## Polynomials

polynomial is the name given to a sum of algebraic terms. The terms are usually arranged in descending order of their powers.

e.g. P(x) = x3 + 2x2 − 5x − 6 is a polynomial. The variable of this polynomial is and thedegree of this polynomial is because that is the highest power of x.

Polynomials can be given in factorised form.

e.g. P(x) = x3 + 2x2 − 5x − 6 = (x + 1)(x − 2)(x + 3)

The value of a polynomial is calculated by substituting a value for x into the polynomial.

e.g. P(4) = 43 + 2 x 42 − 5 x 4 − 6 = 64 + 32 − 20 − 6 = 66

### Polynomial Equations

Polynomial equations, such as quadratic equations can be easily solved if they are given in factorised form.

 example 1 example 2 (x − 3)(x + 1) = 0 x = 3 or x = -1 (x + 1)(x − 6)(x + 5) = 0 x = -1 or x = 6 or x = -5

### Factorising Polynomial Expressions

Quadratic expressions, whose highest power is 2, can often be factorised by inspection.

Cubic expressions, power 3, and above are much harder to factorise. The Factor Theorem can help to factorise any polynomial.

The Factor Theorem states:

 For a polynomial P(x): if P(a) = 0 then (x − a) is a factor of P(x)

Put into words this says that if a value is substituted into a polynomial and the result is 0, then that value forms part of the factor. By a sensible trial and error process the factors can often be found quite quickly.

e.g. Factorise P(x) = x3 + 2x2 − 5x − 6

Now, the factors of the polynomial have to contain factors of 6, the constant at the end.

So, using the Factor Theorem, evaluate P(1), P(-1), P(2), P(-2), P(3), P(-3), P(6), P(-6). The ones which are equal to 0 give the factors.

 P(1) = 13 + 2 x 12 − 5 x 1 - 6 = 1 + 2 − 5 − 6 -8 (x -1) is not a factor P(-1) = (-1)3 + 2 x (-1)2 − 5 x -1 - 6 = -1 + 2 + 5 − 6 0 (x +1) is a factor P(2) = (2)3 + 2 x (2)2 − 5 x 2 - 6 = 8 + 8 − 10 − 6 0 (x − 2) is a factor P(-2) = (-2)3 + 2 x (-2)2 − 5 x- 2 - 6 = -8 + 8 +10 − 6 4 (x − 2) is not a factor P(3) = (3)3 + 2 x (3)2 − 5 x 3 - 6 = 27 + 18 − 15 − 6 24 (x − 3) is not a factor P(-3) = (-3)3 + 2 x (-3)2 − 5 x − 3- 6 = -27 +18 +15 − 6 0 (x + 3) is a factor P(6) There is no need to evaluate these as the three factors have been found.In fact, after the first two had been found the other one could be deduced as they must multiply to -6 P(-6)

So P(x) = x3 + 2x2 − 5x − 6 = (x + 1)(x − 2)(x + 3).