National 5 Maths
Factorising

Before we get onto trinomials and the difference of two squares, this and example 2 below have been included to remind you always to look for a common factor first!

So, in this example:

• 5 is the highest common factor of 10 and 5
• There is a letter \(x\) in the first term, but not in the second term, so \(x\) is not a common factor.

The only common factor is therefore the number 5, so our factorisation is:

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

Note: It can be a good idea to check that your answer expands back into the question. In this case:

\(\overparen{\overparen{5(2x^2}-1}) = 10x^2-5 \)

So we were correct. The factorisation is indeed \(5(2x^2-1)\small.\)

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Following example 1 above, this is our second example of common factor.

In this example:

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

So both the 4 and the \(x\) factorise out, like this:

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

Note: Quickly check for yourself that \(4x(3x-5)\) expands back to \(12x^2-20x\small.\) It is good to get into the habit of checking like this. With practice, you can check quickly and mentally.

Video solution by Clelland Maths 
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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.

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

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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.

Video solution by Clelland Maths 
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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 \(x^2+11x+10=(x+1)(x+10)\)

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

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

Video solution by Clelland Maths 
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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 \(x^2-9x+20=(x-4)(x-5)\)

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

Notice that the terms are in reverse order, and that the coefficient of \(x^2\) is negative.

There are several ways of handling this. Perhaps the best way is just by listing the factor pairs of \(12\) and thinking which of them is most likely to be correct.

\(12\) factorises as \(1\!\times\!12\small,\) \(2\!\times\!6\) or \(3\!\times\!4\small.\)

Only \(2\!\times\!6\) has a sum or difference that equals \(4\small,\) the coefficient of \(x\small.\)

So, without any further working, \(12+4x-x^2=(6-x)(2+x)\small.\)

Video solution by Clelland Maths 
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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 \(3x^2+16x+5=(x+5)(3x+1)\)

Video solution by Clelland Maths 
Detailed breakdown by ChatGPT 

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.

Video solution by Clelland Maths 
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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:

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

Video solution by Clelland Maths 
Detailed breakdown by ChatGPT 

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

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

Video solution by Clelland Maths 
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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.

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

Video solution by Clelland Maths 
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