5 is the highest common factor of 10 and 5. There are no variables common to both terms.

\(10x^2-5\ =\ 5(2x^2-1)\)

4 is the highest common factor of 12 and 20, and \(x\) is a factor of both terms.

\(12x^2-20x\ =\ 4x(3x-5)\)

First, we should always check for common factors. There aren't any here.

So we think about the two terms. 9 = 3², 25 = 5² and \(x^2\) has an even power. So this is a difference of two squares.

\(9x^2-25\ =\ (3x+5)(3x-5)\)

We should always check for common factors first. In this example, we have a common factor of 3.

\(12x^2-27y^2\ =\ 3(4x^2-9y^2)\)

Now we think about the two terms in the bracket. 4 = 2², 9 = 3² and \(x^2\) and \(y^2\) both have even powers. So this is a difference of two squares.

$$ \begin{eqnarray} 12x^2-27y^2 &=& 3(4x^2-9y^2)\\[6pt] &=& 3(2x-3y)(2x+3y) \end{eqnarray} $$

Any exam question of this type will be worth two marks, which is a big clue that you need two steps of working.

First, we should always check for any common factors. In this example, there aren't any.

So is this a difference of two squares? Clearly not, because there are three terms. It's a trinomial.

The coefficient of \(x^2\) is 1 and the other coefficients are positive, so this is the easiest type of trinomial.

We just need to find two numbers that multiply to 10 and add to 11. The factor pairs of 10 are 2\(\times\)5 (but those add to 7) and 1\(\times\)10 (which is the pair we want, as they add to 11).

So \(x^2+11x+10=(x+1)(x+10)\)

This is a trinomial expression with no common factors.

The coefficient of \(x^2\) is 1, so this is a unitary trinomial (also called a monic trinomial).

So we just need to find two numbers that multiply to –8 and add to –2. Clearly one of them will be negative.

The only two numbers that fit these conditions are 2 and –4.

So \(x^2-2x-8=(x+2)(x-4)\)

This is a unitary trinomial expression with no common factors.

So we just need to find two numbers that multiply to –18 and add to 3.

The factor pairs of 18 are 1\(\times\)18, 2\(\times\)9 and 3\(\times\)6.

However, because our two numbers actually have to multiply to –18, not 18, one of the factors will be negative.

The coefficient of \(x\) is positive, so we know that the smaller of the two factors must be negative.

So our factor pairs are actually –1\(\times\)18, –2\(\times\)9 and –3\(\times\)6.

Which of these is correct? Well, –1\(+\)18 \(\neq\) 3 and –2\(+\)9 \(\neq\) 3 but –3\(+\)6 \(=\) 3.

So \(x^2+3x-18=(x-3)(x+6)\)

This is a unitary trinomial expression with no common factors.

So we just need to find two numbers that multiply to 20 and add to –9.

The factor pairs of 20 are 1\(\times\)20, 2\(\times\)10 and 4\(\times\)5.

However, because our two numbers have to multiply to a positive number but add to a negative number, we know that both of them must be negative.

So our factor pairs are actually –1\(\times\)–20, –2\(\times\)–10 and –4\(\times\)–5.

Which of these is correct? Well, –1\(+\)(–20) \(\neq\) –9 and –2\(+\)(–) \(\neq\) –9 but –4\(+\)(–5) \(=\) –9.

So \(x^2-9x+20=(x-4)(x-5)\)

A question of this type has never appeared in a National 5 Maths exam, but it might, so let's be ready for it!

Rearranging the terms:

$$ \begin{eqnarray} 12+4x-x^2 &=& -\!x^2+4x+12 \end{eqnarray} $$

The coefficient of \(x^2\) is –1 so one way of handling this is to take out a common factor of –1.

$$ \begin{eqnarray} -x^2+4x+12 &=& -\!(x^2-4x-12) \end{eqnarray} $$

We can then factorise the unitary trinomial in the brackets in the usual way, looking for two numbers that multiply to –12 and add to –4.

The two numbers are –6 and 2.

Putting this all together:

$$ \begin{eqnarray} 12+4x-x^2 &=& -\!x^2+4x+12\\[6pt] &=& -\!(x^2-4x-12)\\[6pt] &=& -\!(x-6)(x+2)\\[6pt] \end{eqnarray} $$

We think it is fine to leave this with the negative sign at the beginning, but if you don't like that style, you could multiply the \((x-6)\) factor through by –1 and present your answer in the tidier form:

$$ \begin{eqnarray} 12+4x-x^2 &=& (6-x)(2+x) \end{eqnarray} $$

This is another trinomial expression with no common factors.

This time, the coefficient of \(x^2\) is 3, so it's a little more complicated than the previous two examples.

However, this is the easiest such example because both 3 and 5 are prime.

The factorisation must therefore be one of \((x+1)(3x+5)\) or \((x+5)(3x+1)\small.\)

The first of these doesn't work, because expanding the 'outside' and 'inside' pairs gives \(5x\!+\!3x=8x\small,\) not \(16x\small.\)

But the second factorisation does work, because the 'outside' and 'inside' pairs expand to \(x\!+\!15x=16x\small.\)

So \(3x^2+16x+5=(x+5)(3x+1)\)

First check for any common factors. There aren't any.

The coefficient of \(x^2\) is 2. That's good news, because 2 is a prime number. The only possible product is 1\(\times\)2.

The constant term of –4 isn't such good news. 4 isn't prime. It could be 1\(\times\)4 or 2\(\times\)2.

2\(\times\)2 wouldn't work here because the outer pair and inner pair would then be an even number, not –7. So it must be 1\(\times\)4. Checking this both ways around and taking care over the signs gives us:

So \(2x^2-7x-4=(x-4)(2x+1)\)

There are several ways to make this more systematic, and different teachers prefer different methods. We like what is called the "ac method". More on that in the resources below.

First check for any common factors. All of the coefficients are multiples of 3, so we factor that out.

So \(6x^2-21x-12=3(2x^2-7x-4)\)

The trinomial in the brackets is the same as the previous example, so the full factorisation is:

\(6x^2-21x-12=3(x-4)(2x+1)\small.\)

First check for any common factor. In this example, there is a common factor of 3. So:

$$ \begin{eqnarray} 3x^2-48 &=& 3(x^2-16)\\[6pt] &=& 3(x+4)(x-4) \end{eqnarray} $$

This is very similar to the previous example. The answer has two steps. First we remove the common factor and then we factorise the difference of two squares.

$$ \begin{eqnarray} 3a^2-75 &=& 3(a^2-25)\\[6pt] &=& 3(a+5)(a-5) \end{eqnarray} $$