Factorising

Video lesson · from 26:462 worked examples

One lesson video covers all of Polynomials & Quadratics, so it opens at 26:46 for this topic — not from the beginning.

We can use synthetic division (or polynomial long division) alongside the factor theorem to fully factorise polynomials of degree 3 or higher.

Once a linear factor $(x-h)$ is found, we divide the polynomial by $(x-h)$ to obtain a quotient. We then factorise the quotient if possible.

Take out a common factor first: skipping this is extremely common, and it makes the factorisation far harder than it needs to be. Check for one before you start dividing.
Sign of the root: a factor $(x-h)$ corresponds to the root $x = h$ , so you divide by $h$ , not $-h$ . Sign errors are the single biggest cause of lost marks on polynomial questions.
Use the standard layout: synthetic division presented in a non-standard way is hard to credit. Set it out the usual way so the marker can follow it.
Factorise completely: stopping at $(x-h)(\text{quadratic})$ is unfinished if the quadratic factorises further.
Say what you have shown: a correct calculation with no stated conclusion does not score full marks. If the remainder is zero, write that $(x-h)$ is a factor.

Example 1

a) Show that $(x+5)$ is a factor of $x^3+9x^2+23x+15$ .

b) Factorise $x^3+9x^2+23x+15$ fully.

Let $f(x) = x^3+9x^2+23x+15$ .

If $(x+5)$ is a factor, $f(-5)$ must be 0:

f(-5) = (-5)^3 + 9(-5)^2 + 23(-5) + 15

f(-5) = -125 + 9(25) - 115 + 15

f(-5) = -125 + 225 - 115 + 15 = 0

Since the remainder is 0, $(x+5)$ is a factor.

We use synthetic division to divide $x^3+9x^2+23x+15$ by $(x+5)$ to find the quadratic quotient.

The quotient is $x^2+4x+3$ .

x^3+9x^2+23x+15 = (x+5)(x^2+4x+3)

Now factorise the quadratic:

x^2+4x+3 = (x+3)(x+1)

Fully factorised:

x^3+9x^2+23x+15 = (x+5)(x+3)(x+1)

Example 2

Fully factorise $2x^3+5x^2-28x-15$

Let $f(x) = 2x^3+5x^2-28x-15$ .

We test factors of -15 (e.g. $\pm 1, \pm 3, \pm 5$ ) to find a root.

Testing $x = 3$ :

f(3) = 2(3)^3 + 5(3)^2 - 28(3) - 15

f(3) = 54 + 45 - 84 - 15 = 0

Since $f(3) = 0$ , $(x-3)$ is a factor.

Now use synthetic division to find the quadratic factor:

2x^3+5x^2-28x-15 = (x-3)(2x^2+11x+5)

Factorise the quadratic $2x^2+11x+5$ :

2x^2+11x+5 = (2x+1)(x+5)

Fully factorised:

2x^3+5x^2-28x-15 = (x-3)(2x+1)(x+5)