Factorisation
Factorisation of Algebraic Expressions by the Method of Common Factors
We know the identities
(i). a^{2} + 2ab + b^{2} = (a + b)^{2}
(ii). a^{2} − 2ab + b^{2} = (a − b)^{2}
(iii). a^{2} − b^{2} = (a + b) (a − b)
We can use these identities to factorise algebraic expressions as well. Let us discuss each identity one by one.
Application of the identity a^{2} + 2ab + b^{2} = (a + b)^{2} to factorise an algebraic expression
Let us factorise the expression x^{2} + 6x + 9.
In this expression, the first term is the square of x, the last term is the square of 3, and the middle term is positive and is twice the product of x and 3.
Thus, x^{2} + 6x + 9 can be written as
x^{2} + 6x + 9 = (x)^{2} + 2 × x × 3 + (3)^{2}
The right hand side of this expression is in the form of a^{2} + 2ab + b^{2},^{ }where a = x and b = 3.
We know the identity a^{2} + 2ab + b^{2} = (a + b)^{2}.
∴ (x)^{2} + 2 × x × 3 + (3)^{2} = (x + 3)^{2}
Thus, x^{2} + 6x + 9 = (x + 3)^{2}.
Application of the identity a^{2} − 2ab + b^{2} = (a − b)^{2} to factorise an algebraic expression
Let us factorise the expression 9y^{2} − 12y + 4.
In this expression, the first term is the square of 3y, the last term is the square of 2, and the middle term is negative and is twice of the product of 3y and 2.
Thus, 9y^{2} − 12y + 4 can be written as
9y^{2} − 12y + 4 = (3y)^{2} − 2 × 3y × 2 + (2)^{2}
The right hand side of this expression is in the form of a^{2} − 2ab + b^{2},^{ }where a = 3y and
b = 2.
We know the identity a^{2} − 2ab + b^{2} = (a − b)^{2}.
∴ (3y)^{2} − 2 × 3y × 2 + (2)^{2} = (3y − 2)^{2}
Thus, 9y^{2} − 12y + 4 = (3y − 2)^{2}.
Application of the identity a^{2} − b^{2} = (a + b) (a − b) to factorise an algebraic expression
We use this identity when an expression is given as the difference of two squares.
Let us factorise the expression x^{2} − 25.
We can write it as
x^{2} − 25 = (x)^{2} − (5)^{2}
The right hand side of this expression is in the form of a^{2} − b^{2}, where a = x and b = 5
On using the identity a^{2} − b^{2} = (a + b) (a − b), we obtain
x^{2} − 25 = (x + 5) (x − 5)
Thus, (x + 5) and (x − 5) are the factors of x^{2} − 25.
To factorise an algebraic expression, we have to observe the given expression. If it has a form that fits the left hand side of one of the identities mentioned in the beginning, then the expression corresponding to the right hand side of the identity gives the desired factorisation.
Let us discuss some more examples based on what we have discussed so far.
Example 1:
Factorise the given expressions.

25x^{2} + 40xy + 16y^{2}

81x^{3} + x − 18x^{2}

(p + 1)^{2} − (p − 1)^{2}

16a^{2} − 25b^{2} + 60bc − 36c^{2}

al^{2} − bm^{2} − am^{2} + bl^{2}

81x^{4} − 256 y^{4}

16x^{4} − (3a + 5c)^{4}
Solution:
(1)The given expression is 25x^{2} + 40xy + 16y^{2}.
25x^{2} + 40xy + 16y^{2} = (5x)^{2} + 2 × 5x × 4y + (4y)^{2}
= (5x + 4y)^{2} [a^{2} + 2ab + b^{2} = (a + b)^{2}]
∴ 25x^{2} + 40xy + 16y^{2} = (5x + 4y)^{2}
(2) The given expression is 81x^{3} + x − 18x^{2}.
Here, x is a factor common to all terms in the expression.
∴ 81x^{3} + x − 18x^{2} = x (81x^{2} + 1 − 18x)
= x [(9x)^{2} + (1)^{2} − 2 × 9x ×1]
= x [9x − 1]^{2} [a^{2} − 2ab + b^{2} = (a − b)^{2})
∴ 81x^{3} + x − 18x^{2} = x (9x − 1)^{2}
(3) The given expression is (p + 1)^{2} − (p − 1)^{2}.
On using the identity a^{2} − b^{2} = (a + b) (a − b), we obtain
(p + 1)^{2} − (p − 1)^{2} = {(p + 1) + (p − 1)} {(p + 1) − (p − 1)}
= (p + 1 + p − 1) (p + 1 − p + 1)
= (2p) (2) = 4p
∴ (p + 1)^{2} − (p − 1)^{2} = 4p
(4) The given expression is 16a^{2} − 25b^{2} + 60bc − 36c^{2}.
16a^{2} − 25b^{2} + 60bc − 36c^{2}
= 16a^{2} − (25b^{2} − 60bc + 36c^{2})
= (4a)^{2} − {(5b)^{2} − 2 (5b) (6c) + (6c)^{2}}
= (4a)^{2} − (5b − 6c)^{2} [Using the identity a^{2} − 2ab + b^{2} = (a − b)^{2}]
= {(4a) + (5b − 6c)}{(4a) − (5b − 6c)}[Using the identity a^{2} − b^{2} = (a + b) (a − b)]
= (4a + 5b − 6c) (4a − 5b + 6c)
(5) The given expression is al^{2} − bm^{2} − am^{2} + bl^{2}.
al^{2} − bm^{2} − am^{2} + bl^{2}
= al^{2} − am^{2} + bl^{2} − bm^{2} {Regrouping the terms}
= a (l^{2} − m^{2}) + b (l^{2} − m^{2})
= (l^{2} − m^{2}) (a + b)
= (l + m) (l − m) (a + b) [Using the identity a^{2} − b^{2} = (a + b) (a − b)]
(6) The given expression is 81x^{4} − 256y^{4}.
81x^{4} − 256y^{4}
(7) The given expression is 16x^{4} − (3a + 5c)^{4}.
16x^{4} − (3a + 5c)^{4}
How can we factorise the algebraic expression x^{2} + 8x + 15?
Note that we cannot express this expression as (a + b)^{2}, since 15 is not the square of any natural number. What do we do in such a case?
In this case, we can use the identity x^{2} + (a + b) x + ab = (x + a) (x + b).
If we compare x^{2} + 8x + 15 with x^{2} + (a + b) x + ab, then we obtain a + b = 8 and ab = 15.
Hence, we need to find two numbers, a and b, such that their sum is 8 and their product is 15.
The only num…
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