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.

Dividing Polynomials

As promised....

One of the key elements of this is the ability to split off fractions effectively.


Think of a normal, but improper fraction, such as 4/3. This could be expressed as 3/3 + 2/3 (or any other combination but this is clean and tidy and useful for us now). 3/3 is of course 1. We are left with 1 + 1/3.

The same thing can be done algebraically, so as to solve basic division questions.

Take (x + 6)/x. As above, we can think of this as x/x + 6/x. x/x = 1, so the answer is 1 + 6/x.

We use this property, making part of the numerator match the denominator, in dividing polynomials.