When a polynomial is divided by a divisor and the remainder is zero, this means that the divisor is a factor.

Therefore, from the Remainder Theorem, we arrive at the **Factor Theorem.**

The Factor Theorem states:

If f(a) = 0 then (x − a) is a factor of the polynomial f(x) |

**Example**

Show that (x − 3) is a factor of f(x) = x^{3} − 2x^{2} − 5x + 6

Using the Factor Theorem

f(3) = 3

^{3}− 2.3^{2}− 5.3 + 6= 27 − 18 − 15 + 6

= 0

**Therefore (x − 3) is a factor of f(x) = x ^{3} − 2x^{2} − 5x + 6**

**Factorising using the Factor Theorem**

Quadratic expressions can often be factorised by inspection. It is more difficult with cubic expressions. The Factor Theorem makes the factorising of cubics and higher order polynomials a little easier.

**Example**

Factorise f(x) = x^{3} − 2x^{2} − 5x +** 6**

The factors of **6 **are {-1, -2, -3, -6, 1, 2, 3, 6}

So select a couple of these to try with the Factor Theorem.

Try 2 and -2 first:

f(2) = (2)^{3} − 2(2)^{2} − 5.2 + 6 = -4

so (x − 2) is **not **a factor.

f(-2) = (-2)^{3} − 2(-2)^{2} − 5.-2 + 6 = 0

so (x + 2) **is **a factor.

If (x+2) is a factor, we should now try 3 and -3. Because 3 x 2 = **6**

f(3) = (3)^{3} − 2(3)^{2} − 5.3 + 6 =0

so (x − 3) **is **a factor.

So far (x + 2) and (x − 3) are factors so, by inspection the final factor must be (x − 1).

Because 2 x -3 x -1 =** 6**

**Therefore f(x) = (x + 2)(x − 3)(x − 1)**

(Alternatively once a factor is found it can be divided out using long division and the resulting quotient factorised.)