   Factors of Algebraic Expressions

## Factors of natural numbers

You will remember what you learnt about factors in Class VI. Let us take a natural number, say 30, and write it as a product of other natural numbers, say
30 = 2 × 15
= 3 × 10 = 5 × 6
Thus, 1, 2, 3, 5, 6, 10, 15 and 30 are the factors of 30. Of these, 2, 3 and 5 are the prime factors of 30 (Why?)
A number written as a product of prime factors is said to be in the prime factor form; for example, 30 written as 2 × 3 × 5 is in the prime factor form.
The prime factor form of 70 is 2 × 5 × 7.
The prime factor form of 90 is 2 × 3 × 3 × 5, and so on.
Similarly, we can express algebraic expressions as products of their factors. This is what we shall learn to do in this chapter.

## Factors of algebraic expressions

We have seen in Class VII that in algebraic expressions, terms are formed as products of factors. For example, in the algebraic expression 5xy + 3x the term 5xy has been formed by the factors 5, x and y, i.e.,

5xy = 5× x× y

Observe that the factors 5, x and y of 5xy cannot further be expressed as a product of factors. We may say that 5, x and y are ‘prime’ factors of 5xy. In algebraic expressions, we use the word ‘irreducible’ in place of ‘prime’. We say that 5 × x × y is the irreducible form of 5xy. Note 5 × (xy) is not an irreducible form of 5xy, since the factor xy can be further expressed as a product of x and y, i.e., xy = x × y.

Next consider the expression 3x (x + 2). It can be written as a product of factors. 3, x and (x + 2)

3x(x + 2) =3× x×(x + 2)

The factors 3, x and (x +2) are irreducible factors of 3x (x + 2).
Similarly, the expression 10x (x + 2) (y + 3) is expressed in its irreducible factor form
as 10x (x + 2) (y + 3) = 2 × 5 × x × (x + 2) × ( y + 3).

## What is Factorisation?

When we factorise an algebraic expression, we write it as a product of factors. These factors may be numbers, algebraic variables or algebraic expressions.

Expressions like 3xy, 5x2 y , 2x (y + 2), 5 (y + 1) (x + 2) are already in factor form. Their factors can be just read off from them, as we already know.

On the other hand consider expressions like 2x + 4, 3x + 3y, x2 + 5x, x2 + 5x + 6. It is not obvious what their factors are. We need to develop systematic methods to factorise these expressions, i.e., to find their factors. This is what we shall do now.

## Method of common factors

• We begin with a simple example: Factorise 2x + 4.
We shall write each term as a product of irreducible factors;
2x = 2 × x
4 = 2 × 2
Hence                                       2x + 4 = (2 × x) + (2 × 2)
Notice that factor 2 is common to both the terms.
Observe, by distributive law
2 × (x + 2) = (2 × x) + (2 × 2)
Therefore, we can write
2x + 4 = 2 × (x + 2) = 2 (x + 2)
Thus, the expression 2x + 4 is the same as 2 (x + 2). Now we can read off its factors:
they are 2 and (x + 2). These factors are irreducible.
Next, factorise 5xy + 10x.
The irreducible factor forms of 5xy and 10x are respectively,
5xy = 5 × x × y
10x = 2 × 5 × x
Observe that the two terms have 5 and x as common factors. Now,
5xy + 10x = (5 × x × y) + (5 × x × 2)
= (5x × y) + (5x × 2)
We combine the two terms using the distributive law,
(5x× y) + (5x× 2) = 5x × ( y + 2)
Therefore, 5xy + 10x = 5 x (y + 2). (This is the desired factor form.)

Example : Factorise 12a2b + 15ab2
Solution: We have        12a2b = 2 × 2 × 3 × a × a × b
15ab2 = 3 × 5 × a × b × b
The two terms have 3, a and b as common factors.
Therefore, 12a2b + 15ab2 = (3 × a × b × 2 × 2 × a) + (3 × a × b × 5 × b)
= 3 × a × b × [(2 × 2 × a) + (5 × b)]
= 3ab × (4a + 5b)
= 3ab (4a + 5b)                                                             (required factor form)

Example 2: Factorise 10x2 – 18x3 + 14x4
Solution:                 10x2 = 2 × 5 × x × x
18x3 = 2 × 3 × 3 × x × x × x
14x4 = 2 × 7 × x × x × x × x
The common factors of the three terms are 2, x and x.
Therefore, 10x2 – 18x3 + 14x4 = (2 × x × x × 5) – (2 × x × x × 3 × 3 × x)
+ (2 × x × x × 7 × x × x)
= 2 × x × x ×[(5 – (3 × 3 × x) + (7 × x × x)]
= 2x2 × (5 – 9x + 7x2) = 2x2 (7x2 − 9x + 5)

## Factorisation by regrouping terms

Look at the expression 2xy + 2y + 3x + 3. You will notice that the first two terms have common factors 2 and y and the last two terms have a common factor 3. But there is no single factor common to all the terms. How shall we proceed?
Let us write (2xy + 2y) in the factor form:
2xy + 2y = (2 × x × y) + (2 × y)
= (2 × y × x) + (2 × y × 1)
= (2y × x) + (2y × 1) = 2y (x + 1)
Similarly,              3x + 3 = (3 × x) + (3 × 1)
= 3 × (x + 1) = 3 ( x + 1)
Hence, 2xy + 2y + 3x + 3 = 2y (x + 1) + 3 (x +1)
Observe, now we have a common factor (x + 1) in both the terms on the right hand
side. Combining the two terms,
2xy + 2y + 3x + 3 = 2y (x + 1) + 3 (x + 1) = (x + 1) (2y + 3)
The expression 2xy + 2y + 3x + 3 is now in the form of a product of factors. Its
factors are (x + 1) and (2y + 3). Note, these factors are irreducible.

## What is regrouping?

Suppose, the above expression was given as 2xy + 3 + 2y + 3x; then it will not be easy to see the factorisation. Rearranging the expression, as 2xy + 2y + 3x + 3, allows us to form groups (2xy + 2y) and (3x + 3) leading to factorisation. This is regrouping.
Regrouping may be possible in more than one ways. Suppose, we regroup the expression as: 2xy + 3x + 2y + 3. This will also lead to factors. Let us try:
2xy + 3x + 2y + 3 = 2 × x × y + 3 × x + 2 × y + 3
= x × (2y + 3) + 1 × (2y + 3)
= (2y + 3) (x + 1)
The factors are the same (as they have to be), although they appear in different order

Example : Factorise 6xy – 4y + 6 – 9x.
Solution:
Step 1          Check if there is a common factor among all terms. There is none.
Step 2          Think of grouping. Notice that first two terms have a common factor 2y;
6xy – 4y = 2y (3x – 2) (a)
What about the last two terms? Observe them. If you change their order to
– 9x + 6, the factor ( 3x – 2) will come out;
–9x + 6 = –3 (3x) + 3 (2)
= – 3 (3x – 2) (b)
Step 3          Putting (a) and (b) together,
6xy – 4y + 6 – 9x = 6xy – 4y – 9x + 6
= 2y (3x – 2) – 3 (3x – 2)
= (3x – 2) (2y – 3)
The factors of (6xy – 4y + 6 – 9 x) are (3x – 2) and (2y – 3).

## Factorisation using identities

We know that                                (a + b)2 = a2 + 2ab + b2                                                     (I)
(a – b)2 = a2 – 2ab + b2                                                     (II)
(a + b) (a – b) = a2 – b2                                                             (III)
The following solved examples illustrate how to use these identities for factorisation. What we do is to observe the given expression. If it has a form that fits the right hand side of one of the identities, then the expression corresponding to the left hand side of the identity gives the desired factorisation.

Example :          Factorise x2 + 8x + 16
Solution:        Observe the expression; it has three terms. Therefore, it does not fit
Identity III. Also, it’s first and third terms are perfect squares with a positive sign before
the middle term. So, it is of the form a2 + 2ab + b2 where a = x and b = 4
such that               a2 + 2ab + b2 = x2 + 2 (x) (4) + 42
= x2 + 8x + 16
Since                                    a2 + 2ab + b2 = (a + b)2,
by comparison                       x2 + 8x + 16 = ( x + 4)2 (the required factorisation)

Example : Factorise a2 – 2ab + b2 – c2
Solution:     The first three terms of the given expression form (a – b)2. The fourth term is a square. So the expression can be reduced to a difference of two squares.
Thus,                               a2 – 2ab + b2 – c2 = (a – b)2– c                                                  (Applying Identity II)
= [(a – b) – c) ((a – b) + c)]                                                              (Applying Identity III)
= (a – b – c) (a – b + c)                                                                    (required factorisation)
Notice, how we applied two identities one after the other to obtain the required factorisation

## Factors of the form ( x + a) ( x + b)

Let us now discuss how we can factorise expressions in one variable, like x2 + 5x + 6, y2 – 7y + 12, z2 – 4z – 12, 3m2 + 9m + 6, etc. Observe that these expressions are not of the type (a + b) 2 or (a – b) 2, i.e., they are not perfect squares. For example, in
x2 + 5x + 6, the term 6 is not a perfect square. These expressions obviously also do not fit the type (a2 – b2) either.
They, however, seem to be of the type x2 + (a + b) x + a b.We may therefore, try to use Identity IV studied in the last chapter to factorise these expressions:
(x + a) (x + b) = x2 + (a + b) x + ab (IV)
For that we have to look at the coefficients of x and the constant term. Let us see how it is done in the following example.

Example: Find the factors of y2 –7y +12.
Solution: We note 12 = 3 × 4 and 3 + 4 = 7. Therefore,
y2 – 7y+ 12 = y2 – 3y – 4y + 12
= y (y –3) – 4 (y –3) = (y –3) (y – 4)

Note:, this time we did not compare the expression with that in Identity (IV) to identif a and b. After sufficient practice you may not need to compare the given expressions for their factorisation with the expressions in the identities; instead you can proceed directly as we did above

Example : Find the factors of 3m2 + 9m + 6.
Solution: We notice that 3 is a common factor of all the terms.
Therefore,                      3m2 + 9m + 6 = 3(m2 + 3m + 2)
Now,                             m 2 + 3m + 2 = m2 + m + 2m + 2                            (as 2 = 1 × 2)
= m(m + 1)+ 2( m + 1)
= (m + 1) (m + 2)
Therefore,                    3m2 + 9m + 6 = 3(m + 1) (m + 2)

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