Factorisation

Introduction

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.

Factorisation

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× × yx

          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) = (xx +×× 23 )

        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) = \(2\times 5\times x\times (x+2)\times (y+3)\)

Factorisation

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, 5 x ²y , 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, x 2 + 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 1: Factorise 12a ²b + 15ab²

Solution: We have 12a ²b = 2 × 2 × 3 × a × a × b

                          15ab² = 3 × 5 × a × b × b

The two terms have 3, a and b as common factors.

Therefore,12a ²b + 15ab²=(3 × a × b × 2 × 2 × a) + (3 × a × b × 5 × b)

= 3 × a × b × [(2 × 2 × a) + (5 × b)]         (combining the terms)

= 3ab × (4a + 5b)

= 3ab (4a + 5b)                   (required factor form)

Example 2: Factorise 10x ² – 18x ³ + 14x ⁴

Solution: 10x ² = 2 × 5 × x × x

             18x ³ = 2 × 3 × 3 × x × x × x

             14x ⁴ = 2 × 7 × x × x × x × x

The common factors of the three terms are 2, x and x.

Therefore, 10x ² – 18x ³ + 14x ⁴ = (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)]  (combining the three terms)

= 2x ² × (5 – 9x + 7x ² ) = \(2{{x}^{2}}(7{{x}^{2}}-9x+5)\)             

TRY THESE

Factorise: (i) 12x + 36          (ii) 22y – 33z           (iii) 14pq + 35pqr                               

Factorisation

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 3: 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

Factorisation using identities

We know that       (a + b) ² = a ² + 2ab + b ² (I)

                           (a – b) ² = a ² – 2ab + b ² (II)

                           (a + b) (a – b) = a ² – b ² (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 righthand side of one of the identities, then the expression corresponding to the lefthand side of the identity gives the desired factorisation.

Example 4: Factorise x ² + 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 a ² + 2ab + b 2 where a = x and b = 4

such that              a ² + 2ab + b ² = x ² + 2 (x) (4) + 42

                                                  = x ² + 8x + 16

Since a ² + 2ab + b ² = (a + b) ²,

by comparison x ² + 8x + 16 = ( x + 4)² (the required factorisation)

Example 5: Factorise 4y ² – 12y + 9

Solution: Observe 4y ² = (2y) ² , 9 = 3² and 12y = 2 × 3 × (2y)

Therefore, 4y ² – 12y + 9 = (2y) ² – 2 × 3 × (2y) + (3)²

                                     = ( 2y – 3)² (required factorisation)

Example 6: Factorise 49p ² – 36

Solution: There are two terms; both are squares and the second is negative. The expression is of the form (a ² – b² ). Identity III is applicable here;

                                            49p ² – 36 = (7p) ² – ( 6 )²

                                                            = (7p – 6 ) ( 7p + 6) (required factorisation)

Example 7: Factorise a ² – 2ab + b ² – c ²

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,                                a ² – 2ab + b ² – c ² = (a – b) ²– c ²                   (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.

Example 8: Factorise m⁴ – 256

Solution: We note                 m⁴ = (m²) ² and 256 = (16) ²

Thus, the given expression fits Identity III.

Therefore, m⁴ – 256 = (m²) ² – (16)² = (m² –16) (m² +16) [(using Identity (III)]

Now, (m² + 16) cannot be factorised further, but (m² –16) is factorisable again as per Identity III.

                                         m²–16 = m² – 4 ²

                                                   = (m – 4) (m + 4)

Therefore,                      m⁴ – 256 = (m – 4) (m + 4) (m² +16)

Factorisation

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

Let us now discuss how we can factorise expressions in one variable, like x ² + 5x + 6, y ² – 7y + 12, z ² – 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 x ² + 5x + 6, the term 6 is not a perfect square. These expressions obviously also do not fit the type (a ² – b² ) either.

           They, however, seem to be of the type x ² + (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) = x ² + (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 9: Factorise x ² + 5x + 6

Solution: If we compare the R.H.S. of Identity (IV) with x ² + 5x + 6, we find ab = 6, and a + b = 5. From this, we must obtain a and b. The factors then will be (x + a) and (x + b).

           If a b = 6, it means that a and b are factors of 6. Let us try a = 6, b = 1. For these values a + b = 7, and not 5, So this choice is not right.

Let us try a = 2, b = 3. For this a + b = 5 exactly as required.

The factorised form of this given expression is then (x +2) (x + 3).

In general, for factorising an algebraic expression of the type x ² + px + q, we find two factors a and b of q (i.e., the constant term) such that

                             ab = q and a + b = p

Then, the expression becomes x ² + (a + b) x + ab

or                                         x ² + ax + bx + ab

or                                         x(x + a) + b(x + a)

or                                    (x + a) (x + b) which are the required factors.

Example 10: Find the factors of y ² –7y +12.

Solution: We note 12 = 3 × 4 and 3 + 4 = 7. Therefore,

                      y ² – 7y+ 12 = y ² – 3y – 4y + 12

                   = y (y –3) – 4 (y –3) = (y –3) (y – 4)

Example 11: Obtain the factors of z ² – 4z – 12.

Solution: Here a b = –12 ; this means one of a and b is negative. Further, a + b = – 4, this means the one with larger numerical value is negative. We try a = – 4, b = 3; but this will not work, since a + b = –1. Next possible values are a = – 6, b = 2, so that a + b = – 4 as required.

Hence, z ² – 4z –12 = z ² – 6z + 2z –12

                            = z(z – 6) + 2(z – 6 )

                            = (z – 6) (z + 2)

Example 12: 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)

Factorisation

Division of Algebraic Expressions

We have learnt how to add and subtract algebraic expressions. We also know how to multiply two expressions. We have not however, looked at division of one algebraic expression by another. This is what we wish to do in this section.

          We recall that division is the inverse operation of multiplication. Thus, 7 × 8 = 56 gives 56 ÷ 8 = 7 or 56 ÷ 7 = 8.

We may similarly follow the division of algebraic expressions. For example,

(i)                     2x × 3x ² = 6x ³

Therefore,         6x ³ ÷ 2x = 3x ²

and also,           6x ³ ÷ 3x ² = 2x.

(ii)               5x (x + 4) = 5x ² + 20x

Therefore,    (5x ² + 20x) ÷ 5x = x + 4

and also       (5x ² + 20x) ÷ (x + 4) = 5x.

We shall now look closely at how the division of one expression by another can be carried out. To begin with we shall consider the division of a monomial by another monomial.

Factorisation

Division of a monomial by another monomial

Consider 6x ³ ÷ 2x

We may write 2x and 6x ³ in irreducible factor forms,

          2x = 2 × x

         6x ³ = 2 × 3 × x × x × x

Now we group factors of 6x ³ to separate 2x,

6x ³ = 2 × x × (3 × x × x) = (2x) × (3x ² )

Therefore, 6x ³ ÷ 2x = 3x ² .

A shorter way to depict cancellation of common factors is as we do in division of numbers: 77 ÷ 7 = \(\frac{77}{7}=\frac{7\times 11}{7}\)= 11

Similarly, 6x ³ ÷ 2x = \(\frac{6{{x}^{3}}}{2x}\)

                            = \(\frac{2\times 3\times x\times x\times x}{2\times x}\)= 3 × x × x = 3x ²

Example 13: Do the following divisions. (i) –20x ⁴ ÷ 10x ² (ii) 7x ² y ² z ² ÷ 14xyz

Solution: (i) –20x ⁴ = –2 × 2 × 5 × x × x × x × x

                    10x ² = 2 × 5 × x × x

Therefore, (–20x ⁴ ) ÷ 10x ² = \(-\frac{2\times 2\times 5\times x\times x\times x\times x}{2\times 5\times x\times x}\)= –2 × x × x = –2x ²

(ii) 7x ² y ² z ² ÷ 14xyz   = \(\frac{7\times x\times x\times y\times y\times z\times z}{2\times 7\times x\times y\times z}=\frac{x\times y\times z}{2}=\frac{1}{2}xyz\)

TRY THESE

Divide. (i) 24xy² z ³ by 6yz²        (ii) 63a ²b ⁴ c ⁶ by 7a 2b ² c ³

Factorisation

Division of a polynomial by a monomial

Let us consider the division of the trinomial 4y ³ + 5y ² + 6y by the monomial 2y.

                        4y ³ + 5y ² + 6y = (2 × 2 × y × y × y) + (5 × y × y) + (2 × 3 × y)

(Here, we expressed each term of the polynomial in factor form) we find that 2 × y is common in each term. Therefore, separating 2 × y from each term. We get

4y ³ + 5y ² + 6y =2 × y × (2 × y × y) + 2 × y \((\frac{5}{2}\times y)\)+ 2 × y × 3

= 2y (2y ² ) + 2y \((\frac{5}{2}y)\)+ 2y (3)

= \(2y\left( 2{{y}^{2}}+\frac{5}{2}y+3 \right)\) (The common factor 2y is shown separately.

Therefore, \(\frac{4{{y}^{3}}+5{{y}^{2}}+6y}{2y}=\frac{2y(2{{y}^{2}}+\frac{5}{2}y+3)}{2y}=2{{y}^{2}}+\frac{5}{2}y+3\)

Alternatively, we could divide each term of the trinomial by the monomial using the cancellation method.

(4y ³ + 5y ² + 6y) ÷ 2y = \(\frac{4{{y}^{3}}+5{{y}^{2}}+6y}{2y}=\frac{4{{y}^{3}}}{2y}+\frac{5{{y}^{2}}}{2y}+\frac{6y}{2y}=2{{y}^{2}}+\frac{5}{2}y+3\)

Example 14: Divide 24(x ² yz + xy² z + xyz ² ) by 8xyz using both the methods.

Solution: 24 (x ² yz + xy² z + xyz² ) = 2 × 2 × 2 × 3 × [(x × x × y × z) + (x × y × y × z) + (x × y × z × z)] = 2 × 2 × 2 × 3 × x × y × z × (x + y + z) = 8 × 3 × xyz × (x + y + z)

Therefore, 24 (x ² yz + xy² z + xyz² ) ÷ 8xyz

= \(\frac{8\times 3\times xyz\times (x+y+z)}{8\times xyz}\)= 3 × (x + y + z) = 3 (x + y + z)

Alternately,  24(x ²yz + xy² z + xyz² ) ÷ 8xyz \(\frac{24{{x}^{2}}yz}{8xyz}+\frac{24x{{y}^{2}}z}{8xyz}+\frac{24xy{{z}^{2}}}{8xyz}\) = 3x + 3y + 3z = 3(x + y + z)

Factorisation

Division of Algebraic Expressions Continued (Polynomial ÷ Polynomial)

• Consider (7x ² + 14x) ÷ (x + 2)

We shall factorise (7x ² + 14x) first to check and match factors with the denominator:

7x ² + 14x = (7 × x × x) + (2 × 7 × x)

= 7 × x × (x + 2) = 7x(x + 2)

Now (7x ² + 14x) ÷ (x + 2)

\(\frac{7{{x}^{2}}+14x}{x+2}=\frac{7x(x+2)}{x+2}\)=7x (Cancelling the factor (x + 2))

Example 15: Divide 44(x ⁴ – 5x ³ – 24x ² ) by 11x (x – 8)

Solution: Factorising 44(x ⁴ – 5x ³ – 24x ² ), we get

44(x ⁴ – 5x ³ – 24x ² ) = 2 × 2 × 11 × x ² (x ² – 5x – 24)

            (taking the common factor x ² out of the bracket)

= 2 × 2 × 11 × x ² (x ² – 8x + 3x – 24)

= 2 × 2 × 11 × x ² [x (x – 8) + 3(x – 8)]

= 2 × 2 × 11 × x ² (x + 3) (x – 8)

Therefore, 44(x ⁴ – 5x ³ – 24x ² ) ÷ 11x(x – 8)

= \(\frac{2\times 2\times 11\times x\times x\times (x+3)\times (x-8)}{11\times x\times (x-8)}\)

We cancel the factors 11, x and (x – 8) common to both the numerator and denominator.

= 2 × 2 × x (x + 3) = 4x(x + 3)

Example 16: Divide z(5z ² – 80) by 5z(z + 4)

Solution: Dividend = z(5z ² – 80)

= z[(5 × z ² ) – (5 × 16)]

= z × 5 × (z ² – 16)

= 5z × (z + 4) (z – 4)

[using the identity a ² – b ² = (a + b) (a – b)]

Thus, z(5z ² – 80) ÷ 5z(z + 4) = \(\frac{5z(z-4)(z+4)}{5z(z+4)}=(z-4)\)