Algebra Review
ClementRadcliffe, Contributor

As we continue to review algebra, I wish to remind you of the following:

The concepts included in algebra are fairly routine and, with effort, you all should be able to do them well.

Many areas were done in the lower forms and must be effectively revised.

Algebra should be selected as one of the compulsory topics in Section Two.

We will now review last week's homework:
1. Solve : x/4 + 16 = 2x

Solution

The appropriate method is to multiply both sides by 4.

... 4 x x/4 + 4 x 16 = 4 x 2x

... x + 64 = 8x

... 7x = 64

... x = 64/7

N.B. You may also simplify the left- hand side prior to equating both sides. You may wish to try this approach on your own.

2. Solve: (2x - 3)/2 - (x + 4)/4 = 1

Solution

In this case, the method recommended above may also be used.

As the L.C.M. of 2 and 4 is 4, simplify the left hand side:

(2x - 3)/2 - (x + 4)/4

2(2x - 3) - (x + 4)/4 = (4x - 6 - x - 4)/4 = (3x - 10)/4

Equating both sides:

... (3x - 10)/4 = 1

... 3x - 10 = 4

... 3x = 14 or x = 14/3

3. Solve (5y - 4)/4 - (3y - 7)/2 = y

You are reminded that you may:

Simplify the left-hand side and then equate it to y.

Multiply both sides by 4 which is the L.C.M. of 4 and 2.

In this case, the latter is recommended.

... 4 x (5y - 4)/4 - 4 x (3y - 7)/2 = 4 x y

... 5y - 4 - 2(3y - 7) = 4y

... 5y - 4 - 6y + 14 = 4y

... -5y = -10

... y = -10/-5 = 2

Answer is y = 2.

We will now continue algebra with the topic factorisation.

Note that an algebraic expression is factorised when it is expressed as the product of its simplest factors. The usual methods are:

(a) Common factor

(b) Grouping

(c) Factorising of quadratic expressions

(d) Difference of two squares

The methods are adequately explained in the text books and you should use them to aid you as you revise for your exams.

It is important that you do the following in all cases:

(a) Bring each factor to its simplest form, for example, a factor 12x + 9 should be expressed as 3(4x + 3).

(b) Check your answers, if you have the time, by expanding and comparing the result with the original expression.

The following examples are presented for your benefit.

EXAMPLES OF COMMON FACTOR METHOD

1. Factorise: 8x2 -12x

The common factor method is used, as 4x is the factor which is common to both terms. Both terms are divided by 4x for us to obtain the second factor.

Answer: 4x(2x - 3)

2. Factorise: 15x2y -10xy3

Note that the common factor is 5xy

... Answer is 5xy(3x - 2y2)

EXAMPLES OF GROUPING METHOD

3. Factorise ax + ay + bx + by

Note that a is the common factor of ax + ay and b the common factor of bx + by

... ax + ay + bx + by = a(x + y) + b(x + y)

Do you realise that (x + y) is common to both expressions?

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

This method could therefore be described as repeated common factor method.

4. Factorise 2ax - 6ay + bx - 3by

2a(x - 3y) + b (x - 3y)

= (x - 3y)(2a+ b)

EXAMPLES OF METHOD OF FACTORISING QUADRATIC EXPRESSIONS

5. Factorise x2 + 8x + 15

This method is based on the principle that (x + b)(x + c) = x2 + (b + c) x + bc. Do you see the relationship between (b + c) which is the coefficient of x, bc which is the constant term, and b and c which are the values in the brackets on the left- hand side? This relationship and the 'trial and error' play an important role in this method.

Using the above:

x2 + 8x + 15 = (x + 5)(x +3)

If you have not realised the relationship mentioned, then please note that:

5 + 3 = 8 (coefficient of x)

5 x 3 = 15 (the constant term)

You may use 'trial and error' to identify 5 and 3, the values which satisfy the relationship.

6. Factorise: 2x2 + 5x -12

Despite the coefficient of x2 being 2, a method similar to that of example 5 is used.

... 2x2 + 5x - 12 = (2x - 3)(x + 4)

EXAMPLES OF METHOD OF DIFFERENCE OF TWO SQUARES

7. Factorise: 16 - x2

This is based on the fact that a2 - b2 = (a - b)(a + b). The critical problem is therefore to find the square root of each term.

As square root of 16 = 4 and square root of x square = x

... 16 - x? = (4 - x)(4 + x).

We will try another example.

8. Factorise: 9x2 - 25

By using the method of difference of two squares, you can show that since:

square root of 9X2 = 3x and square root of 25 = 5, then

9X2 - 16 = (3x - 5)(3x + 5).

Competence is developed in the solution of these problems if you practise extensively. Remember to check your answers by expanding the factors.

Now, please attempt the following for homework:

Factorise:

(a) x2 + 7x + 12

(b) x2 - 4x - 21

(c) 3x2 - 7x -6

(d) 2x2 + 5x -12

(e) 3x -8y - 4xy + 6

(f) x2 - y2 - 4x + 4y

(g) 9a2 - b2

(h) 16x2 -1

Clement Radcliffe is the principal of Glenmuir High School in May Pen.