Select Board & Class

Factorisation

Factorisation of Algebraic Expressions by the Method of Common Factors

Suppose we need to find the product of the numbers 79 and 81. Instead of multiplying these two numbers, we can use the identity (a + b) (ab). This identity is very important and is applicable in various situations.

Let us first understand this identity.

(a + b) (a − b) = a (a − b) + b (a − b) (By distributive property)

= a2ab + bab2

= a2ab + abb2 (ab = ba)

= a2b2

 ∴ (a + b) (a − b) = a2 − b2

Deriving the identity geometrically:

This identity can be derived from geometric construction as well.

For this, let us consider a square AEGD whose each side measures a units.

It can be seen that, a segment BC is drawn outside the square AEGD such that BC is parallel to EG and at a distance of b unit from G.

Another segment HF is drawn inside the square AEGD such that HF is parallel to GD and at a distance of b unit from G.

From the figure, it can be observed that

Area of rectangle ABFH = Area of rectangle ABCD – (Area of rectangle HKGD + Area of square KFCG)

⇒ (a + b) (a − b) = a(a + b) − [(a × b) + (b × b)]

⇒ (a + b) (a − b) = a2 + ababb2

 ⇒ (a + b) (a − b) = a2 − b2

Now, let us solve some examples in which the above identity can be applied.

Example 1:

Simplify the following expressions.

(a) (x + 3) (x − 3)

(b) (11 + y) (11 − y)

Solution:

(a) (x + 3) (x − 3)

This expression is of the form (a + b) (a − b).

Hence, we can use the identity (a + b) (a − b) = a2b2 .

(x + 3) (x − 3) = x2 − 32 = x2 − 9

(b) (11 + y) (11 − y)

This expression is of the form (a + b) (a − b).

Hence, we can use the identity (a + b) (a − b) = a2b2 .

(11 + y) (11 − y) = 112y2 = 121y2

Example 2:

Simplify the following expressions.

(a)

(b)

Solution:

(a) The given expression is .

Using the identity (a + b) (a b) = a2b2, we get

(b)

Using the identity (a + b) (a b) = a2b2, we get

Example 3:

Find the values of the following expressions using suitable identities.

(a) 195 × 205

(b) (993)2 − (7)2

(c) 24.5 × 25.5

Solution:

(a) 195 = 200 − 5
and, 205 = 200 + 5

∴ 195 × 205 = (200 − 5) × (200 + 5)

= (200)2 − (5)2                       [$\because$ (a + b) (a − b) = a2b2]

= 40000 − 25

= 39975

(b) (993)2 − (7)2
= (993 + 7) (993 − 7)          [$\because$ (a + b) (a − b) = a2b2]

= (1000) (986)

= 986000

(c) 24.5 × 25.5

= (25 − 0.5) (25 + 0.5)

= (25)2 − (0.5)2                  [$\because$ (a + b) (a − b) = a2b2]

= 625 − 0.25

= 624.75

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 t…

To view the complete topic, please

What are you looking for?

Syllabus