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Created: 20 Sep, 2010; Last Modified: 07 Oct, 2016

Algebra II - 05

Factorisation

Factorisation of an algebraic expression implies writing the expression as a product of simpler algebraic expressions, each of which is called a factor. Factorisation is also termed resolution into factors.

Principle Rules of Factorisation

Taking out the common factor in polynomials

Simple factor

When each of the terms in a polynomial contains a common factor, which is basically a monomial, it can be easily separated out. Such a common factor is called a simple factor. Simple factors are easily spotted by inspection.

Factorise (i) 3 a 2 6 a b (ii) 5 a 2 b x 3 15 a b x 2 20 b 3 x 2 . (Hall & Knight, Elementary Algebra, p. 125 Example 1, 2)

(i) The terms of the expression have 3a as the common factor, which can be separated out. Thus,

3 a 2 6 a b = 3 a ( a 2 b )

(ii) The common factor is 5 b x 2 . So,

5 a 2 b x 3 15 a b x 2 20 b 3 x 2 = 5 b x 2 ( a 2 x 3 a 4 b 2 )

Compound factor

Sometimes, the common factor itself is a polynomial, and can be identified and fished out only after rearranging the terms of the original polynomial into suitable groups. Such a common factor is called a compound factor.

Resolve into factors (i) x 2 a x + b x a b (ii) 6 x 2 9 a x + 4 b x 6 a b . (Hall & Knight, Elementary Algebra, p. 126 Example 1, 2)

(i) Noticing that the first two terms contain a common factor x, and the last two terms a common factor b, we group the terms as follows and continue:

x 2 a x + b x a b = ( x 2 a x ) + ( b x a b ) = x ( x a ) + b ( x a ) = ( x + b ) ( x a )

(ii) Using a similar thinking process, we have

6 x 2 9 a x + 4 b x 6 a b = ( 6 x 2 9 a x ) + ( 4 b x 6 a b ) = 3 x ( 2 x 3 a ) + 2 b ( 2 x 3 a ) = ( 3 x + 2 b ) ( 2 x 3 a )

Figuring out the compound factors in complex expressions becomes easier with practice.

Splitting the middle term of trinomials of degree 2

Factorisation of (x2+bx+c)

Let the factors of x 2 + b x + c be of the form ( x + p ) ( x + q ) . Then,

x 2 + b x + c = ( x + p ) ( x + q ) = x 2 + p x + q x + p q = x 2 + ( p + q ) x + p q

Comparing the coefficients of like terms on both sides, we get ( p + q ) = b and p q = c .

Thus, to factorise a trinomial of the form x 2 + b x + c ,

  1. Find two number p and q such that ( p + q ) = b and p q = c
  2. The trinomial now stands factorised as ( x + p ) ( x + q ) .

Resolve into factors (i) x 2 + 11 x + 24 (ii) x 2 10 x + 24 . (Hall & Knight, Elementary Algebra, p. 127 Example 1, 2)

(i) We need to find p and q such that ( p + q ) = 11 and p q = 24 . Since the product is positive, as well as the sum, both the numbers must be positive. Clearly, p = 8 and q = 3 satisfy the conditions. Thus,

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

(ii) We need to find p and q such that ( p + q ) = 10 and p q = 24 . This implies that both must be negative, since their sum is negative but the product is positive. Clearly, p = 6 and q = 4 satisfy the conditions. Thus,

x 2 10 x + 24 = ( x 4 ) ( x 6 )

Resolve into factors (i) x 2 + 2 x 35 (ii) x 2 3 x 54 . (Hall & Knight, Elementary Algebra, p. 129 Example 1, 2)

(i) For ( p + q ) = 2 and p q = 54 to be true, the numbers must have opposite signs, and the greater must be positive for the sum to be positive. Clearly, p = 7 and q = 5 satisfy the conditions.

x 2 + 2 x 35 = ( x + 7 ) ( x 5 )

(ii) For ( p + q ) = 3 and p q = 54 to be true, the numbers must have opposite signs, and the greater must be negative for the sum to be negative. p = 7 and q = 5 fit the bill.

x 2 3 x 54 = ( x 9 ) ( x + 6 )

Factorisation of (ax2+bx+c)

Let factors of a x 2 + b x + c be of the form ( r x + s ) ( t x + u ) . Then,

a x 2 + b x + c = ( r x + s ) ( t x + u ) = r t x 2 + r u x + s t x + s u = r t x 2 + ( r u + s t ) x + s u

Comparing the coefficients of like terms on both sides, we get a = rt, b = ru + st and c = su.

If we let p = ru and q = st, then

pq = rtsu = ac and ( p + q ) = r u + s t = b

Thus, we should find p and q such that ( p + q ) = b and pq = ac.

So, to factorise a trinomial of the form a x 2 + b x + c ,

  1. Find two number p and q such that

    ( p + q ) = b and pq = ac.

  2. Rewrite the expression a x 2 + b x + c with the middle term split as:

    ( a x 2 + p x + q x + c )   or   ( a x 2 + p x + q x + p q a )

  3. In this form, the compound factors can then be taken out as follows:

    a x 2 + p x + q x + p q a = x ( a x + p ) + q ( x + p a ) = x ( a x + p ) + q a ( a x + p ) = ( a x + p ) ( x + q a ) = 1 a ( a x + p ) ( a x + q )

Resolve into factors (i) 6 x 2 + 7 x 3 (ii) 15 x 2 26 x + 8 . (Aggarwal & Aggarwal, Mathematics Class IX, p. 35 Ex.2 1, 2)

(i) We need to find p and q such that ( p + q ) = 7 and pq = 18. Clearly, one of them is negative, while the positive number has a larger absolute value. p = 9 and q = 2 satisfy the conditions. Thus,

6 x 2 + 7 x 3 = 6 x 2 + 9 x 2 x 3 = 3 x ( 2 x + 3 ) 1 ( 2 x + 3 ) = ( 3 x 1 ) ( 2 x + 3 )

(ii) We need to find p and q such that ( p + q ) = 26 and pq = 120. Clearly, both of them are negative. p = 20 and q = 6 satisfy the conditions. Thus,

15 x 2 26 x + 8 = 15 x 2 20 x 6 x + 8 = 5 x ( 3 x 4 ) 2 ( 3 x 4 ) = ( 5 x 2 ) ( 3 x 4 )

The difference of two squares

As we have seen in the monograph on Expansions for the simplification of ( a + b ) ( a b ) :

( a + b ) ( a b ) = a 2 a b + b a b 2 = a 2 b 2

Thus,

a 2 b 2 = ( a + b ) ( a b )

.......(1)

In words, the difference of the squares of two numbers is equal to the product of the sum and the difference of the two numbers.

Resolve into factors (i) 25 x 2 16 y 2 (ii) 1 49 c 6 . (Hall & Knight, Elementary Algebra, p. 133 Example 1, 2)

(i) 25 x 2 16 y 2 = ( 5 x ) 2 ( 4 y ) 2 = ( 5 x + 4 y ) ( 5 x 4 y )

(ii) 1 49 c 6 = ( 1 ) 2 ( 7 c 3 ) 2 = ( 1 + 7 c 3 ) ( 1 7 c 3 )

Resolve into factors (i) x 2 + x y 2 y 4 (ii) ( 1 a 2 ) ( 1 b 2 ) + 4 a b . ICSE

(i) x 2 + x y 2 y 4 = ( x 2 4 ) + x y 2 y (Rearranging the terms) = ( x + 2 ) ( x 2 ) + y ( x 2 ) = ( x + y + 2 ) ( x 2 )

(ii) ( 1 a 2 ) ( 1 b 2 ) + 4 a b = 1 b 2 a 2 + a 2 b 2 + 4 a b = 1 ( a 2 + b 2 ) + a 2 b 2 + 2 a b + 2 a b = 1 + 2 a b + a 2 b 2 ( a 2 + b 2 2 a b ) = ( 1 + a b ) 2 ( a b ) 2 = ( ( 1 + a b ) + ( a b ) ) ( ( 1 + a b ) ( a b ) ) = ( 1 + a b + a b ) ( 1 a + b + a b )

The sum and difference of two cubes

Dividing a 3 + b 3 by ( a + b ) gives the quotient a 2 a b + b 2 , while dividing a 3 b 3 by ( a b ) gives the quotient a 2 + a b + b 2 . This leads us to the following identities:

a 3 + b 3 = ( a + b ) ( a 2 a b + b 2 )

.......(2)

a 3 b 3 = ( a b ) ( a 2 + a b + b 2 )

.......(3)

Hence, any expression which can be written as a sum or a difference of two cubes can be resolved into factors as above.

Resolve 64 a 3 + 1 into factors. (Hall & Knight, Elementary Algebra, p. 136 Example 2)

64 a 3 + 1 = ( 4 a ) 3 + ( 1 ) 3 = ( 4 a + 1 ) ( ( 4 a ) 2 4 a 1 + ( 1 ) 2 ) = ( 4 a + 1 ) ( 16 a 2 4 a + 1 )

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List of References

Aggarwal, RS & Aggarwal, V, Secondary School Mathematics for Class IX, Patna, India: Bharati Bhawan, 1999.
Hall, HS & Knight, SR, Elementary Algebra for Schools, metric edn, Agra, India: AK Publications, 1966.

Bibliography

Aggarwal, RS & Aggarwal, V, Secondary School Mathematics for Class IX, Patna, India: Bharati Bhawan, 1999.
Gupta, SD & Banerjee, A, ICSE Mathematics for Class 9, Patna, India: Bharati Bhawan, 2003.
Hall, HS & Knight, SR, Elementary Algebra for Schools, metric edn, Agra, India: AK Publications, 1966.
Schultze, A, Elements of Algebra, NY, USA: The Macmillan Company, 1910.