The algebra of the four basic operations  (Page 4/5)

1. (a + b) (a – b) = a2 – ab + ab – b2 = a2 – b2 (when simplified)

2. (a + 2) (a + 3) = a2 + 3a + 2a + 6 = a2 + 5a + 6

3. (a + b) (a + b) = a×a +ab + ba + b×b = a2 + ab + ab + b2 = a2 + 2ab + b2 (in the simplest form)

4. (a + b) (c + d) = ac + ad + bc + bd (and this answer can not be simplified)

The answer to the type of problem represented by number 1 above is a difference of squares.

The answers to numbers 2 and 3 are trinomials, and we will now see how to factorise them

The first fact that you must always remember is that not all trinomials can be factorised.

Work backwards through problem 2:

a2 + 5a + 6 = a2 + 3a + 2a + 6 = (a + 2) (a + 3).

So that you can see clearly where the a2 came from, and the 5a and the 6.

Now try to factorise a2 + 7a + 12 = ( …………. ) ( …………… ) by filling in two correct binomials in the brackets.

You can check your answer by multiplying the binomials in your answer as you were taught in activity 2. Keep trying and checking your answer until you have it right. Do the same in the following three exercises:

Match up the two columns:

A. a2 – 5a – 6 1. (x + 2)(x + 3)

B. a2 – a – 6 2. (x – 2)(x + 3)

C. a2 – 5a + 6 3. (x + 1)(x – 6)

D. a2 + 7a + 6 4. (x – 2)(x – 3)

E. a2 + 5a + 6 5. (x + 1)(x + 6)

F. a2 + 5a – 6 6. (x – 1)(x + 6)

G. a2 + a – 6 7. (x + 2)(x – 3)

H. a2 – 7a + 6 8. (x – 1)(x – 6)

Now factorise the following trinomials by using the same techniques you have just learnt. The last two are more difficult than the first four!

a2 + 3a + 2

a2 + a – 12

a2 – 4a + 3

a2 – 9a + 20

a2 + ab – 12b2

2a2 – 18a + 40

Finally, in groups of 3, 4 or 5, work out exactly how one should go about factorising these trinomials, and write down a strategy that will get you to the answer accurately and quickly.

Activity 4

To use factorising in simplifying fractions, and in multiplying, dividing and adding fractions

[LO 1.2, 1.6, 2.9]

A. Simplifying algebraic fractions

Two of the following four fractions can be simplified and the others can’t. Which is which?

$\frac{2+a}{2-a}$

$\frac{3\left(a+b\right)}{a+b}$

$\frac{4+x}{x+4}$

As you have seen in the previous activity, factorising is a lot of trouble. So, why do we do it?

The following expression cannot be simplified as it is $\frac{{\mathrm{6a}}^{2}b-\mathrm{6b}}{\mathrm{2a}-2}$ because we are not allowed to cancel terms. If we can change the sum expressions into product expressions (by factorising) then we will be able to cancel the factors, and simplify.

6a2b – 6b = 6b (a2 – 1) = 6b (a + 1) (a – 1) and 2a – 2 = 2(a – 1)

So, the reason we factorise is that it allows us to simplify expressions better.

Now: $\frac{{\mathrm{6a}}^{2}b-\mathrm{6b}}{\mathrm{2a}-2}$ = $\frac{6b\left(a+1\right)\left(a-1\right)}{2\left(a-1\right)}$ = $\frac{\mathrm{3b}\left(a+1\right)}{1}$ = 3b(a + 1) .

It is very important that you factorise completely.

Exercise:

Factorise the numerator and denominator, cancel factors and write in the simplest form:

1 $\frac{\text{12}a+\mathrm{6b}}{\mathrm{2a}+b}$

2 $\frac{x^2-9}{x+3}$

3 $\frac{2\left(a+1\right)\left(a-1\right)}{6{\left(a+1\right)}^2}$

4 $\frac{{\mathrm{5a}}^2-5}{\mathrm{5a}+5}$

B. Multiplying and dividing fractions

Our normal rules for multiplication and division of fractions are still the same. Study the following examples, taking special note of the factorising and cancelling.