C2: Algebra and Functions
Algebra was a huge part of C1. In C2 we build on those ideas by looking mainly at factorisation of higher order polynomials. That means things like cubics.
The problem with factorising cubics is that we don't have an easy method. For quadratics (in C1) the method is relatively simple and even when we can't figure out how to do it, we can always just chuck the coefficients into the quadratic formula.
Unfortunately no (simple) formula exists for factorising cubics. That's the bad news.
The good news is that all the cubics we need to factorise must be pretty simple. If they are too hard then it's just not possible to factorise them at this level.
The method for factorising cubics will involve finding one of the factors by some method or other and then we should be left with a quadratic which we can easily factorise from C1.
Algebraic Division
The best way to explain algebraic division is with a clip. We'll see how to do long division with numbers first using 457 ÷ 13, and then we'll look at (x3 + 3x2 - 2x + 4) ÷ (x + 2)
Factor Theorem
Wouldn't it be good if we could tell whether (x + 2) was a factor of (x3 + 3x2 - 2x + 4) without doing all that long division stuff?
Well, there is a very simple way. And it's called the Factor Theorem.
If (x + 2) is a factor then we could write (x3 + 3x2 - 2x + 4) as (x + 2)(ax2 + bx + c). So
(x3 + 3x2 - 2x + 4) = (x + 2)(ax2 + bx + c)
If this was correct then we could substitute x = -2 in the right hand side (RHS) of the equation (actually it's an identity but never mind about that for now). This would automatically make the RHS = 0.
This leads to the Factor Theorem: (x - a) is a factor of a polynomial if (and only if) we get zero when we substitue x = a into the polynomial.
Or to put it another way: (x - a) is a factor of a polynomial f(x) ⇔ f(a) = 0.
So, to see whether (x + 2) is a factor of (x3 + 3x2 - 2x + 4) we just put -2 in and see what we get.
f(-2) = ((-2)3 + 3(-2)2 - 2(-2) + 4)
= 12, so it isn't a factor.
Remainder Theorem
Back to our example in the clip.
Another way of writing 457 ÷ 13 = 35 rem 2 is 437 = 13×35 + 2
The algebraic example could be written as (x3 + 3x2 - 2x + 4) = (x + 2)(x2 + x - 4) + 12
Notice that 12 is the remainder. Now, how could we find the remainder without all of that calculation?
Easy! Just substitute -2 in to the RHS and we get the remainder. This leads to the Remainder Theorem
To find the remainder when we divide f(x) by (x - a), just chuck x = a into f(x).
In our example f(-2) = 12, so that's the remainder.
Now for three exam questions from Jan 2005, May 2006 and Jan 2009. Have a bash at them before watching the clip.