Factorising polynomials

Why factorise?

To factorise is to write a polynomial as a product of polynomials of lower degree, in the same way that we write $12 = 3 \times 4$. Factorising is useful because a product is zero only when one of its factors is zero. So if $(x - 2)(x - 3) = 0$, then $x = 2$ or $x = 3$. Factorising solves equations, simplifies fractions and shows the zeroes of a polynomial at once.

Step one: a common factor

Always look first for a factor common to all the terms.

• $6x^2 + 9x = 3x(2x + 3)$.
• $x^3 - 4x = x(x^2 - 4)$, and then $x^2 - 4$ factorises further.

Step two: the identities

Some polynomials fit an identity that you know. Use it at once.

Examples: $x^2 - 25 = (x + 5)(x - 5)$. $x^2 + 6x + 9 = (x + 3)^2$. $8x^3 - 27 = (2x - 3)(4x^2 + 6x + 9)$.

Factorising a quadratic

Splitting the middle term. To factorise $ax^2 + bx + c$:

1. Find the product $a \times c$.
2. Find two numbers whose product is $ac$ and whose sum is $b$.
3. Split the middle term into these two numbers, and group the terms.

Example 1. Factorise $x^2 + 7x + 12$.

$ac = 12$. The numbers with a product of $12$ and a sum of $7$ are $3$ and $4$.

Example 2. Factorise $x^2 - 5x + 6$.

$ac = 6$, and we need a sum of $-5$. The numbers are $-2$ and $-3$.

Example 3. Factorise $6x^2 + 17x + 5$.

$ac = 6 \times 5 = 30$. The numbers with a product of $30$ and a sum of $17$ are $2$ and $15$.

Hints for the signs. If $c$ is positive, the two numbers have the same sign as $b$. If $c$ is negative, the two numbers have different signs, and the larger one has the sign of $b$. For $x^2 - x - 12$: $c = -12$, so one is positive and one is negative. The numbers are $-4$ and $3$, with sum $-1$.

Start from $3x^2 + 10x + 8$, so $ac = 24$. Move $m$ until the message appears: $4$ and $6$ multiply to $24$ and add up to $10$.

Factorising a cubic

For a cubic polynomial, we use the factor theorem.

Example 4. Factorise $x^3 - 23x^2 + 142x - 120$.

1. The constant term is $-120$. Try $x = 1$: $p(1) = 1 - 23 + 142 - 120 = 0$. So $x - 1$ is a factor.
2. Divide by $x - 1$. The quotient is $x^2 - 22x + 120$.
3. Factorise the quadratic. We need a product of $120$ and a sum of $-22$. The numbers are $-10$ and $-12$. So $x^2 - 22x + 120 = (x - 10)(x - 12)$.

Check. The zeroes are $1, 10, 12$. Their sum is $23$, which matches the coefficient $-23$.

A way to check

Multiply the factors back. For $(x + 3)(x + 4)$: $x^2 + 4x + 3x + 12 = x^2 + 7x + 12$. ✓. This takes only a few seconds, and it finds almost every mistake.

Practise with new numbers

Try it in Python

This program finds a zero, divides it out, and repeats, until all the factors are found.

Common mistakes

• Missing the common factor. Take it out first.
• Stopping too early. $x^4 - 16 = (x^2 - 4)(x^2 + 4)$ is not complete. $x^2 - 4$ factorises more.
• Using a wrong sign in the sum. The two numbers must add to $b$, and multiply to $ac$.
• Not checking. Multiply the factors back.

Remember these

• Splitting the middle term: find two numbers with product $ac$ and sum $b$.
• For a cubic, find one zero by the factor theorem, divide, and factorise the quadratic that is left.
• Look first for a common factor, then for an identity.
• Check a factorisation by multiplying the factors back, or by comparing the zeroes.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Factorise $x^2 + 5x + 6$.

2. Factorise $x^2 - x - 12$.

3. Factorise $4x^2 - 9$.

4. Factorise $2x^2 + 7x + 3$.