Dividing Polynomials (ii)

Alright then, let's have a crack at this properly.

We are armed with some fraction revision and a little bit of knowledge of polynomials, including factorising using the system of comparing coefficients.

We only need to divide a polynomial by a linear factor for C1.

Division has this nasty habit of not being exact, but of leaving remainders. Unfortunately, 5 does not go easily into 24 but leaves a remainder. Polynomials are also subject to this problem. A linear expression might not be a factor of P(x) as such, but might go into something else, leaving factors and a remainder when asked to be slotted into P(x).

Remember that QUADRATIC EQUATIONS sometimes factorise into TWO LINEAR expressions,
while CUBICS factorise into LINEAR + QUADRATIC (and sometimes three linear).

So if you divide a quadratic by a linear, this might happen:

x² + 2x + 4 divided by x-2.

If we are to use the concept of comparing coefficients, which is very efficient, we first rewrite the equation into the form it would take expressed as factors and remainders:

x² + 2x + 4 = (x+p)(x+q) + r.

Got it?

We need to divide it by x-2 though, so we already have one desired factor (THIS DOES NOT MEAN THAT x-2 WILL BE A FACTOR - just that we are trying to express the quadratic in terms of how much x-2 goes into it.

So:

x² + 2x + 4 = (x-2)(x+q) + r.

Then we collect the different terms in this form. We are not completely, expanding the brackets, we are re-arranging them in terms of their own factors in order to look for coefficients which will tell us the other factors and remainders.

x² + 2x + 4 = x² + (-2 + q)x + (-2q+r)

F(x) is in this case equal to x² and (-2+q) lots of x, and then the two constants, with no x variable, (-2q and r, the possible remainder).

Therefore 2= (-2+q) so q= 4.

4 = (-2x4 + r)

4=8+r

r = -4

x² + 2x + 4 = (x-2)(x+4) -4.

This only expresses the quadratic as factors and a remainder. The division isn't over yet.

Now let's actually divide the quadratic by (x-2).

x² + 2x + 4
-------------
    x-2

=

(x+4) - 4
            ----
            (x-2)



I'll explain it later. Proper work beckons.

Why Doesn't Completing the Square Always Work?

I was meant to be going to bed but then I saw that someone had googled the above question and come to me.

The honest answer is "it does always work" but I imagine the problem my putative reader has is that some quadratics are really inconsiderate and don't have real roots.


Well imaginary and complex numbers are Further Pure nowadays but I will hazard an answer.

A quadratic is lovely because it has its famous formula: -b +/- (sqt b²-4ac)/2a

The b² - 4ac bit is called the discriminant and determines the kinds of solutions the equation can have.

I can't really draw a graph tonight but we see it in terms of the points the quadratic crosses the x-axis of the graph.

It goes a little bit like this:

b² - 4ac = a square number (36, say) - quadratic factorises (ie can be reduced to (x+1)(x-2) say).

b² - 4ac = +ve numbers, including fractions - quadratic cuts x-axis twice but can't factorise.

b² - 4ac = 0 - quadratic hits x-axis once (x-axis is a tangent to quadratic)

b² - 4ac = -ve number, quadratic does not touch x-axis at all.


If b² - 4ac is less than 0 it means we need a square root of a negative number. At c1 we can't do that at all. Later on one learns that numbers which square to give -ve numbers are called imaginary numbers and have all kinds of key roles. For now we just don't really want them,

so for our A/S Level purposes the quadratic does not have real roots (ie the values would be complex numbers - an imaginary with a real number) and we simply cannot calculate it at all.

You just end up with a complex number statement like (-2+sqrt -1)/2 or something. As I say, probably great for FP1, no use for C1.

However, that does not mean that completing the square doesn't work - it can ALWAYS be used to solve a quadratic. It just means that it can produce imaginary numbers.