Polynomials

These are a very common sight in almost any study of mathematics. It’s worth taking a moment to define them properly and think about how they can be used.

A polynomial is a mathematical expression involving a whole, positive power of x. The type of polynomial depends on the highest power of x in the expression. The different bits of the expression are known as terms, and the polynomial as a whole is the sum of the terms it contains.

Take a quadratic expression: 4x² + 2x - 2. It has three powers of x: 2, 1 and 0. These are whole numbers and positive (except 0). This is therefore a polynomial. The same applies to cubic expressions: x³ + 4x² + 5x + 2 is a polynomial too.

In fact, the same applies, whatever the power of x, as long as it is not a fractional or negative number. A polynomial does not need to have several terms, like most cubics and quadratics do. So, 1 is also a polynomial, being a term in x raised to the power of zero. The same applies to any constant (ie any number). 4x + 1 is a polynomial too, as x is x raised to the power of one.

Polynomials are written in descending powers of x, though they do not have to be.

The numbers in front of the x variables are called coefficients, and can take any value, including fractional or negative values, as can constants.

The highest power of x in a polynomial defines the degree of that polynomial. So 1 is a polynomial of degree 0, because the highest power of x here is 0. 4x + 1 is a polynomial of degree 1, and quadratics are polynomials of degree 2.

Polynomials can be created by expanding brackets. (x+2)² gives x² + 4x + 4 when expanded. In fact, both expressions here are polynomials. If the brackets were to give us an expression with a fractional power, then of course this would not be the case.

When you make a polynomial like a quadratic equal to 0, then you have a polynomial equation. In the case of a quadratic, such as 4x² + 2x – 2, the resulting equation is usually written as: 4x² + 2x – 2 = 0. As an equation, it is now a description of a parabolic curve which intersects the x-axis at (-1,0) and (½, 0).

Polynomials can be added, subtracted, multiplied and divided. The Factor and Remainder Theorems can be used to find out additional information about them. For example, if a number assigned to the variable x causes the polynomial to equal zero, that number is a factor of the polynomial (a root, a value of x where the curve of the equation crosses the x-axis). So we can demonstrate that -1 is a factor of 4x² + 2x – 2 by putting it in place of x. 4(-1)² + (-1x2) – 2 = 0. Therefore -1 is a factor of this polynomial. This is the essence of the Factor Theorem. The Remainder Theorem is a little more involved but states that if we divide the polynomial by x-a then the remainder is f(a) – in other words, the remainder is the sum of the polynomial when a is put in place of the variable x.

These are the absolute basics. Of course it does get a lot more complicated! But it is worth familiarising yourself with these principles before moving on.

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.

Rounding Up Factorisation

...so after all that you now have the technology (all right, the skills) to factorise a polynomial completely. You first do trial and error to find the first linear factor, then you multiply it by ax² + bx + c and compare coefficients to find the quadratic. Then you check if the quadratic factorises further.

Onto dividing polynomials proper....

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.

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.