The algebra of the four basic operations  (Page 3/5)

6x2 + 12x = 3x (2x + 4)

5x3 – 2x2 = x2 (5x – 2)

–4x3 + 12x2 = –4x (x2 – 3x)

9a2 + 9a3 = (3a + 3a2) (3a)

Look back at the exercise in section B of the previous activity – did you recognise the problems?

C Finding common factors of polynomials

In exactly the same way we can find the common factors of more than two terms. Here are some examples:

Examples:

6x3 – 3x2 + 6x = 3x (2x2 – x + 2)

ab3c – 3a2b3c + a3b2c = ab2c (b – 3ab + a2)

3a + 24a2 + 6a3 = 3a ( 1 + 8a + 2a2)

20x – 8x2 + 16x3 – 12x4 +4x5 = 4x (5 – 2x + 4x2 – 3x3 + x4)

You will notice that the terms remaining in the brackets don’t have any more common factors left. This is because they have been fully factorised. You must always take out the highest common factor from all the terms.

Exercise:

Fully factorise the following expressions by taking out the highest common factor:

12abc + 24ac

15xy – 21y

3abc + 18ab2c3

8x2y2 – 2x

2a2bc2 + 4ab2c – 7abc

12a(bc)2 – 8(abc)3 + 4(ab)2c3 – 20bc + 4a

Pair activity:

Did you notice that in each case the number of terms in the brackets after factorising was the same as the number of terms in the original expression?

Explain to your partner why you think this will always happen.

D Factorising difference of squares

In section D of the previous activity you had to multiply these three pairs of binomials:

(a + b) (a – b) ,

(2y + 3) (2y – 3) and

(2a2 + 3b) (2a2 – 3b)

Here are the solutions:

(a + b) (a – b) = a2 – b2

(2y + 3) (2y – 3) = 4y2 – 9

(2a2 + 3b) (2a2 – 3b) = 4a4 – 9b2

You will have noticed that the answers have a very special pattern: square minus square.

This is called a difference of squares and this is how it is factorised:

First–square minus second–square

= ( $\sqrt{\text{first}-\text{square}}$ plus $\sqrt{\text{sec}\text{ond}-\text{square}}$ ) ( $\sqrt{\text{first}-\text{square}}$ minus $\sqrt{\text{sec}\text{ond}-\text{square}}$ )

Examples:

x2 – 25 = (x + 5) (x – 5)

4 – b2 = (2 + b) (2 – b)

9a2 – 1 = (3a + 1) (3a – 1)

YOU HAVE TO BE VERY SURE OF THE MOST COMMON SQUARES AND THEIR ROOTS.

These are a few important ones – you must add to this list.

22 = 4 32 = 9 (a2)2 = a4

(a3)2 = a6

(½)2 = ¼ 12 = 1

Exercise:

Factorise fully:

1. a2 – b 2

4y2 – 9

4a4 – 9b2

1 – x2

25 – a6

a8 – ¼

4a2b 2 – 81

0,25 – x2y6

9. 2a2 – 2b2 (take care!)

E Combining common factors with differences of squares

As you saw in the last exercise (number 9), it is essential to check for common factors first and then to factorise the bracketed polynomial if possible.

Another example:

Factorise 12ax2 – 3ay2

First recognise that there is a common factor of 3a, before saying that this can’t be a difference of squares.

12ax2 – 3ay2 = 3a (4x2 – y2) Now we recognise 4x2 – y2 as the difference of two squares.

12ax2 – 3ay2 = 3a (4x2 – y2) = 3a(2x + y)(2x – y).

Exercise:

Factorise completely:

1. ax2 – ay4

2. a3 – ab2

3. 0,5a2x – 4,5b2x

4. a5b3c – abc

F Successive differences of squares

Try factorising this binomial completely by keeping your eyes open: a4 – b4

Now do this exercise – as usual factorise as far as possible.

1. x6 – 64

2. 1 – m8

3. 3a4 – 24b8

4. x – x9

G Factorising trinomials

If you study the answers to the following four problems (they are from a previous activity), you will notice that the answers, after simplifying, sometimes have two terms, sometimes three terms and sometimes four terms. Discuss what you see happening (with a partner) and decide why they are different.