Products and factorization

Products and factorization

There are many occasions when we are required to determine the factors and products of algebraic expressions. Literal numbers are used in expressions and formulae to provide a precise, technically accurate way of generalizing laws and statements associated with mathematics, science and engineering, as mentioned previously. When manipulating such expressions, we are often required to multiply them together (determine their product) or carry out the reverse process, that of factorization. You will see, in your later studies, that these techniques are very useful when it comes to changing the subject of a particular algebraic formula, in other words when you are required to transpose a formula, in terms of a particular variable. We begin by considering the products of some algebraic expressions. Once we are familiar with the way in which these expressions are `built–up' we can then look at the rather more difficult inverse process, that of factorization.

Consider the two factors (1 + a) and (l + b), noting that each factor consists of a natural number and a literal number. Suppose we are

required to find (1 + a)(1 + b), in other words, their product, providing we follow a set sequence, obeying the laws of multiplication of arithmetic, then the process is really quite simple!

In order to describe the process accurately, I need to remind you of some basic terminology. In the factor (1 + a) the natural number

1 is considered to be a constant because it has no other value, on the other hand the literal number a, can be assigned any number of values; therefore, it is referred to as a variable. Any number or group of numbers, whether natural or literal, separated by a +, – or = sign

is referred to as a term. So, for example, the expression (1 + a) has two terms.

When multiplying (1 + a) by (1 + b) we start the multiplication process from the left and work to the right, in the same manner as reading a book. We multiply each term in the left hand bracket by each of the terms in the right hand bracket, as follows: (l+a)(I+b)=(1–1)+(l–b)+(a–l)+(a–b)=1+b+a+ab =1+a+b+ab

Note:   1. The `dot' notation for multiplication has been used to avoid confusion with the variable x.

2. It does not matter in which order the factors are multiplied refer back to the commutative law of arithmetic, if you do not understand this fact.

Example 4.8

Determine the product of the following algebraic factors:

(a) (a + b) (ca – b)     and     (b) (2a – 3)(a – 1)

(a) In this example we proceed in the same manner as we did above, that is, (a + b) (a – b) = (a ^. a) + (a)(–b) + (b ^. a) + (b)(–b) = a2 + (–ab) + (ba) + (–b²), which by the laws of signs = a² – ob + bo – b² and by the commutative law this can be written as, a² – ab + ab – b² or (a + b)(a – b) = a² – b². I hope you have followed this process and recognize the notation for multiplying two bracketed terms.

The product a² – b² is a special case and is known as the product of two squares. This enables you to write down the product of any two factors that take the form (x + y) (x – y) as equal to x^Z – y², where x and y are any two variables.

(b) Again for these factors, we follow the process, where we get:

(2a – 3)(a – 1) = 2a ^. a + (2a)(–1) + (–3)(a) + (–3)( –1) = 2a² – 2a – 3a + 3 and so,

(2a–3)(a–1)=2a²–5a+3

I hope you are getting the idea of how to multiply factors to produce products. So far we have restricted ourselves to just two factors. Can we adopt the process for three or more factors? Well, if you did not know already, you will be pleased to know that we can!