In the same way, algebraic expressions also have factors that divide the expression exactly. Given an algebraic expression, the factors of that algebraic expression are two or more expressions whose product is the given expression.

**Factorisation using identities**

A factor of a number is the number that
exactly divides a given number without leaving any remainder. For example, the factors
of 4 are 1, 2 and 4; the factors of 12 are 1, 2, 3, 4, 6 and 12. And, the number
can be expressed as product of its factors such as 12 = 1 × 12 or 2 × 6 or 3 × 4.

In the same way, algebraic expressions
also have factors that divide the expression exactly. Given an algebraic expression,
the **factors of that algebraic expression**
are two or more expressions whose product is the given expression. For example,
*x*^{2} *y* = *x* × *x
*×*y* , hence the factors of the algebraic
expression *x*^{2} *y* are *x*, *x* and *y*.

Consider the algebraic expression *x*^{2 }*– *4 .

Now, *x* ^{2} −
4 =
*x*^{2} −2^{2} (*x* + 2)(*x*
− 2) [because *a* ^{2} − *b* ^{2} =(*a* + *b*)(*a* −*b*) ]

Therefore, (*x* +2) and (*x*
− 2) are the factors of *x*^{2} − 4 .

Obviously, 1 is also a factor of *x*^{2}
− 4.

We know that, *x*^{2} =
*x* × *x*. Similarly, (*x* + 5)^{2} =(*x*
+
5)×(*x*
+5)
.

Thus, the factors of (*x* +
5)^{2} are (*x* + 5) and (*x* + 5).

The process of writing an algebraic expression
as the product of its factors is called factorisation.

We shall learn more to factorise algebraic
expressions, using the basic identities.

** **

__Example 3.5____ __

Express the following algebraic expressions
as the product of its factors:

(i) *ab*^{2} �(ii)
−3*pq*^{3}
(iii)
12*m*^{2}*n*^{2} *p*

**Solution:**

(i) *ab*^{2}* *=* a *×*b*×*b
*.

(ii) − 3 *pq*^{3} =− 3 ×*p*× *q* × *q* ×*q* .

(iii) 12 *m*^{2}*n*^{2}*p*
= 4
× 3 **×** *m* *×**m* *×**n* *×**n* *×**p*_{ }= 2 ×
2× 2 × 3 *×**m* *×**m* *×**n*
*×**n* *×**p*.

** **

__Example 3.6__

Factorise by using the identity: *a ^{2
}*−

(i) a^{2} − 1 (ii) 9*k*^{2} −^{ }25 (iii) 64*x*^{2} − 81*y*^{2 }

**Solution:**

(i)* a*^{2}* *−* *1* *=*a*^{2}* *−1^{2 }= (*a* + 1)(*a*
− 1)^{ }[since 1^{2=}1×1=1]^{}

Therefore, the factors of a^{2 }−
1 are (a + 1) and (a − 1).

(ii) 9*k*^{2} − 25 = (3^{2} × *k*^{2} ) − 5^{2} = (3k)^{2 }− 5^{2} [since *a ^{m}
*×

= (3*k*+ 5)(3*k* − 5) [ since, *a*^{2}
− *b*^{2} = (*a* + *b*)(*a* − *b*)]

Therefore, the factors 9k^{2 }−
25 are (3*k*+5)and(3*k*-5).

(iii) 64*x*^{2} −
81*y*^{2} =(8 ^{2} ×*x* ^{2} )−
(9
^{2} × *y*^{2} )= (8*x*)^{2} −(9
*y*)^{2}

= (8*x* +9
*y*)(8*x* −9 *y*)

Therefore, the factors of 64*x*
^{2} − 81*y*^{2} are (8*x* + 9*y*) and (8*x*
– 9*y*).

** **

__Example 3.7__

** **Factorise** **9*x*^{2}** **+** **30*xy*** **+25*y*^{2}

**Solution:**

9x^{2}+30xy+25y^{2}=(3^{2}×*x*^{2})+(2×3×5)
×*xy*+(5^{2}×*y*^{2})

=(3*x*)^{2}+2×(3*x*) ×(5*y*) ×(5*y*)^{2}

This expression is in the form of identity
*a*^{2} + 2*ab* + *b*^{2} = (*a+b*)^{2}.

Hence, (3*x*)^{2} +2×
(3*x*)×(5*y*) +(5*y*)^{2} =
(3*x* +5*y*)^{2}

=(3*x*+5*y*)(3*x*+5*y*).

Therefore, the factors of 9*x*^{2}
+
30*xy* +25*y*^{2} are (3*x*+5*y*) and (3*x*+5*y*).

** **

__Example 3.8__

Factorise
4*x*^{2} − 4 *xy* +
*y*^{2}

**Solution:**

4*x*^{2} −
4 *xy* + *y*^{2} = 2 ^{2} ×*x* ^{2} −
(2×2)×
*xy* + *y*^{2}

= (2*x* )^{2} −2×
(2*x*)×*y* + *y*^{2}

This expression is in the form of identity
*a*^{2 }– 2*ab* + *b*^{2} = (*a–b*)^{2}.

Say 2*x= a* and *y =b*, thus we have,

(2*x* )^{2} −2×
(2*x*)×*y* + *y*^{2} =
(2*x* − *y*)^{2}

= (2*x* − *y*)(2*x* −
*y*)

Therefore,are
the factors of 4*x*^{2} − 4*xy* +
*y*^{2 }are (2*x* − *y*) and (2*x* − *y*).^{}

** **

__Think____ __

**Can we factorise the following expressions using any basic identities?
Justify your answer.**

**(i) x**^{2}** ****+**** ****5 x **

**Solution:**

**(i)** *x*^{2} + 5*x* + 4 = *x*^{2} + (1 + 4)*x*
+ (1 × 4)

Which is of the form *x*^{2}
+ (*a + b*)*x* + *ab*

= (*x *+ *a*) (*x*
+* b*)

*x*^{2} + (1+4)*x* +(1 × 4) = (*x *+ 1) (x +
4)

∴ *x*^{2} + 5*x* + 4 = (*x* + 1) (*x* + 4)

**(ii) x^{2} **

*x*^{2 }– 5*x* + 4 = *x*^{2} + ((–1) + (–4))*x* + (–1)(–4)

Which is of the form *x*^{2}
+ (*a + b*) *x *+ *ab*

= (*x + a*) (*x + b*)

*x*^{2} + ((–1) + (–4))*x* + ((–1)(–4)) = (*x* + (–1))
(*x* + (–4)) = (*x* –1) (*x* – 4)

*x*^{2} – 5*x* + 4 = (*x* – 1) (*x* – 4))

** **

__Example 3.9 __

Factorise: *x* ^{2} −
2 *xy* + *y*^{2} −*z*^{2}

**Solution:**

*x *^{2}* *−* *2* xy *+*
y*^{2}* *−*z*^{2 }= ( *x* ^{2} −2*xy* + *y*^{2} ) −*z*^{2} =( *x* −
*y* )^{2} −*z*^{2} [by identity-3]

Let, (*x* − *y*) *a *and
*z = b.*

Therefore, ( *x *−*
y *)^{2}* *−*z
*^{2}* *=* a *^{2}* *−*b*^{2}

We know that, *a*^{2}*
*− *b*^{2 }= ( *a* + *b*)(*a* − *b*) [Identity-4]

Hence,* x*^{2} − 2 *xy + y ^{2}
−z^{2}* = (

Therefore, the factors of *x*^{2 }– 2*xy* +y^{2 }− z^{2} are (*x* − *y* + *z*) and (*x*
− *y* − *z*).

** **

**Activity**

Here is an interesting
number game to thrill your friends. Follow the steps.

1. Ask your friend to think a number in mind. Let the number be in
between 1 to 10. (for example, let the number be 5).

2. Double the number and add to its square. (So, 10 + 25 = 35).

3. Add one to the result. (So, 35 + 1 = 36).

4. Ask the final resultant number.

5. The number will be a perfect square. Consider its base in the
exponent form and reduce 1 from it. The result is the number in your friend’s mind.
( 36 = 6^{2} ⇒ 6 is the base; so, 6 − 1 =5 ).

Remember the perfect square
numbers and their corresponding base numbers 1→1, 4 → 2^{2}, 9 → 3^{2}, 16 → 4^{2}, 25 → 5^{2}, 36 → 6^{2}, 49 → 7^{2}, 64 → 8^{2}, 81→9^{2}, 100 → 10^{2}

^{}

Tags : Algebra | Term 3 Chapter 3 | 7th Maths , 7th Maths : Term 3 Unit 3 : Algebra

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7th Maths : Term 3 Unit 3 : Algebra : Factorisation using identities | Algebra | Term 3 Chapter 3 | 7th Maths

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