ and ‘B’, which is surprisingly straightforward. To be honest, the only tricky thing from this point on is keeping track of minus signs and parentheses. Lock that down and you’re golden.

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