Polynomials

A polynomial is a function of the form axn where a is a constant (this means that it has a fixed value) and n is a positive integer.

Examples of polynomials are: (x2 + 3x) or (2x69x2) or (x − 6)3

The 'degree of order' of the polynomial is the highest power of x.

For example:

2x53x2 + 6x is a polynomial of degree 5.

Polynomials can also be part of fractional functions (a function, that includes fractions). An example is shown below:

Polynomial

There are two types of this kind of equation.

'Proper' fractions - ones where the degree of the numerator (the equation on the top) is less than the degree of the denominator (the equation on the bottom). The equation above is an example of this.

'Improper' fractions - ones where the degree of the numerator is greater than or equal to the degree of the denominator. An example of this is:

Improper fraction

This is a nice simple method that helps you find the remainder when a polynomial is divided by a linear function. You can find a teacher here to help teach you this common theorem.

The theorem states that when f(x) is divided by (x-a) the remainder is f(a). This means that when you are given the equations to be divided, then the remainder is the value of the equation when x = a.

For example: if we want to find the remainder when 2x3x2 + 2 is divided by x − 3, we write:

 

f(x) = 2x3 - x2 + 2
f(3) = 2(33) - 32 + 2
= (2 × 27) - 9 + 2 = 47.

 

This theorem is linked to the remainder theorem in that if (x − a) is a factor of the polynomial, there will be no remainder. So, f(a) = 0.

Therefore, if you want to find if (x - a) is a factor of f(x), just check that f(a) = 0.

For example:

to check if x − 2 is a factor of x36x2 + 6x − 2

We substitute x = 2 into the function:

 

f(2) = 23 − 6(22) + 6×2 − 2
= 8 − 24 + 12 − 2
= -6

 

So f(2) is not equal to 0, and we conclude that x − 2 is not a factor of the function x36x2 + 6x − 2.

The factor theorem can also be used to factorise polynomials of greater degree than 2 and therefore helps us solve some cubic, quartic, etc. equations.

Example:

Factorise as far as possible: x3 + 2x2 − x − 2

Firstly choose sensible values as possible factors - i.e. numbers that are factors of 2 (the constant). i.e. try -2, -1, 1, 2. (Can you see why?)

 

F(2) = 8 + 8 − 2 − 2 ≠ 0.so (x − 2) is not a factor.
F(1) = 1 + 2 − 1 − 2 = 0so (x − 1) is a factor.
F(-1) = -1 + 2 + 1 − 2 = 0so (x + 1) is a factor
F(-2) = -8 + 8 + 2 − 2 = 0so (x + 2) is a factor

 

Therefore,

x3 + 2x2 − x − 2 = (x − 1)(x + 1)(x + 2)

Note: If you cannot find all the factors using the factor theorem, but have found one, then we can find the others by multiplication, (or division), as follows:

If we know that (x − 1) is a factor of x3 + 2x2 − x − 2, then,

x3 + 2x2 − x − 2 = (x − 1)(x2 + bx + c) (cubic = linear × quadratic)

x3 + 2x2 − x − 2 = x3 + (b − 1)x2 + (c − b)x − c

Now match the terms to get:

c = -2 and b = 3 and so,

x3 + 2x2 − x − 2 = (x − 1)(x2 + 3x + 2)

We can now factorise the quadratic part (or use the formula when solving the equation f(x) = 0), to get:

(x2 + 3x + 2) = (x + 2)(x + 1)

Therefore,

x3 + 2x2 − x − 2 = (x − 1)(x + 2)(x + 1)

And,

f(x) = 0 when x = 1, x = -1, and x = -2, which are the points where the graph y = f(x) crosses the x-axis.

Note: A useful piece of information when solving these problems is that,

 

a3 − b3 = (a − b)(a2 + ab + b2) and
a3 + b3 = (a + b)(a2 − ab + b2)

S-cool Exam Slip-Ups

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