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Factorisation

Factorisation of Algebraic Expressions by the Method of Common Factors

Let us start with a simple question.

What is the value of (203)3?

Yes, its value is 8365427.

For sure, you would have used the identity (a + b)3 = a3 + b3 + 3ab (a + b)

We can write (203)3 as (200 + 3)3 and then using the identity for (a + b)3, we can find its value as this is an easier method as compared to multiplication.

However, we have many more applications of this identity. We can also factorise algebraic expressions using this identity.

For this, we have to rewrite the identity as follows:

This form of the identity is used to factorise the expressions of the form a3 + b3.

In the same way, we can write the identity for a3b3 as follows:

To understand how to use these identities to factorise expressions, let us see an example.

Let us factorise the expression x6 − 729y6.

x6 − 729y6

= (x3)2 − (27y3)2

= (x3 + 27y3) (x3 − 27y3)[Using a2b2 = (a + b) (ab)]

= [(x)3 + (3y)3] [(x)3 − (3y)3]

Using identities (1) and (2), we obtain

(x + 3y) (x2 + 9y2 − 3xy) (x − 3y)(x2 + 9y2 + 3xy)

This is the factorised form of the given expression.

To understand this method more clearly, let us solve some more examples.

Example 1:

Factorise the expression: 125a6 − 343

Solution:

125a6 − 343 = (5a2)3 − (7)3

= (5a2 − 7) (25a4 + 49 + 35a2) [Using a3b3 = (ab) (a2 + b2 + ab)]

This is the factorised form of the given expression.

Example 2:

Factorise the expression: x3 + 64y3 + 12x2y + 48xy2 − (4x + y)3

Solution:

x3 + 64y3 + 12x2y + 48xy2 − (4x + y)3

= x3 + (4y)3 + 3 (x) (4y) (x + 4y) − (4x + y)3

= (x + 4y)3 − (4x + y)3 [Using (a + b)3 = a3 + b3 + 3ab (a + b)]

= (x + 4y − 4xy) [(x + 4y)2 + (4x + y)2 + (x + 4y) (4x + y)]

[Using a3b3 = (a b) (a2 + b2 + ab)]

= (− 3x + 3y) [x2 + 16y2 + 8xy + 16x2 + y2 + 8xy + (4x2 + 17xy + 4y2)

= 3 (yx) [21x2 + 21y2 + 33xy]

= 3 (yx) × 3 (7x2 + 7y2 + 11xy)

= 9 (yx) (7x2 + 7y2 + 11xy)

This is the factorised form of the given expression.

Example 3: 

Factorise the following expressions.

(1)

(2)

(3)

(4)  

 

Solution:

 

Factorisation of Quadratic Polynomials

We know that 7 × 6 = 42. Here, 7 and 6 are factors of 42. Now, consider the linear polynomials
x − 2 and x + 1. On multiplying the two, we get: x (x + 1) − 2 (x + 1) =
x2 + x − 2x − 2 = x2x − 2, which is a quadratic polynomial. So, x − 2 and x + 1 are factors of the quadratic polynomial
x2x − 2. A quadratic polynomial can have a maximum of two factors.

In the above example, we found the quadratic polynomial from its two factors. We can also find the factors from the quadratic polynomial. This process of decomposing a polynomial into a product of its factors (which when multiplied give the original expression) is called factorisation.

There are two ways of finding the factors of quadratic polynomials viz., by applying the factor theorem and by splitting the middle term. We will discuss these methods of factorisation in this lesson and also solve some examples based on them.

Factorisation of Quadratic Polynomials Using the Factor Theorem

The factor theorem states that: For a polynomial p(x) of a degree greater than or equal to 1 and for any real number a, if p(a) = 0, then xa will be a factor of p(x).

Consider the quadratic polynomial, p(x) = x2 − 5x + 6. To find its factors, we need to ascertain the value of x for which the value of the polynomial comes out to be zero. For this, we first determine the factors of the…

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