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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