7th Maths : Term 3 Unit 3 : Algebra : Introduction to Identities, Geometrical Approach to Multiplication of Monomials, Geometrical proof of Identities, Factorisation using identities : Exercise 3.1 : Text Book Back Exercises Questions with Answers, Solution

**Exercise
3.1**

**1.
Fill in the blanks**

**(i) ( p **

**(ii) The product of ( x + 5) and
(x − 5) is ----------------------. **

**(iii) The factors of x^{2}
**

**(iv) Express 24 ab^{2}c^{2}
as product of its factors is ---------. **

** **

**2. Say whether the following statements
are True or False.**

**(i) (7 x + 3)(7x − 4) **

**(ii) ( a **

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

**(iv) 2 p is the factor of 8pq.**

** **

**3. Express the following as the product
of its factors.**

**(i) 24 ab^{2}c^{2}.**

**(ii) 36 x^{3}y^{2}z.**

**(iii) 56 mn^{2}p^{2}.**

**Solution: **

**(i) **24*ab*^{2}*c*^{2} = 2 × 2 × 2 × 3 × *a* × *b*
× *b* × *c* × c

**(ii) **36 *x*^{3}*y*^{2}*z* = 2 × 2 × 3 × 3 × *x* × *x* × *x*
× *y* × *y* × *z*

**(iii)** 56 *mn*^{2}*p*^{2 }= 2 × 2 × 2 × 7 × *m* × *n*
×* n* × *p* × *p*

** **

**4.
Using the identity ( x +a)(x +b) **

**(i)
( x + 3)(x + 7)**

**(ii)
(6 a + 9)(6a − 5)**

**(iii)
(4 x + 3y)(4x + 5y)**

**(iv)
(8 + pq)(pq + 7)**

**Solution: **

**(i)** **( x + 3)(x + 7)**

Let *a* = 3;* b*
= 7, then

(*x* + 3)(*x* + 7) is of the form *x*^{2} + *x* (*a + b*) + *ab*

(*x *+ 3)(*x* + 7) = *x*^{2} + *x*(3 + 7)
+ (3 × 7) = *x*^{2}+ 10*x *+ 21

**(ii) (6a + 9)(6 a – 5)**

Substituting *x* = 6*a* ;
*a* = 9 and *b *= –5

In (*x + a*) (*x + b*) = *x*^{2} + *x* (*a + b*) + *ab*, we get

(6*a* + 9)(6*a* – 5) = (6*a*)^{2} + 6*a* (9 +
(–5)) + (9 × (–5))

= 6^{2} *a*^{2}
+ 6*a* (4) + (–45) = 36*a*^{2 }+ 24*a* – 45

(6*a* + 9) (6*a* –5) = 36*a*^{2} + 24*a* – 45

**(iii) (4 x + 3y)(4x + 5y)**

Substituting *x* = 4*x* ; *a*
= 3*y* and *b* = 5*y* in

(*x* + *a*) (*x
+ b*) = *x*^{2} + *x* (*a*
+ *b*) + *ab*, we get

(4*x* + 3*y*)(4*x*
– 5*y*) = (4*x*)^{2} + 4*x* (3*y* + 5*y*)
+ (3*y*) (5*y*)

= 4^{2} *x*^{2}
+ 4*x* (8*y*) + 15*y*^{2} = 16*x*^{2} + 32*xy* + 15*y*^{2}

(4*x* + 3*y*)(4*x*
+ 5*y*) = 16*x*^{2} + 32*xy* + 15*y*^{2}

**(iv) (8 + pq ) ( pq + 7)**

Substituting *x* = *pq* ; *a*
= 8 and *b* = 7 in

(*x* + *a*) (*x
+ b*) = *x*^{2} + *x* (*a
+ b*) + *ab*, we get

(*pq* + 8)(*pq* + 7) = (*pq*)^{2}+ *pq* (8 +
7) + (8) (7)

= *p*^{2} *q*^{2} + *pq* (15) + 56

(8 + *pq*)(*pq* + 7) = *p*^{2} +*q*^{2}
+ 15*pq* + 56

** **

**5.
Expand the following squares, using suitable identities.**

**(i)
(2 x **

**(ii)**** ****( b **

**(iii)
( mn **

**(iv)
( xyz **

**Solution: **

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

Comparing (2*x* + 5)^{2}
with (*a + b*)^{2} we have *a* = 2*x*
and *b* = 5

*a* = 2*x* and *b* = 5,

(*a + b*)^{2} = *a*^{2} + 2*ab* + *b*^{2}

(2*x* + 5)^{2} =
(2*x*)^{2} + 2(2*x*) (5) + 5^{2} = 2^{2} *x*^{2} + 20*x* + 25

= 2^{2} *x*^{2}
+ 20*x* + 25

(2*x* + 5)^{2} =
4*x*^{2} + 20*x* + 25

**(ii) ( b – 7)^{2}**

Comparing (*b* – 7)^{2}
with (*a* – *b*)^{2} we have *a*
= *b* and *b* = 7

(*a* – *b*)^{2} = *a*^{2} – 2*ab* + *b*^{2}

(*b* – 7)^{2} = *b*^{2} – 2(*b*)(7) + 7^{2}

(*b* – 7)^{2} = *b*^{2} – 14*b* + 49

**(iii) ( mn + 3p)^{2}**

Comparing (*mn *+ 3*p*)^{2} with (*a + b*)^{2} we have

(*a + b*)^{2} = *a*^{2} + 2*ab* + *b*^{2}

(*mn *+ 3*p*)^{2} = (*mn*)^{2} + 2(*mn*)
(3*p*) + (3*p*)^{2}

(*mn *+ 3*p*)^{2} = *m*^{2} *n*^{2}
+ 6*mnp* + 9*p*^{2}

**(iv) ( xyz – 1)^{2}**

Comparing (*xyz* – l)^{2}
with (*a – b*)^{2} we have = *a* + *xyz*
and *b* = 1

*a* = *xyz* and *b* = 1

(*a – b*)^{2} = *a*^{2} – 2*ab* + *b*^{2}

(*xyz* – 1)^{2}
= (*xyz*)^{2} – 2 (*xyz*) (1) + 1^{2}

(*xyz* – 1 )^{2}
= *x*^{2}*y*^{2}*z*^{2}
– 2 *xyz* + 1

** **

**6. Using the identity ( a + b)(a
− b) **

**(i) ( p + 2)(p − 2)**

**(ii) (1 + 3 b)(3b − 1)**

**(iii) (4 − mn)(mn + 4)**

**(iv) (6 x + 7y)(6x
– 7y)**

**Solution:**

**(i)** **( p + 2)(p – 2)**

Substituting *a* = *p* ; *b*
= 2 in the identity (*a + b*) (*a– b*) = *a*^{2} – *b*^{2},
we get

(*p* + 2)(*p* – 2) = *p*^{2} – 2^{2}

**(ii) (1 + 3 b)(3b
– 1)**

(1 + 3*b*) (3*b* –1) can be written as (3*b* + 1) (3*b* –1)

Substituting *a* = 3*b* and *b* = 1 in the identity

(*a + b*) (*a – b*) = *a*^{2} – b^{2}, we get

(3*b* + 1) (3*b* – 1) = (3*b*)^{2} – 1^{2} = 3^{2} × *b*^{2} – 1^{2}

(3*b* + 1) (3*b* – 1) = 9*b*^{2} – 1^{2}

**(iii) (4 – mn)(mn
+ 4)**

(4 – *mn*) (*mn *+ 4) can be written as (4 – *mn*) (4 + *mn*) = (4 + *mn*) (4 – *mn*)

Substituting *a* = 4 and
*b* = *mn* is

(*a + b*) (*a – b*) = *a*^{2} – *b*^{2},
we get

(4 + *mn*) (4 – *mn*) = 4^{2} – (*mn*)^{2} = 16 – *m*^{2} *n*^{2}

**(iv) (6 x + 7y)(6x – 7y)**

Substituting *a* = 6*x* and *b* = 7*y* in

(*a + b*) (*a – b*) = *a*^{2} – *b*^{2},
we get

(6*x* + 7*y*) (6*x*
– 7*y*). = (6*x*)^{2} – (7*y*)^{2
}= 6^{2}*x*^{2} –
7^{2}*y*^{2}

(6*x* + 7*y*) (6*x*
– 7*y*) = 36*x*^{2} – 49*y*^{2}

** **

**7.
Evaluate the following, using suitable identity.**

**(i)
51 ^{2}**

**(ii)
103 ^{2}**

**(iii)
998 ^{2}**

**(iv)
47 ^{2}**

**(v)
297 × 303**

**(vi)
990 × 1010**

**(vii)
51 × 52**

**Solution: **

**(i). 51 ^{2}
**

51^{2 }= (50+ l)^{2}

Taking *a* = 50 and *b* – 1 we get

(*a + b*)^{2 }= *a*^{2}
+ 2*ab* + *b*^{2}

(50+ l)^{2} = 50^{2}
+ 2 (50) (1) + 1^{2} = 2500 + 100 +1

51^{2} = 2601

**(ii) 103 ^{2}
**

103^{2} = (100+
3)^{2}

Taking *a *= 100 and *b* = 3

(*a + b*)^{2} = *a*^{2}
+ 2*ab* + *b*^{2} becomes

(100+ 3)^{2} = 100^{2}
+ 2 (100) (3) + 3^{2} = 10000 + 600 + 9

103^{2} = 10609

**(iii) 9982 **

998^{2} = (1000–2)^{2}

Taking *a *= 1000 and *b* = 2

(*a – b*)^{2} = *a*^{2}
+ 2*ab *+ *b*^{2 }becomes

(1000 – 2)^{2} = 1000^{2}
– 2 (1000) (2) + 2^{2}

= 1000000 – 4000 + 4

998^{2} = 10,04,004

**(iv) 47 ^{2}
**

47^{2} = (50 – 3)^{2}

Taking *a* = 50 and *b* = 3

(*a – b*)^{2} = *a*^{2}
– 2*ab* + *b*^{2} becomes

(50 – 3)^{2} = 50^{2}
– 2 (50) (3) + 3^{2}

= 2500 – 300 + 9 = 2200 +
9

47^{2} = 2209

**(v) 297 **×** 303 **

297 × 303 = (300–3) (300+
3)

Taking *a *= 300 and *b* = 3, then

(*a + b*) (*a – b*) = *a*^{2} – *b*^{2} becomes

(300+ 3) (300–3) = 300^{2}
– 3^{2}

303 × 297 = 90000 – 9

297 × 303 = 89,991

**(vi) 990 **×** 1010 **

990 × 1010 = (1000– 10)
(1000+ 10)

Taking *a* = 1000 and *b* = 10 , then

(*a – b*) (*a + b*) = *a*^{2} – *b*^{2} becomes

(1000 – 10) (1000 + 10) =
1000^{2 }– 10^{2}

990 × 1010 = 1000000 –
100

990 × 1010 = 999900

** (vii)** **51 **×** 52**

51 × 52 = (50+ 1) (50 + 2)

Taking *x* = 50 , *a* = 1 and *b* = 2

then (*x + a*) (*x + b*) = *x*^{2} + (*a + b*) *x* + *ab*
becomes

(50 +1) (50 + 2) = 50^{2} + (1 + 2) 50 + (l × 2)

= 2500 + (3) 50 + 2 = 2500 + 150 + 2

51 × 52 = 2652

** **

**8.
Simplify: ( a **

**Solution: ** (*a + b*)^{2}
– 4*ab* = *a*^{2} + *b*^{2 }+
2*ab* – 4*ab* = *a*^{2} + *b*^{2} – 2*ab* = (*a – b*)^{2}

** **

**9.
Show that ( m **

**Solution: **

Taking the LHS = (*m – n*)^{2}
+ (*m + n*)^{2}

= *m*^{2} – 2*mn* + *n*^{2}
+ *m*^{2} + 2*mn* + *n*^{2}
= *m*^{2} + *n*^{2} + *m*^{2} + *n*^{2}

= 2*m*^{2} + 2*n*^{2}

= 2(*m*^{2} + *n*^{2}) = RHS

∴ (*m – n*)^{2} +
(*m + n*)^{2 }= 2(*m*^{2} + *n*^{2})

** **

**10.
If a + b **

**Solution: **

We have (*a + b)*^{2}
= *a*^{2} + 2*ab* + *b*^{2}

(*a + b)*^{2} = *a*^{2} + *b*^{2} + 2*ab*

given *a + b* = 0 and *ab *= 18

10^{2} = *a*^{2}
+ *b*^{2} + 2(18)

100 = *a*^{2} + *b*^{2} + 36

100 – 36 = *a*^{2}
+ *b*^{2}

*a*^{2} + *b*^{2} = 64

** **

**11.
Factorise the following algebraic expressions by using the identity a ^{2}
**

**(i)
z^{2} **

**(ii)
9 ****− ****4 y^{2}**

**(iii)
25 a^{2} **

**(iv) x **

**Solution:**

**(i)** *z*^{2 }– 16

*z*^{2 }– 16 = *z*^{2} – 4^{2}

We have *a*^{2} –
*b*^{2} = (*a + b*) (*a – b*)

let *a* = *z * and *b* =
4,

*z*^{2} – 4^{2} = (*z* + 4)(*z* – 4)

**(ii) 9 – 4 y^{2}**

9 – 4*y*^{2 }= 3^{2}
– 2^{2} *y*^{2} = 3^{2}
– (2*y*)^{2}

let *a* = 3 and *b* = 2*y*,
then

*a*^{2 }– *b*^{2} = (*a + b*) (*a – b*)

∴ 3^{2} – (2*y*)^{2}
= (3 + 2*y*)(3 – 2*y*)

9 – 4*y*^{2} =
(3 + 2*y*) (3 – 2*y*)

**(iii) 25 a^{2 }– 49b^{2}**

25*a*^{2}– 49*b*^{2 }= 5^{2} *a*^{2}
– 7^{2} *b*^{2} = (5*a*)^{2} – (7*b*)^{2}

let A = 5*a* and B = 7*b*

A^{2} – B^{2 }= (A + B) (A – B) becomes

(5a)^{2} – (7*b*)^{2}
= (5*a* + 7*b*)(5*a* –7*b*)

**(iv) x^{4} – y^{4}**

Let *x*^{4} – *y*^{4} = (*x*^{2})^{2} – (*y*^{2})^{2}

We have *a*^{2} –
*b*^{2} = (*a + b*) (*a – b*)

(*x*^{2})^{2 }– (*y*^{2})^{2}
= (*x* ^{2}+ *y*^{2})(*x*^{2} – *y*^{2})

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

Again we have *x*^{2}
– *y*^{2} = (*x + y*) (*x – y*)

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

** **

**12.
Factorise the following using suitable identity.**

**(i) x**

**(ii) y**

**(iii)
36 m^{2} **

**(iv)
64 x^{2} **

**(v) a**

**Solution:**

**(i)** **(i) x^{2 }– 8x +16**

*x*^{2 }– 8*x *+16** **= *x*^{2} – (2 × 4 × *x*) + 4^{2}

This expression is in the form of identity

*a*^{2} – 2*ab* + *b*^{2}** **= (*a
– b*)^{2}

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

*x*^{2} – 8*x* + 16 = (*x *– 4) (*x *– 4)

**(ii) y^{2} + 20y +100**

*y*^{2} + 20*y *+ 100** **= *y*^{2} + (2 × (10))*y* + (10 × 10)

= *y*^{2} + (2 ×
10 ×* y*) + 10^{2}

This is of the form of identity

*a*^{2} + 2*ab* + *b*^{2}** **= (*a
+ b*)^{2}

*y*^{2} + (2 × 10 × *y*) + 10^{2} = (*y *+
10)^{2}

*y*^{2} + 20*y* +100 = (*y* +10)^{2}

*y*^{2} + 20y + 100 = (*y* + 10) (*y* + 10)

**(iii) 36 m^{2} + 60m + 25**

36*m*^{2} + 60*m* + 25** **= 6^{2} *m*^{2}
+ 2 × 6*m* × 5 + 5^{2}

This expression is of the form of identity

*a*^{2 }+ 2*ab* + *b*^{2} = (*a + b*)^{2}

(6m)^{2} + (2 × 6*m*
× 5) + 5^{2 }= (6*m* + 5)^{2}

36* m*^{2} + 60*m* + 25 = (6*m* + 5) (6*m* + 5)

**(iv) 64 x^{2} – 112xy + 49y^{2}**

64*x*^{2} – 112*xy* + 49*y*^{2 }= 8^{2}
*x*^{2} – (2 × 8*x* × 7*y*)
+ 7^{2}*y*^{2}

This expression is of the form of identity

*a*^{2} – 2*ab* + *b*^{2} = (*a – b*)^{2}

(8*x*)^{2} – (2
× 8*x* × 7*y*) + (7*y*)^{2} =
(8*x* – 7*y*)^{2}

64*x*^{2} – 112*xy* + 49*y*^{2} = (8*x* – 7*y*) (8*x*
– 7*y*)

**(v) a^{2} + 6ab + 9b^{2} – c^{2}**

*a*^{2} + 6*ab* + 9*b*^{2} – *c*^{2} = *a*^{2} + 2 × *a* × 3*b* + 3^{2} *b*^{2} – *c*^{2}

= *a*^{2} + (2 ×
*a* × 3*b*) + (3*b*)^{2} – *c*^{2}

This expression is of the form of identity

[*a*^{2} + 2*ab* + *b*^{2}]
– *c*^{2} = (*a + b*)^{2} – *c*^{2}

*a*^{2} + (2 × *a* × 3*b*) + (3*b*)^{2} – *c*^{2} = (*a* + 3*b*)^{2} – *c*^{2}

Again this RHS is of the form of identity

*a*^{2} – *b*^{2} = (*a + b*) (*a – b*)

(*a* + 3*b*)^{2} – *c*^{2} = [(*a* + 36)
+ *c*] [(*a* + 3*b*) – *c*]

*a*^{2} + 6*ab* + 9*b*^{2} – *c*^{2} = (*a* +3*b* + *c*) (*a*
+ 3*b* – *c*)

** **

__Objective Type Questions__

**13.
If a + b = 5 and a^{2} **

(i) 12

(ii) 6

(iii) 5

(iv) 13

**[Answer: (ii) 6]**

**Solution: **(*a + b*)^{2} =
25

13 + 2*ab* = 25

2*ab* =12

* ab* = 6

**14.
(****5**** +****20****)(−****20****−****5****)=****?**

(i) −425

(ii) 375

(iii) −625

(iv) 0

**[Answer: (iii) –625]**

**Solution: **(50 + 20) (–20 – 5) = – (5 + 20)^{2} = – (25)^{2}
= – 625

**15.
The factors of x^{2} **

(i) (*x* − 3)(*x* − 3)

(ii) (*x* − 3)(*x* + 3)

(iii) (*x* + 3)(*x* + 3)

(iv) (*x* − 6)(*x* + 9)

**[Answer: (i) ( x – 3)(x –3)]**

**Solution: ***x*^{2} – 6*x* + 9 = *x*^{2} –
2(*x*) (3) + 3^{2}

*a*^{2 }– 2*ab* + *b*^{2} = (*a *– *b*)^{2}
= (*x* – 3)^{2} = (*x* – 3) (*x* – 3)

**16.
The common factors of the algebraic expressions ax^{2} y ,
bxy^{2} and cxyz is**

(i) *x*^{2} *y*

(ii) *xy*^{2}

(iii) *xyz*

(iv) *xy*

**[Answer: (iv) xy]**

**Solution: ***ax*^{2}*y* = *a* × * x* ×

*bxy*^{2} *=
b *× * x *×

*cxyz* = *c *×* x *×

Common factor = *xy*

**Exercise 3.1**

1. (i) p^{2}–2pq+q^{2}
(ii) *x*^{2}–25 (iii) (*x*–2) and (*x*–2) (iv) 2 × 2 × 2 × 3 × a × b × b × c × c

2. (i) True (ii) False
(iii) True (iv) True.

3. (i) 2 × 2 × 2 × 3 ×
*a × b × b × c × c*

(ii) 2 × 2 × 3 × 3 × *x
× x × x × y* × *y* × *z*

(iii) 2 × 2 × 2 × 7 × *m*
× *n × n × p × p*

4. (i) *x*^{2} +10*x* + 21 (ii) 36*a*^{2}
+ 24*a* – 45 (iii) 16*x*^{2} + 32*xy*+15*y*^{2} (iv) *p*^{2}*q*^{2} + 15*pq* + 56

5. (i) 4*x*^{2} + 20*x* + 25 (ii) *b*^{2}
– 14*b* + 49 (iii) *m*^{2}*n*^{2}
+ 6 *mnp* + 9*p*^{2} (iv) *x*^{2}*y*^{2}*y*^{2} – 2 *xyz* + 1

6. (i) *p*^{2}– 4 (ii) 9*b*^{2}–1 (iii) 16 – *m*^{2}*n*^{2} (iv) 36*x*^{2}–
49*y*^{2}

7. (i) 2, 601 (ii) 10,
609 (iii) 9,96,004 (iv) 2, 209 (v) 89,991 (vi) 9,99,900 (vii) 2,652

8. (a–b)^{2}

10. 64

11. (i) (z + 4) (z –
4) (ii) (3 + 2y) (3 – 2y) (iii) (5a + 7b) (5a – 7b) (iv) (*x*^{2} + *y*^{2})
(*x + y*) (*x – y*)

12. (i) (x–4) (x–4)
(ii) (y + 10) (y + 10) (iii) (6m + 5) (6m + 5) (iv) (8x – 7y) (8x – 7y) (v) (a
+ 3b + c) (a + 3b – c)

** Objective type questions**

13. (ii) 6

14. (iii) – 625

15. (i) (x – 3) (x–3)

16. (iv) xy

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