Dividing Polynomials (1)

This can actually be a bit tricky but it's not impossible.

Any polynomial of order 3 - which is what you deal with at C1 - is the product of either linear factors, or a linear and a quadratic factor.

Finding a factor like this, you don't need to worry about remainders but these crop up soon enough, and we'll come onto them in due course.

A basic example of what I mean is: x(x² + 2x - 4). Multiplying these two factors gives the cubic equation x³ + 2x² - 4x.

But what do you do if you know one factor and the polynomial?

Say you know the linear factor to be (x-2) and the polynomial to be x³ + x² - 11x + 10

The other factor must be a quadratic.

(x-2)(ax²+ bx + c) = x³ + x² - 11x + 10

expand the two brackets as you would normally:

ax³ + bx² + cx - 2ax² -2bx - 2c

and then find common factors to tidy this expression up and make it usable:

= ax³ + (b-2a)x² + (c-2b)x -2c

That bit is worth checking. You're simply collecting together the different values of x² or whatever, as you would normally in an equation. You're only putting brackets round them because you don't know their values and you need to understand them as together making x².

This expression is now completely identical to our original polynomial:
= ax³ + (b-2a)x² + (c-2b)x -2c
x³ + x² - 11x + 10


We now go through the fairly simple but again easy to cock up process of comparing the co-efficients. Co-efficients are of course the numbers in front of variables (3 is the coefficient of x in the expression 3x).

Comparing the co-efficients here means we look at our original polynomial and compare it with the unknown result - which we can see is a cubic - of our algebraic multiplication. The co-efficients in the known polynomial will tell us the missing co-efficients of our unknown quadratic.

In this case: a = 1

(b-2a) = 1
b - 2 = 1
b= 3

(c-2b) = -11
c-6 = -11
c = -5

We can check this last one by seeing -2c at the end of our algebraic expression. If c is right it will match -2 x -5 = 10, so yes it's right.

This process has given us the three co-efficients for our missing quadratic: 1, 3 and -5

so the two factors of our polynomial are (x-2) and (x² + 3x -5).

Just be careful at this point: can the quadratic factor be factorised itself, to give three linear factors? Give it a quick b² - 4ac test: 9 - (4x1x-5) = 9 - (-20) = 29. NO. It has to stay as a quadratic.

Often it will. You'll find that you're often - but not always - asked to find the three factors of p(x) or to factorise p(x) completely when this is the case.