Factor Theorem (ii)

Well that's all very neat and tidy, you might be thinking, but what good is this theorem?

It gives you a lovely and easy way of trying to find linear factors by substituting values into the polynomial you are given. Sometimes this does not work easily but it is often worth a try for a minute or so.

It also enables you to do slightly trickier things with polynomials.

You often get questions like this (we'll use the cubic from the previous post for this). A polynomial, P(x) = x³ + ax² + bx + 6, has (x+1) and (x+2) as factors. Find the values of a and b.

This looks tricky but it is where the factor theorem comes in.

P(-1) = 0. You know this from the theorem.

that means that (-1) + a(1) + b(-1) + 6 = 0

so -1 + a - b + 6 = 0

therefore a - b = -5

P(-2) = 0.

So -8 + 4a -2b + 6 = 0

therefore 4a - 2b = 2

You now have two equations to solve SIMULTANEOUSLY!!

a - b = -5
4a - 2b = 2

Times the first equation by 2:

2a - 2b = -10
4a - 2b = 2


SUBTRACT THE TWO TO ELIMINATE b

-2a = -12

a = 6

Plug this back into one of the equations:

2(6) - 2b = -10
-2b = -22

b = 11

a = 6, b = 11

There is nothing that hard here and yet in just writing this post I have made LOADS of mistakes, which is why it's taken me an hour or so to write.

Here are the errors I made:

1) Using the wrong constant value - ie reading the cubic wrong;
2) Forgetting to cube or square the values in the cubic equation;
3) subtracting with minus signs incorrectly;
4) multiplying incorrectly.

All because I was surfing the blogs while writing this post!

The Factor Theorem

Well that stuff in the previous post leads us nicely on to the factor theorem. Although this sounds like a low budget 70s sci-fi drama, it is in fact a key part of A Level maths.

The factor theorem is dead simple. If you have a polynomial x³ + 6x² + 11x + 6, it has factors, namely a linear and a quadratic.

Only this time, the quadratic itself factorises, so that the cubic has 3 linear factors, namely (x+1)(x+2)(x+3).

If I now did something a bit sneaky, I could prove that these are factors of this cubic.

Watch. P(-1) = -1³ + 6(-1²) + 11(-1) + 6
= -1 + 6 - 11 + 6
= 0

When x=-1 the cubic equation = 0.

This proves that (x+1) is a factor of x³ + 6x² + 11x + 6

In fact, if (x-a) is a factor of P(x) then P(a) = 0. Always and everywhere.

Why?

Well, look at (x+1)(x+2)(x+3).

These are all factors of x³ + 6x² + 11x + 6.

So if you put x= -1 into the first factor, it will equal 0 (-1+1) and therefore the whole polynomial will equal 0. The same goes for all the other factors.

If a value, a, put into P(x) does not equal 0, then (x-a) IS NOT a factor.

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.

Multiplying polynomials

This is fairly easy too: it is, at its basic level, expanding brackets, though you can make use of the distributive law of algebra to help you do it more systematically.

For example.

(x+2)(x-5) can either be expanded bit by bit (ac+ad+bc+bd) or you can write it like this: x(x-5) + 2(x-5). Instead of adding four separate products individually, you are adding two sets of them. It's basically the same thing, just a little more organised. You could also write it x(x+2) -5(x+2).

You can use this principle to help you multiply polynomials.

(x+9)(3x³ -4x² + 3)

x(3x³ -4x² + 3)
+ 9(3x³ -4x² + 3)

= (3x⁴ - 4x³ + 3x)
+ (27x³ - 36x² + 27)

using the knowledge of collecting polynomials from the previous post:

= 3x⁴ + 23x³ - 36x² + 3x + 27.


It gets fiddlier than this but usually the problem is signs. Pay really close attention to them because they are an easy ways of losing lots of marks quickly.

Collecting Polynomials

This is not difficult in principle, but it can become messy if you are not careful with your signs.

You can only add or subtract values with the same powers of x.

Be careful: this means that you CANNOT add 3x⁴ and 3x³ because they have different powers of x and are therefore completely different values.

You also CANNOT add 3x² and 3r² because they are different variables and so different values.

Here is a fairly basic example: add 3x³ + 5x to 2x³ -4x

It helps to collect the two expressions into brackets first:

(3x³ + 5x) + (2x³ -4x)

Now you can sort out the signs more easily:

3x³ + 5x + 2x³ -4x

As it happens there is no sort out needed here!

Now collect the like terms:

(3x³ + 2x³) + (5x - 4x)

= 5x³ + x

It does get a lot more fiddly than this sometimes:

(4x⁴ + 5x³ - 3x²) - (2x⁴ - 6x³ + x² + 9)

But use the same principles. Here the two expressions are already in brackets so we can SORT THE SIGNS OUT:

4x⁴ + 5x³ - 3x² - 2x⁴ + 6x³ - x² - 9

What we've done here is simply say "all the values in the second polynomial are being subtracted and we all know what to do with a - and a + or a - and another -". We've effectively subtracted the two expressions here by reversing the signs in the second expression.

You might want to look back at that again.

OK. Now we can collect the like terms.

4x⁴ - 2x⁴ + 5x³ + 6x³ - 3x² - x² - 9

= 2x⁴ + 11x³ - 4x² - 9

Polynomials - The Basics

I've got fed up with doing straight line geometry, as it needs diagrams, and I am having real difficulty drawing & importing them.

So to polynomials. Firstly, what is a polynomial? Well it is an expression involving numbers and powers of x. Such as:

2x + 1

3x² + 4x - 6

2x⁴ - 4x³ + 2x² +2


These are all polynomials. They are usually written, as here, in descending powers of x, from highest to lowest.

Sometimes they are called polynomials of degree x, depending on the highest power of x. So a cubic expression is a polynomial of degree 3, a quadratic of degree 2, a linear of degree 1 and a constant (ie just a number on its own) of degree 0 (because the highest power of x here is 0, ie x⁰ - ie 1).

Polynomials DO NOT have fractional or negative powers.

2x^-2 + 1

is not a polynomial.


That's the basics of it.

Bit Brighter

Thought it was a bit dark and dismal around here! This looks better to me.