Taking out common factors

Before we get on to taking out common factors, we will look at expanding brackets, because it is closely linked to factorising. You could even say that taking out a common factor is exactly the opposite of expanding brackets.

Expanding brackets

Expanding brackets is a simple operation, which we will show with an example: let’s work out the expression 3 · (5 + 8). We have two ways to go about it. We can add up the numbers in the bracket and get 3 · 13, because 5 + 8 = 13.

The second way is to expand the bracket. This means we take the number in front of the bracket, the three, and multiply both numbers in the bracket by it. The plus sign in the bracket stays. So we get:

$$3\cdot (5+8)=(3\cdot 5+3\cdot 8)$$

Both methods lead to the same result: (3 · 5 + 3 · 8) = 15 + 24 = 39.

Why expanding brackets works

Why are we allowed to do this, and why does expanding work? Think of a product as a written-out sum. If we write 3 · 7, for example, we are saying that we want to add the number seven three times, that is, 3 · 7 = 7 + 7 + 7.

We can write out an example with a bracket in the same way: 3 · (5 + 8) is the same as (5 + 8) + (5 + 8) + (5 + 8). Since the order does not matter in addition, we can remove the brackets and move the fives and eights next to each other like this: 5 + 5 + 5 + 8 + 8 + 8. And here we can already see that this expression can be rewritten using multiplication as 3 · 5 + 3 · 8, which is what we did when expanding.

Examples of expanding brackets

1. Expand 5 · (4 + 7). Again we simply multiply the numbers in the bracket by five and get 5 · (4 + 7) = 5 · 4 + 5 · 7.
2. Expand 6 · (x + 4). Let’s not be put off by the unknown, and simply multiply it by six too. We get 6 · (x + 4) = 6x + 6 · 4.
3. Expand x · (4 + 173). Now we have the unknown in front of the bracket, so we multiply both numbers inside the bracket by x. We get x · (4 + 173) = 4x + 173x.
4. Expand 4x · (2x + 7). Now we have the unknown both in front of the bracket and inside it, and a more complicated expression in front of the bracket. Again, that is no problem: we multiply the contents of the bracket by 4x like this: 4x · (2x + 7) = 4x · 2x + 4x · 7 = 8x² + 28x.

Simple factorising

Taking out a common factor is the opposite procedure to expanding a bracket. At the start we have an expression such as 10x + 5. If we can find a number that both terms can sensibly be divided by, we can take that number out. In our expression it is the number 5, because 10x / 5 = 2x and 5 / 5 = 1. So we can write

$$10x+5=5\cdot (2x+1).$$

If we now expand the bracket 5 · (2x + 1) again, we get 10x + 5 back. The aim of taking out a common factor is usually to simplify the expression, or to get it into the form of a product so that we can cancel it in a fraction.

Taking out the unknown

Often we take out the unknown itself, not just a number. So if we have the expression 3x² + 7x, we can take out x: we divide both terms by x, add brackets and multiply the bracket by x. We get 3x² + 7x = x · (3x + 7).

We usually try to take out as “much” as possible. If we had the expression 8x² + 12x, we could take out just x and get 8x² + 12x = x · (8x + 12), but we can see that we could still take the number 4 out of the bracket. We can do that now, afterwards, and get: x · (4 · (2x + 3)). We can remove the outer brackets and get: 4x · (2x + 3).

We would get the same result, though, if we took out 4x straight away from the original expression 8x² + 12x. Written out, we would get

$$8x^2+12x=4x\cdot (8x^2/4x+12x/4x)=4x\cdot (2x+3).$$

Factorising more complicated expressions

So far we have only taken common factors out of expressions with two terms. But we can also do it with longer expressions. For example, from 7x³ + 5x² + 2x we can take out x and get x · (7x² + 5x + 2).

Nor do we always have to take out a nice number: we can take out anything. For example, from the expression 6x + 7 we can take out the number 2, but we get an ugly result, because we have to divide both terms by two. For 6x it works out fine, because 6x / 2 = 3x, but for the number 7 it does not work out so neatly: 7 / 2 = 3.5. We get the resulting expression 6x + 7 = 2 · (3x + 3.5).

Worked examples

• Factorise 25x + 45. We can see straight away that we can take out the number 5, so we get 5 · (5x + 9).

• Simplify the fraction

  $$\frac{3x^2+7x}{x}.$$

  First we take out x in the numerator of the fraction. So the numerator becomes x · (3x + 7). Now we can cancel x in the whole fraction.

  $$\frac{3x^2+7x}{x}=\frac{x\cdot (3x+7)}{x}=3x+7$$

• Factorise −2x² − 8x. Here we have negative numbers for the first time. That does not matter at all: just as we took out positive factors, we can take out negative ones. So we can take out −2x straight away. We get

  $$-2x^2-8x=-2x(x+4).$$