Factorisation

Factorisation

Highest common factor (HCF)

An expression is made up of terms which can be algebraic (letters or symbols), numbers or a combination of both. These usually have arithmetic operators between them or to put simply a plus or minus sign which makes up the expression.

e.g. 6x + 3

he important first step in factorisation is to recognise the highest common factor HCF. The HCF is the biggest number that each term in the expression can be divided by. So you have to look at each term starting with the first to work out what is common to each term. The first term is 6x which is made up of a symbol and a number and the second term is 3 which is just a number. The symbol x cannot be part of the common factor because it does not appear in both terms. A number can be the common factor because it appears in both terms but to fulfil the criteria of having a HCF both numbers must be divisible by a common number.

So 6 and 3 can be divided by 3 which is the HCF as opposed to 2 which is not the HCF. Once the HCF is found it will appear outside the bracket and inside the bracket will be terms that when multiplied by the HCF will give the original expression as shown below:

HCF = 3

Answer: 3(2x + 1)

As can be seen expanding the above expression will result in the original expression which is a quick way of checking if the factorised expression is correct (try this).

The HCF of an expression can be made up of symbols only.

e.g. x³ + x²

Here we have an x term that appears in all terms so that x must be in the HCF for the expression.

Which power of x is the HCF?

Both terms are divisible by x² but not by x³ as x² cannot be divided by x³. So the general rule is if a symbol appears in all terms but has different powers then the lowest power of that term/symbol is the HCF.

HCF = x²

Answer: x²(x +1)

The HCF of an expression can be made up of a combination of numbers and symbols. For this to be the case the numbers and symbols must appear in all terms in the expression.

e.g. 4x² + 2x

Again it is important to look at each term in turn, looking at numbers first then symbols, to determine what makes up the HCF. So 4 appears in one term and 2 in the other give a HCF of 2 for the numbers. For symbols x² appears in one term and x in the other so x is the HCF for the symbols.

HCF = 2x

Answer: 2x(2x + 1)

Expand the above expression to see if it is correct (If you are not sure how to expand brackets see section: expanding brackets).

For the following exercise:

Look at each term in expression and find the HCF
Put the HCF outside the bracket and appropriate terms inside bracket that will give original expression
Check to see if expression has been factorised (Expand out in your head).

Exercise 1

Factorising quadratic expressions

There are two main categories of expression that can be factorised although the number of methods that can be used to factorise an expression is much greater and this depends on the expression itself.

The first category we have already looked at above and this is simple factorisation which results in the HCF outside a single bracket. The other category of expression is a quadratic expression which has the form as shown below:

ax² + bx + c

where a, b and c are constants (numbers).

Factorising a quadratic expression will result in an answer with a double bracket and the method of factorisation is quite different from the previous method.

Where the coefficient of the squared term is unity (multiple of x² = 1)

This may sound complicated but it simply means that the multiple (coefficient) of the x² term is one so nothing appears in front of it. Have a look at the following example:

x² + 7x + 12

In the above expression the first term in each bracket will be x because this will give x² when expanding (see section expanding brackets). The second term in each bracket will be the factors of the last term (12) that when added together give the multiple of the x term (7).

So the factors of 12 when added together give 7 are 3 and 4.

Answer: (x + 3)(x + 4)

Exercise 2

Factorise the following:

1. \(x^2 + 6x + 5\)

2. \(x^2 + 10x + 16\)

3. \(x^2 + 4x - 5\)

4. \(x^2 + 3x - 10\)

5. \(x^2 - 8x + 15\)

Where the coefficient of the squared term is not unity (multiple of x² > 1)

This just means that the x² term has a multiple e.g. 2x², 3x² etc.

Factorising x² terms that have a multiple is carried out differently from the previous type of expression where there is no multiple. It is slightly more difficult to factorise but should be straightforward using the following method.

e.g. 2x² + 5x - 3

List all the possible factors of the first term and the last term as shown below:

2x² -3

2x and x 3 and -1

-3 and 1

Ensure that when there is a negative term all combinations are listed, as in the case of -3 where the minus sign can either be on 1 or 3 giving 2 possible combinations.

Once all factors have been listed then it is simply a case of using the correct combinations. So for the first term there are only 1 pair of factors so each one is the first term in each bracket.

(2x )(x )

Next see which pair of factors of the second term gives the correct factorised expression.

(2x + 3)(x - 1) = 2x² - 2x + 3x - 3 = 2x² + x - 3

As you can see this is not the correct factorised expression. However, by switching the factors of the last terms will give the correct factorisation as shown below.

(2x - 1)(x + 3) = 2x² + 6x - x - 3 = 2x²+ 5x - 3

Remember to try all possible combinations of factors but also you may have to switch pairs of factors around until the expression is factorised correctly. As you practice and improve you will recognise factors quickly so a lot of the processing will be carried out in your head.

Exercise 3

Try factorising the following using the method shown above:

1. \(2x^2 + 5x - 3\)

2. \(12x^2 - 35x + 8\)

3. \(9x^2 + 18x + 8\)

Factorisation by grouping

To factorise the expression ax + ay + bx + by group the terms in pairs before

factorising so that each pair of terms has a common factor. Then simply put the common factor outside the bracket.

ax + ay + bx + by = (ax + ay) + (bx + by)

= a(x + y) + b(x + y)

Notice that the two terms have the common factor (x + y) so the expression can be further factorised to:

(x + y)(a + b)

Exercise 4

1. ay + bz + by + bz

2. mp + np - mq - nq

Where the factors form a perfect square

\((a + b)^2\) = \((a + b)(a + b)\)

= \(a^2 + ab + ab + b^2\)

= \(a^2 + 2ab + b^2\)

The square of a binomial expression of the type shown above consists of the following: (the square of the 1st term) + (twice the product of both terms) + (the square of the last term). Factorising an expression that forms a perfect square is simply to recognise that it does and then put the 2 terms into the single squared bracket as shown in the example below.

To factorise \(9a^2 + 12ab + 4b^2\) you can see that \(\sqrt{9a^2} = 3a\) and \(\sqrt{4b^2} = 2b\) and that \(2\times3a\times2b = 12ab\). So \(9a^2 + 12ab + 4b^2 = (3a + 2b)^2\).

Exercise 5

Factorise the following:

1. \(16x^2 - 40x + 25\)

2. \(25x^2 + 20x + 4\)

3. \(36x^2 + 48x +16\)

Factorising the difference of two squares

(a + b)(a - b) = a² - ab + ba - b² = a² - b²

As can be seen from the expansion of the two brackets that if you have two squared terms with a minus sign between them then simply put the square root of the two terms in a bracket with a plus sign and then into another bracket with a minus sign.

e.g. x² - 25 = (x + 5)(x - 5)

Exercise 6

Factorise the following:

1. x² - 4

2. x² - 16

3. x² - 36

4. 64x² - 9

5. 4x² - 25y²

6. 64y² - 16z²

Exercise 7

For this final exercise you need to carefully look at the expression to be factorised in order to determine which method of factorisation will be appropriate. You will be using all methods given on this sheet.

1. a²c² + acd + acd + d²

2. 2pr - 4ps + qr - 2qs

3. 4ax - 6ay - 4bx - 6by

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