Factorising Polynomials
Factor Theorem

Quadratics and cubics are particular examples of polynomial functions.

a quadratic function is a polynomial of degree 2

a cubic has degree 3.

In general a polynomial of degree n has the form

a_n x^n + a_{n-1} x^{n-1} + ... + a_2 x^2 + a_1 x + a_0

where a_0, a_1, ... a_n are real numbers, and a_n ≠ 0 .

When solving cubic equations, the spotting of a root leads immediately to a simple linear factor. For example,

3 gives a zero value for the cubic x^3 - 5x^2 + 11x - 15

⇒ (x - 3) is a factor of x^3 - 5x^2 + 11x - 15

⇒ x^3 + 5x^2 + 11x - 15 = (x - 3) × (\text{quadratic})

The same technique can be applied to polynomials of any degree. For example,

2 gives a zero value for the function x^4 + 3x^2 - 17x + 6

⇒ (x - 2) is a factor of x^4 + 3x^2 - 17x + 6

⇒ x^4 + 3x^2 - 17x + 6 = (x - 2) × (\text{cubic})

In general

If P(x) is a polynomial of degree n and has a zero at x = a,
that is P(a) = 0, then (x - a) is a factor of P(x) and

P(x) = (x - a) × Q(x)

where Q(x) is a polynomial of degree (n - 1).

This is known as the factor theorem.

Worked Examples

1

Show that (x - 5) is a factor of the polynomial

x^5 - 4x^4 - x^3 - 21x^2 + 25

Show me Denote the polynomial by P(x) so
P(5) = 5^5 - 4 × 5^4 - 5^3 - 21 × 5^2 + 25
= 3125 - 2500 - 125 - 525 + 25
= 0
Hence 5 gives a zero value for P(x) and so by the factor theorem, (x - 5) is a factor of this polynomial.
2

Show that x = \frac{1}{2} gives a zero value to the 2x^3 - 3x^2 - 3x + 2.

Also show that 2x^3 - 3x^2 - 3x + 2 has two linear factors other than x = \frac{1}{2}.

Show me

We write the function as P(x) = 2x^3 - 3x^2 - 3x + 2

We can see that

P\left(\frac{1}{2}\right) = 2\times\left(\frac{1}{2}\right)^3-3\times\left(\frac{1}{2}\right)^2-3\times\left(\frac{1}{2}\right)+2
= \frac{1}{4}-\frac{3}{4}-\frac{3}{2}+2
= 0

We know that \left(x - \frac{1}{2}\right) is a factor so (2x - 1) is also a factor and we can write

2x^3 - 3x^2 - 3x + 2 = (2x - 1) × (ax^2 + bx + c)

Here a, b and c are constants and so a = 1 and c = -2. Comparing coefficients of x on each side of the identity, -3 = -b + 2c, giving b = -1.

You can check this by comparing coefficients of x^2.

So

P(x) = (2x - 1) × (x^2 - x - 2)
= (2x - 1) × (x - 2) × (x + 1), after factorising the quadratic.

Exercises

Choose a liear factor for each of these polinomials.

(a)

x^4 + 3x^3 - 2x^2 + 5x – 7

(x + 1)
(x + 10)
(x - 1)
(x - 2)
(x - 3)
(x - 8)
(b)

x^5 - 7x^4 - 8x^3 + 2x^2 + x – 1

(x + 1)
(x + 10)
(x - 1)
(x - 2)
(x - 3)
(x - 8)
(c)

x^4 - 2x^3 - 7x – 6

(x + 1)
(x + 10)
(x - 1)
(x - 2)
(x - 3)
(x - 8)
(d)

2x^6 + 5x^4 - 27x^3 + 8

(x + 1)
(x + 10)
(x - 1)
(x - 2)
(x - 3)
(x - 8)
(e)

5x^5 + 23x^4 - 269x^3 - 90x + 100

(x + 1)
(x + 10)
(x - 1)
(x - 2)
(x - 3)
(x - 8)
(f)

x^{10} - 9x^9 + 8x^8 - x^2 + x + 56

(x + 1)
(x + 10)
(x - 1)
(x - 2)
(x - 3)
(x - 8)

Find a linear factor for each of these polynomials.

(a)

x^3+4x^2+x-6

()
(b)

x^4-6x-4

()
(c)

x^4 - 6x^3 - 8x^2 + 6x + 7

()
(d)

2x^4+x^3-6x^2-x+1

()
(e)

3x^3 + 10x^2 - 11x + 2

()