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# NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities | Class 8th Maths

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# NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities class 8th maths

## NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities class 8th maths Ex 9.1

Exercise 9.1

1. Identify the terms, their coefficients for each of the following expressions.

 (i) 5xyz2 − 3zy (ii) 1 + x + x2 (iii) 4x2y2 − 4x2y2z2 + z2 (iv) 3 − pq + qr − rp (v) $$\displaystyle \frac{x}{2}+\frac{y}{2}-xy$$ (vi) 0.3a − 0.6ab + 0.5b

Solutions :

 Terms Coefficients (i) 5xyz2 − 3zy There are two terms in this expressions. 5xyz2 and 3zy 5 and − 3 (ii) 1 + x + x2 There are two terms in this expressions. 1, x and x2 1, 1 and 1 (iii) 4x2y2 − 4x2y2z2 + z2 There are three terms in this expressions. 4x2y2 , − 4x2y2z2 and z2 4, − 4 and 1 (vi) 3 − pq + qr – rp There are four terms in this expressions. 3, − pq, qr and − rp 3, − 1, 1 and − 1 (v) $$\displaystyle \frac{x}{2}+\frac{y}{2}-xy$$ There are three terms in this expressions. $$\displaystyle \frac{x}{2}$$ , $$\displaystyle \frac{y}{2}$$ and − xy $$\displaystyle \frac{1}{2}$$ , $$\displaystyle \frac{1}{2}$$ and − 1 (vi) 0.3a − 0.6ab + 0.5b There are three terms in this expressions. 0.3, − 0.6ab and 0.5b 0.3, − 0.6 and 0.5

2. Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?
x + y, 1000, x + x2 + x3, 7 + y + 5x, 2y − 3y2, 2y − 3y2 + 4y3, 5x − 4y + 3xy, 4z − 15z2, ab + bc + cd + da, pqr, p2q, 2p + 2q
Solutions :
Monomials :- 1000, pqr
Binomials :- x + y, 2y − 3y2, 4z − 15z2, p2q + pq2, 2p + 2q
Trinomials :- 7 + y + 5x, 2y − 3y2 + 4y3, 5x − 4y + 3xy
These polynomials do not fit in any of these three categories :- x + x2 + x3 + x4, ab + bc + cd + da

 (i) ab − bc, bc − ca, ca − ab (ii) a − b + ab, b − c + bc, c − a + ac (iii) 2p2q2 − 3pq + 4, 5 + 7pq − 3p2q2 (iv) l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + 2nl

Solutions :
(i) ab − bc, bc − ca, ca − ab
on writing similar terms together –
= ab − bc + bc − ca + ca − ab

= ab − ab + bc − bc + ca − ca
= 0

(ii) a − b + ab, b − c + bc, c − a + ac
= a − b + ab + b − c + bc + c − a + ac
on writing similar terms together –
= a − a + b − b + c − c + ab + bc + ac

by adding and substracting same terms –
= ab + bc + ac

(iii) 2p2q2 − 3pq + 4, 5 + 7pq − 3p2q2
= 2p2q2 − 3pq + 4 + 5 + 7pq − 3p2q2

on writing similar terms together –
= 2p2q2 − 3p2q2 + 7pq − 3pq + 4 + 5
= − p2q2 + 4pq + 9

(iv) l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + 2nl
= l2 + m2 + m2 + n2 + n2 + l2 + 2lm + 2mn + 2nl
on writing similar terms together –
= l2 + l2 + m2 + m2 + n2 + n2 + 2lm + 2mn + 2nl
= 2l2 + 2m2 + 2n2 + 2lm + 2mn + 2nl
= 2(l2 + m2 + n2 + lm + mn + nl)

4. (a) Subtract 4a − 7ab + 3b + 12 from 12a − 9ab + 5b − 3
(b) Subtract 3xy + 5yz − 7zx from 5xy − 2yz − 2zx + 10xyz
(c) Subtract 4p2q − 3pq + 5pq2 − 8p + 7q − 10 from 18 − 3p − 11p + 5pq − 2pq2 + 5p2q
Solutions :
(a) Subtract 4a − 7ab + 3b + 12 from 12a − 9ab + 5b − 3
= 12a − 9ab + 5b − 3 − (4a − 7ab + 3b + 12)
remove the bracket –
= 12a − 9ab + 5b − 3 − 4a + 7ab − 3b − 12  (If there is a negative sign outside the bracket, the symbols of all the sign inside will change.)
on writing similar terms together –
= 12a − 4a + 5b − 3b + 7ab − 9ab − 4 − 12

= 8a + 2b − 2ab − 16

(b) Subtract 3xy + 5yz − 7zx from 5xy − 2yz − 2zx + 10xyz
= 5xy − 2yz − 2zx + 10xyz − (3xy + 5yz − 7zx)

remove the bracket –
= 5xy − 2yz − 2zx + 10xyz − 3xy − 5yz + 7zx (If there is a negative sign outside the bracket, the symbols of all the sign inside will change.)

on writing similar terms together –
= 5xy − 3xy − 2yz − 5yz + 7zx − 2zx + 10xyz

= 2xy − 7yz + 5zx + 10xyz

(c) Subtract 4p2q − 3pq + 5pq2 − 8p + 7q − 10 from 18 − 3p − 11p + 5pq − 2pq2 + 5p2q
= 18 − 3p − 11p + 5pq − 2pq2 + 5p2q − (4p2q − 3pq + 5pq2 − 8p + 7q − 10)
remove the bracket –
= 18 − 3p − 11q + 5pq − 2pq2 + 5p2q − 4p2q + 3pq − 5pq2 + 8p − 7q + 10)  (If there is a negative sign outside the bracket, the symbols of all the sign inside will change.)

on writing similar terms together –
= − 3p + 8p − 11q − 7q + 5pq + 3pq − 2pq2 − 5pq2 + 5p2q − 4p2q

= 5p − 18q + 8pq − 7pq2 + p2q

## NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities class 8th maths Ex 9.2

Exercise 9.2

1. Find the product of the following pairs of monomials.

 (i) 4, 7p (ii) − 4p, 7p (iii) − 4p, 7pq (iv) 4p3, − 3p (v) 4p, 0

Solutions :
(i) 4, 7p
= 4 × 7p
= 28p

(ii)  − 4p, 7p
= − 4p × 7p
= − 28p2

(iii) − 4p, 7pq
= − 4p × 7pq
= − 28p2q

(iv) 4p3, − 3p
= 4p3 × − 3p
= − 12p4

(v) 4p, 0
= 4p × 0
= 0

2. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.
(p, q); (10m, 5n); (20x2, 5y2); (4x, 3x2); (3mn, 4np)
Solutions :
(p, q)
length = p
area of rectangle = length × breadth
= p × q
= pq unit square

(10m, 5n)
length = 10m
area of rectangle = length × breadth
= 10m × 5n
= 50mn unit square

(20x2, 5y2)
length = 20x2
area of rectangle = length × breadth
= 20x2 × 5y2
= 100x2y2 unit square

(4x, 3x2)
length = 4x
area of rectangle = length × breadth
= 4x × 3x2
= 12x3 unit square

(3mn, 4np)
length = 3mn
area of rectangle = length × breadth
= 3mn × 4np
= 12mn2p unit square

3. Complete the table of products.

 First monomial → Second monomial ↓ 2x − 5y 3x2 − 4xy 7x2y − 9x2y2 2x 4x2 − 5y − 15x2y 3x2 − 4xy 7x2y − 9x2y2

Solutions :

 First monomial → Second monomial ↓ 2x − 5y 3x2 − 4xy 7x2y − 9x2y2 2x 4x2 − 10xy 6×3 − 8x2y 14x3y − 18x3y2 − 5y − 10xy 25y2 − 15x2y 20xy2 − 35x2y2 45x2y3 3x2 6x3 − 15x2y 9x4 − 12x3y 21x4y − 27x4y2 − 4xy − 8x2y 20xy2 − 12x3y 16x2y2 − 28x3y2 36x3y3 7x2y 14x3y − 35x2y2 21x4y − 28x3y2 49x4y2 − 63x4y3 − 9x2y2 − 18x3y2 45x2y2 − 27x4y2 36x3y3 − 63x4y3 81x4y4

4. Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

 (i) 5a, 3a2, 7a4 (ii) 2p, 4q, 8r (iii) xy, 2x2y, 2xy2 (iv) a, 2b, 3c

Solutions :
(i) 5a, 3a2, 7a4
volume of rectangular box = length × breadth × height
= 5a × 3a2 × 7a4
= 5 × 3 × 7 × a × a2 × a4
= 105a7

(ii) 2p, 4q, 8r
volume of rectangular box = length × breadth × height
= 2p × 4q × 8r
= 2 × 4 × 8 × p × q × r
= 64pqr

(iii) xy, 2x2y, 2xy2
volume of rectangular box = length × breadth × height
= xy × 2x2y × 2xy2
= 1 × 2 × 2 × xy × x2y × xy2
= 4x4y4

(iv) a, 2b, 3c
volume of rectangular box = length × breadth × height
= a × 2b × 3c
= 1 × 2 × 3 × a × b × c
= 6abc

5. Obtain the product of

 (i) xy, yz, zx (ii) a, − a2, a3 (iii) 2, 4y, 8y2, 16y3 (iv) a, 2b, 3c, 6abc (v) m, − mn, mnp

Solutions :
(i) xy, yz, zx
= xy × yz × zx
= x2y2z2

(ii) a, − a2, a
= a × (− a2) × a3
= − a6

(iii) 2, 4y, 8y2, 16y3
= 2 × 4y × 8y2 × 16y3
= 2 × 4 × 8 × 16 × y × y2 × y3
= 1024y3

(iv) a, 2b, 3c, 6abc
= 2 × 3 × a × b × c × abc
= 6a2b2c2

(v) m, − mn, mnp
= m × (− mn) × mnp
= − m2n2p

### NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities class 8th maths Ex 9.3

Exercise 9.3

1. Carry out the multiplication of the expressions in each of the following pairs

 (i) 4p, q + r (ii) ab, a − b (iii) a + b, 7a2b2 (iv) a2 − 9, 4a (v) pq + qr + rp, 0

Solutions :
(i) 4p, q + r
= 4p × (q + r)
= 4p × q + 4p × r
= 4pq + 4pr

2. Complete the table.

 First expression Second expression Product (i) a b + c + d ____ (ii) x + y − 5 5xy ____ (iii) p 6p2 − 7p + 5 ____ (iv) 4p2q2 p2 − q2 ____ (v) a + b + c abc ____

Solutions :

 First expression Second expression Product (i) a b + c + d ab + ac + ad (ii) x + y − 5 5xy 5x2y + 5xy2 − 25xy (iii) p 6p2 − 7p + 5 6p3 − 7p2 + 5p (iv) 4p2q2 p2 − q2 4p4q2 − 4p2q4 (v) a + b + c abc a2bc + ab2c + abc2

(i) a × (b + c + d)
= ab + ac + ad

(ii) (x + y − 5) × 5xy
= 5x2y + 5xy2 − 25xy

(iii) p × (6p2 − 7p + 5)
= 6p3 − 7p2 + 5p

(iv) 4p2q2 × (p2 − q2)
= 4p4q2 − 4p2q4

(v) (a + b + c) × abc
= a2bc + ab2c + abc2

3. Find the product.

 (i) (a2) × (2a22) × (4a26) (ii) $$\displaystyle \left( {\frac{2}{3}xy} \right)\times \left( {\frac{{-9}}{{10}}{{x}^{2}}{{y}^{2}}} \right)$$ (iii) $$\displaystyle \left( {-\frac{{10}}{3}p{{q}^{3}}} \right)\times \left( {\frac{6}{5}{{p}^{3}}q} \right)$$ (iv) x × x2 × x3 × x4

Solutions :
(i) (a2) × (2a22) × (4a26)
= 2 × 4 × a2 × a22 × a26
= 8a50

(ii) $$\displaystyle \left( {\frac{2}{3}xy} \right)\times \left( {\frac{{-9}}{{10}}{{x}^{2}}{{y}^{2}}} \right)$$

$$\displaystyle \begin{array}{l}=\frac{2}{3}\times \left( {\frac{{-9}}{{10}}} \right)\left( {xy\times {{x}^{2}}{{y}^{2}}} \right)\\=\frac{{-18}}{{30}}{{x}^{3}}{{y}^{3}}\\=\frac{{-3}}{5}{{x}^{3}}{{y}^{3}}\end{array}$$

(iii) $$\displaystyle \left( {-\frac{{10}}{3}p{{q}^{3}}} \right)\times \left( {\frac{6}{5}{{p}^{3}}q} \right)$$
$$\displaystyle \begin{array}{l}=\left( {-\frac{{10}}{3}} \right)\times \frac{6}{5}\times p{{q}^{3}}\times {{p}^{3}}q\\=-\frac{{60}}{{15}}{{p}^{4}}{{q}^{4}}\\=-4{{p}^{4}}{{q}^{4}}\end{array}$$

(iv) x × x2 × x3 × x4
= x10

4. (a) Simplify 3x (4x – 5) + 3 and find its values for (i) x = 3 (ii) $$\displaystyle x=\frac{1}{2}$$
(b) Simplify a (a2 + a + 1) + 5 and find its value for (i) a = 0, (ii) a = 1
(iii) a = – 1.

Solutions :
(a) (i) 3x (4x − 5) + 3
= 3x × 4x − 3x × 5 + 3
= 12x2 − 15x + 3 ………………………….. (i)
putting x = 3 equ. in (i)
= 12 × (3)2 − 9 × 5 + 3
= 12 × 9 − 45 + 3
= 108 + 3 − 45
= 111 − 45
= 66

(ii) Putting $$\displaystyle x=\frac{1}{2}$$ in equ (i) –
$$\displaystyle \begin{array}{l}=12\times {{\left( {\frac{1}{2}} \right)}^{2}}-15\times \frac{1}{2}+3\\=12\times \frac{1}{4}-\frac{{15}}{2}+3\\=3-\frac{{15}}{2}+3\\=\frac{{6-15+6}}{2}\\=\frac{{12-15}}{2}\\=\frac{{-3}}{2}\end{array}$$

5. (a) Add: p (pq), q (qr) and r (rp
(b) Add: 2x (zxy) and 2y (zyx
(c) Subtract: 3l (l − 4m + 5n) from 4l (10n − 3m + 2l
(d) Subtract: 3a (a + b + c ) − 2b (ab + c) from 4c (− a + b + c
Solutions :
(a) Add: p (pq), q (qr) and r (rp
= p (p − q) + q (qr) + r (rp)
= p2pq + q2qr + r2 − rp
= p2 + q2 + r2pqqrrp

(b) Add: 2x (zxy) and 2y (zyx
= 2x (zxy) + 2y (zyx)
= 2xz − 2x2 − 2xy + 2yz − 2y2 − 2xy
= − 2x2 − 2y2 − 2xy − 2xy + 2yz + 2xz
= − 2x2 − 2y2 − 4xy + 2yz + 2xz

(c) Subtract: 3l (l − 4m + 5n) from 4l (10n − 3m + 2l
= 4l (10n − 3m + 2l) − 3l (l − 4m + 5n)
= 4l × 10n − 4l × 3m + 4l + 2l − 3l × l − 3l × (− 4m) − 3l × 5n
= 40ln − 12lm + 8l2 − 3l2 + 12lm − 15ln
= 40ln − 15ln − 12lm + 12lm + 8l2 − 3l2
= 25ln − 0lm + 5l2
= 5l2 + 25ln

(d) Subtract: 3a (a + b + c ) − 2b (ab + c) from 4c (− a + b + c
= 4c (− a + b + c) − [3a (a + b + c) − 2b (ab + c)]
= − 4ac + 4bc + 4c2 − [3a2 + 3ab + 3ac − 2ab + 2a2 – 2bc]
= − 4ac + 4bc + 4c2 − 3a2 − 3ab − 3ac + 2ab − 2b2 + 2bc
= − 4ac − 3ac + 4bc + 2bc 3ab + 2ab + 4c2 − 3a2 − 2b2
= − 7ac + 6bcab + 4c2 − 3a2 − 2b2
=  − 3a2 − 2b2 + 4c2ab + 6bc − 7ac

## NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities class 8th maths Ex 9.4

Exercise 9.4

1. Multiply the binomials.

 (i) (2x + 5) and (4x − 3) (ii) (y − 8) and (3y − 4) (iii) (2.5l − 0.5m) and (2.5l + 0.5m) (iv) (a + 3b) and (x + 5) (v) (2pq + 3q2) and (3pq − 2q2) (vi) $$\displaystyle \left( {\frac{3}{4}{{a}^{2}}+3{{b}^{2}}} \right)$$ and $$\displaystyle \left( {{{a}^{2}}-\frac{2}{3}{{b}^{2}}} \right)$$

Solutions :
(i) (2x + 5) and (4x − 3)
= 2x (4x − 3) + 5 (4x − 3)
= 2x × 4x − 2x × 3 + 5 × 4x − 5 × 3
= 8x2 − 6x + 20x − 15
= 8x2 + 14x − 15

(ii) (y – 8) and (3y − 4)
= y (3y − 4) − 8 (3y − 4)
= y × 3y − y × 4 − 8 × 3y − 8 × (− 4)
= 3y2 − 4y − 24y + 32
= 3y2 − 28y + 32

(iii) (2.5l − 0.5m) and (2.5l + 0.5m)
= 2.5l (2.5l + 0.5m) − 0.5m (2.5l + 0.5m)
= 2.5l × 2.5l + 2.5l × 0.5m – 0.5m × 2.5l − 0.5m × 0.5m
= 2.25l2 + 1.25lm − 1.25lm − 0.25m2
= 2.25l2 − 0.25m2

(iv) (a + 3b) and (x + 5)
= a (x + 5) + 3b (x + 5)
= a × x + a × 5 + 3b × x + 3b × 5
= ax + 5a + 3bx + 15b
= ax + 5a + 3bx + 15b

(v) (2pq + 3q2) and (3pq – 2q2)
= 2pq (3pq − 2q2) + 3q2 (3pq − 2q2)
= 2pq × 3pq − 2q2 × 2pq + 3q2 × 3pq − 3q2 × 2q2
= 2p2q2 − 4pq3 + 9pq3 − 6q4
= 6p2q2 + 5pq3 − 6q4

2. Find the product.

 (i) (5 − 2x) (3 + x) (ii) (x + 7y) (7x − y) (iii) (a2 + b) (a + b2) (iv) (p2 – q2) (2p + q)

Solutions :
(i) (5 − 2x) (3 + x)
= 5 × (3 + x) − 2x × (3 + x)
= 15 + 5x − 6x − 2x2
= 15 − x − 2x2

(ii) (x + 7y) (7xy)
= x × (7xy) + 7y × (7xy)
= 7x2xy + 49xy − 7y2
= 7x2 + 48xy − 7y2

(iii) (a2 + b) (a + b2)
= a2 × (a + b2) + b × (a + b2)
= a3 + a2b2 + ab + b2

(iv) (p2 − q2) (2p + q)
= p2 × (2p + q) − q2 × (2p + q)
= 2p3 + p2q − 2pq2 − q3

3. Simplify.

 (i) (x2 − 5) (x + 5) + 25 (ii) (a2 + 5) (b3 + 3) + 5 (iii) (t + s2) (t2 − s) (iv) (a + b) (c − d) + (a − b) (c + d) + 2 (ac + bd) (v) (x + y) (2x + y) + (x + 2y) (x − y) (vi) (x + y) (x2 − xy + y2) (vii) (1.5x − 4y) (1.5x + 4y + 3) − 4.5x + 12y (viii) (a + b + c) (a + b − c)

Solutions :
(i) (x2 – 5) (x + 5) + 25
= x2 × (x + 5) − 5 × (x + 5) + 25
= x3 + 5x2 − 5x − 25 + 25
= x3 + 5x2 − 5x

(ii) (a2 + 5) (b3 + 3) + 5
= a2 × (b3 + 3) + 5 × (b3 + 3) + 5
= a2b3 + 3a2 + 5b3 + 15 + 5
= a2b3 + 3a2 + 5b3 + 20

(iii) (t + s2) (t2 − s)
= t × (t2 − s) + s2 × (t2 − s)
= t3 − st + s2t2 − s3

(iv) (a + b) (c − d) + (a − b) (c + d) + 2 (ac + bd)
= a × (c − d) + b × (c − d) + a × (c + d) − b × (c + d) + 2ac + 2bd
= ac − ad + bc − bd + ac + ad − bc − bd + 2ac + 2bd
= ac + ac + 2ac + bc − bc − bd − bd + 2bd − ad + ad
= 2ac + 2ac − 2bd + 2bd
= 4ac

(v) (x + y) (2x + y) + (x + 2y) (x – y)
= x × (2x + y) + y × (2x + y) + x × (x − y) + 2y × (x − y)
= 2x2 + xy + 2xy + y2 + x2 − xy + 2xy − 2y2
= 2x2 + x2 + xy + 2xy − xy + 2xy + y2 − 2y2
= 3x2 + 5xy − xyy2
= 3x2 + 4xyy2

(vii) (1.5x − 4y) (1.5x + 4y + 3) − 4.5x + 12y
= 1.5x × (1.5x + 4y + 3) − 4y × (1.5x + 4y + 3) − 4.5x + 12y
= 2.25x2 + 6.0xy + 4.5x – 6.0xy − 16y2 − 12y − 4.5x − 12y
= 2.25x2 + 6.0xy − 6.0xy + 4.5x − 4.5x − 16y2 − 12y − 12y
= 2.25x2 − 16y2

(viii) (a + b + c) (a + b − c)
= a × (a + b − c) + b × (a + b – c)  + c × (a + b − c)
= a2 + ab − ac + ab + b2 − bc + ac + bc − c2
= a2 + ab + ab − ac + ac + b2 − bc + bc − c2
= a2 + 2ab + b2 − c2
= a2 + b2 − c2 + 2ab

## NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities class 8th maths Ex 9.5

Exercise 9.5

1. Use a suitable identity to get each of the following products.

 (i) (x + 3) (x + 3) (ii) (2y + 5 ) (2y + 5) (iii) (2a – 7) (2a – 7) (iv) (v) (1.1m – 0.4) (1.1m + 0.4) (vi) (a2 + b2) (− a2 + b2) (vii) (6x – 7) (6x + 7) (viii) (– a + c) (– a + c) (ix) (x) (7a – 9b) (7a – 9b)

Solutions :
(i) (x + 3) (x + 3)
= (x + 3)2
[Use the Identity (a + b)2 = a2 + 2ab + b2 ]
= x2 + 2 × x × 3 + 32
= x2 + 6x + 9

(ii) (2y + 5 ) (2y + 5)
= (2y + 5)2
[Use the Identity (a + b)2 = a2 + 2ab + b2 ]
= (2y)2 + 2 × 2y × 5 + 52
= 4y2 + 20y + 25

(iii) (2a – 7) (2a – 7)
= (2a – 7)2
[Use the Identity (a − b)2 = a2 − 2ab + b2 ]
= (2a)2 − 2 × 2a × 7 + 72
= 4a2 − 28a + 49

(v) (1.1m – 0.4) (1.1m + 0.4)
= (1.1m)2 − (0.4)2
[Use the Identity (a)2 − (b)2 = (a + b) (a − b) ]
= (1.1m)2 − (0.4)2
= 1.21m2 − 0.16

(vi) (a2 + b2) (− a2 + b2)
= (b2 + a2) (b2a2)
= (b2)2 − (a2)2
[Use the Identity (a)2 − (b)2 = (a + b) (a − b) ]
= b4a4

(vii) (6x – 7) (6x + 7)
= (6x)2 − (7)2
[Use the Identity (a)2 − (b)2 = (a + b) (a − b) ]
= 36x2 − 49

(viii) (– a + c) (– a + c)
= (ca) (ca)
= (ca)2
[Use the Identity (a − b)2 = a2 − 2ab + b2 ]
= c2 − 2 × c × a + a2
= c2 − 2ca + a2

(x) (7a − 9b) (7a – 9b)
= (7a – 9b)2
[Use the Identity (a − b)2 = a2 − 2ab + b2 ]
= (7a)2 − 2 × 7a × 9b + (9b)2
= 49a2 − 126ab + 81b2

2. Use the identity (x + a) (x + b) = x2 + (a + b) x + ab to find the following products.

 (i) (x + 3) (x + 7) (ii) (4x + 5) (4x + 1) (iii) (4x – 5) (4x − 1) (iv) (4x + 5) (4x – 1) (v) (2x + 5y) (2x + 3y) (vi) (2a2 + 9) (2a2 + 5) (vii) (xyz – 4) (xyz – 2)

Solutions :
(i) (x + 3) (x + 7)
by identity (x + a) (x + b) = x2 + (a + b) x + ab
(x + 3) (x + 7) = x2 + (3 + 7) x + 3 × 7
= x2 + 10x + 21

(ii) (4x + 5) (4x + 1)
by identity (x + a) (x + b) = x2 + (a + b) x + ab
(4x + 5) (4x + 1) = (4x)2 + (5 + 1) 4x + 5 × 1
= 16x2 + 6 × 4x + 5
= 16x2 + 24x + 5

(iii) (4x – 5) (4x – 1) = 16x − 24x + 5
by identity (x − a) (x − b) = x2 − (a + b) x + ab
(4x – 5) (4x – 1) = (4x)2 − (5 + 1) 4x + 5 × 1
= 16x2 − 6 × 4x + 5
= 16x2 − 24x + 5

(iv) (4x + 5) (4x − 1) = 16x2  − 4x + 20x − 5
by identity (x + a) (x − b) = x2 + (a − b) x − ab
(4x + 5) (4x – 1) = (4x)2 + (5 − 1) 4x − 5 × 1
= 16x2 + 4 × 4x − 5
= 16x2 + 16x − 5

(v) (2x + 5y) (2x + 3y)
by identity (x + a) (x + b) = x2 + (a + b) x + ab
(2x + 5y) (2x + 3y) = (2x)2 + (5y + 3y) 2x + 5y × 3y
= 4x2 + 8y × 2x + 15y2
= 4x2 + 16xy + 15y2

(vi) (2a2 + 9) (2a2 + 5)
by identity (x + a) (x + b) = x2 + (a + b) x + ab
(2a2 + 9) (2a2 + 5) = (2a2)2 + (9 + 5) 2a2 + 9 × 5
= 4a4 + 14 × 2a2 + 45
= 4a4 + 28a2 + 45

(vii) (xyz – 4) (xyz – 2)
by identity (x − a) (x − b) = x2 − (a + b) x + ab
(xyz – 4) (xyz – 2) = (xyz)2 − (4 + 2) xyz + 4 × 2
= x2y2z2 − 6xyz + 8

3. Find the following squares by using the identities.

 (i) (b − 7)2 (ii) (xy + 3z)2 (iii) (6x2 − 5y)2 (iv) (v) (0.4p − 0.5q)2 (iv) (2xy + 5y)2

Solutions :
(i) (b − 7)2
by identity (a − b)2 = a2 − 2ab + b2
(b − 7)2 = (b)2 − 2 × b × 7 + (7)2
= b2 − 14b + 49

(ii) (xy + 3z)2
by identity (a + b)2 = a2 + 2ab + b2
(xy + 3z)2 = (xy)2 + 2 × xy × 3z + (3z)2
= x2y2 + 6xyz + 9z2

(iii) (6x2 − 5y)2
by identity (a − b)2 = a2 − 2ab + b2
(6x2 − 5y)2 = (6x2)2 − 2 × 6x2 × 5y + (5y)2
= 36x4 – 60x2y + 25y2

(v) (0.4p − 0.5q)2
by identity (a − b)2 = a2 − 2ab + b2
(0.4p − 0.5q)2 = (0.4p)2 − 2 × 0.4p × 0.5q + (0.5q)2
= 0.16p2 − 0.4pq + 0.25q2

(iv) (2xy + 5y)2
by identity (a + b)2 = a2 + 2ab + b2
(2xy + 5y)2 = (2xy)2 + 2 × 2xy × 5y + (5y)2
= 4x2y2 + 20xy + 25y2

4. Simplify.

 (i) (a2 − b2)2 (ii) (2x + 5)2 − (2x − 5)2 (iii) (7m – 8n)2 + (7m + 8n)2 (iv) (4m + 5n)2 + (5m + 4n)2 (v) (2.5p − 1.5q)2 − (1.5p − 2.5q)2 (vi) (ab + bc)2 − 2ab2c (vii) (m2 − n2m)2 + 2m3n2

Solutions :
(i) (a2b2)2
by identity (a − b)2 = a2 − 2ab + b2
(a2b2)2 = (a2)2 − 2 × a2 × b2 + (b2)2
= a4 − 2a2b2 + b4

(ii) (2x + 5)2 − (2x − 5)2
[by identity (a)2 − (b)2 = (a + b) (a − b) ]
(2x + 5)2 − (2x − 5)2 = (2x + 5 + 2x − 5) {2x + 5 − (2x − 5)}
= 4x (2x + 5 − 2x + 5)
= 4x × 10
= 40x

(iii) (7m – 8n)2 + (7m + 8n)2
by identity (a − b)2 = a2 − 2ab + b2 and (a + b)2 = a2 + 2ab + b2
= [(7m)2 − 2 × 7m × 8n + (8n)2] + [(7m)2 + 2 × 7m × 8n + (8n)2]

= 49m2 − 112mn + 64n2 + 49m2 + 112mn + 64n2
= 49m2 + 64n2 + 49m2 + 64n2
= 98m2 + 128n2

(iv) (4m + 5n)2 + (5m + 4n)2
by identity (a + b)2 = a2 + 2ab + b2
= [(4m)2 + 2 × 4m × 5n + (5n)2] + [(5m)2 + 2 × 5m × 4n + (4n)2]
= 16m2 + 40mn + 25n2 + 25m2 + 20mn + 16n2
= 16m2 + 25m2 + 80mn + 25n2 + 16n2
= 41m2 + 80mn + 16n2

(v) (2.5p − 1.5q)2 − (1.5p − 2.5q)2
by identity (a − b)2 = a2 − 2ab + b2
= [(2.5p)2 − 2 × 2.5p × 1.5q + (1.5q)2] − [(1.5p)2 – 2 × 1.5p × 2.5q + (2.5q)2]
= 6.25p2 − 7.8pq + 2.25q2 − [2.25p2 − 7.8pq + 6.25q2]
= 6.25p2 − 7.8pq + 2.25q2 − 2.25p2 + 7.8pq − 6.25q2
= 6.25p2 − 2.25p2 + 2.25q2 − 6.25q2
= 4p2 − 4q`2

(vi) (ab + bc)2 − 2ab2c
by identity (a + b)2 = a2 + 2ab + b2
= (ab)2 + 2 × ab × bc + (bc)2 − 2ab2c
= a2b2 + 2ab2c + b2c2 − 2ab2c
= a2b2 + b2c2 + 2ab2c − 2ab2c
= a2b2 + b2c2

(vii) (m2n2m)2 + 2m3n2
by identity (a − b)2 = a2 − 2ab + b2
= (m2)2 − 2 × m2 × n2m + (n2m)2 + 2m3n2
= m4 − 2m3n2 + n4m2 + 2m3n2
= m4 + n4m2

5. Show that.

 (i) (3x + 7)2 − 84x = (3x − 7)2 (ii) (9p − 5q)2 + 180pq = (9p + 5q)2 (iii) (iv) (4pq + 3q)2 − (4pq − 3q)2 = 48pq2 (v) (a − b) (a + b) + (b − c) (b + c) + (c − a) (c + a) = 0

Solutions :
(i) (3x + 7)2 − 84x = (3x − 7)2
L.H.S.
(3x + 7)2 − 84x
by identity (a + b)2 = a2 + 2ab + b2
= (3x)2 + 2 × 3x × 7 + 72 − 84x
= 9x2 + 42x + 49 − 84x
= 9x2 − 42x + 49
(because a2 − 2ab + b2 = (a − b)2
therefor 9x2 − 42x + 49 = (3x)2 − 2 × 3x × 7 + (7)2
= (3x − 7)2
L.H.S. = R.H.S.

(ii) (9p − 5q)2 + 180pq = (9p + 5q)2
L.H.S.
(9p − 5q)2 + 180pq
by identity (a − b)2 = a2 − 2ab + b2
= (9p)2 − 2 × 9p × 5q + (5q)2 + 180pq
= 81p2 − 90pq + 25q2 + 180pq
= 81p2 + 90pq + 25q2
(because a2 + 2ab + b2 = (a + b)2
therefor 81p2 + 90pq + 25q2 = (9p)2 − 2 × 9p × 5q + (5q)2
= (9p + 5q)2
L.H.S. = R.H.S.

(iv) (4pq + 3q)2 − (4pq − 3q)2 = 48pq2
L.H.S.
(4pq + 3q)2 − (4pq − 3q)2
by identity (a + b)2 = a2 + 2ab + b2 and identity (a − b)2 = a2 − 2ab + b2 से –
= (4pq)2 + 2 × 4pq × 3q + (3q)2 − [{4pq)2 − 2 × 4pq × 3q + (3q)2]
= 16p2q2 + 24pq2 + 9q2 − [16p2q2 − 24pq2 + 9q2]
= 16p2q2 + 24pq2 + 9q2 − 16p2q2 + 24pq2 − 9q2
= 16p2q2 − 16p2q2 + 9q2 − 9q2 + 24pq2 + 24pq2
= 48pq2
L.H.S. = R.H.S.

(v) (ab) (a + b) + (bc) (b + c) + (ca) (c + a) = 0
L.H.S.
(ab) (a + b) + (bc) (b + c) + (ca) (c + a)
(by identity (a)2 − (b)2 = (a + b) (a − b)
= (a2b2) + (b2c2) + (c2a2)
= a2b2 + b2c2 + c2a2
= a2a2b2 + b2c2 + c2
= 0
L.H.S. = R.H.S.

6. Using identities, evaluate.

 (i) 712 (ii) 992 (iii) 1022 (iv) 9982 (v) 5.22 (vi) 297 × 303 (vii) 78 × 82 (viii) 8.92 (ix) 10.5 × 9.5

Solutions :

(i) 712
= (70 + 1)2
by identity (a + b)2 = a2 + 2ab + b2
(70 + 1)2 = (70)2 + 2 × 70 × 1 + (1)2

= 4900 + 140 + 1
= 5041

(ii) 992
= (100 − 1)2
by identity (a − b)2 = a2 − 2ab + b2
(100 − 1)2 = (100)2 − 2 × 100 × 1 + (1)2
= 10000 − 200 + 1
= 10001 − 200
= 9801

(iii) 1022
= (100 + 2)2
by identity (a + b)2 = a2 + 2ab + b2
(100 + 2)2 = (100)2 + 2 × 100 × 2 + (2)2
= 10000 + 400 + 4
= 10404

(iv) 9982
= (1000 − 2)2
by identity (a − b)2 = a2 − 2ab + b2
(1000 − 2)2 = (1000)2 − 2 × (1000) × 2 + (2)2
= 1000000 − 4000 + 4
= 1000004 − 4000
= 996004

(vi) 297 × 303
= (300 − 3) (300 + 3)
by identity (a)2 − (b)2 = (a + b) (a − b)
= (300)2 − (3)2
= 90000 − 9
= 89991

(vii) 78 × 82
= (80 − 2) (80 + 2)
by identity (a)2 − (b)2 = (a + b) (a − b)
= (80)2 − (2)2
= 6400 − 4
= 6396

(viii) 8.92
= (10.0 − 1.1)2
by identity (a − b)2 = a2 − 2ab + b2
(10.0 − 1.1)2 = (10)2 − 2 × 10 × 1.1 + (1.1)2

= 100 − 22 + 1.21
= 100.21 − 22
= 79.21

(ix) 10.5 × 9.5
= (10.0 + 0.5) (10.0 – 0.5)
by identity (a)2 − (b)2 = (a + b) (a − b)
= (10)2 − (0.5)2
= 100 − 0.25
= 99.75

7. Using a2 − b2 = (a + b) (a − b), find

 (i) 512 − 492 (ii) (1.02)2 − (0.98)2 (iii) 1532 − 1472 (iv) 12.12 − 7.92

Solutions :
(i) 512 − 492
by identity a2 − b2 = (a + b) (a − b)
512 − 492 = (51 + 49) (51 − 49)
= 100 × 2
= 200

(ii) (1.02)2 − (0.98)2
by identity a2 − b2 = (a + b) (a − b)
(1.02)2 − (0.98)2 = (1.02 + 0.98) (1.02 − 0.98)
= 2 × 0.04
= 0.08

(iii) 1532 − 1472
by identity a2 − b2 = (a + b) (a − b)
1532 − 1472 = (153 + 147) (153 − 147)

= 300 × 6
= 1800

(iv) 12.12 − 7.92
by identity a2 − b2 = (a + b) (a − b)
12.12 − 7.92 = (12.1 + 7.9) (12.1 − 7.9)

= 20 × 4.2
= 84

8. Using (x + a) (x + b) = x2 + (a + b) x + ab, find

 (i) 103 × 104 (ii) 5.1 × 5.2 (iii) 103 × 98 (iv) 9.7 × 9.8

Solutions :
(i) 103 × 104
= (100 + 3) (100 + 4)
by identityसर्वसमिका (x + a) (x + b) = x2 + (a + b) x + ab
= (100)2 + (3 + 4) 100 + 3 × 4

= 10000 + 700 + 12
= 10712

(ii) 5.1 × 5.2
= (5.0 + 0.1) (5.0 + 0.2)
by identity (x + a) (x + b) = x2 + (a + b) x + ab
= (5.0)2 + (0.1 + 0.2) 5.0 + 0.1 × 0.2

= 25 + 1.5 + 0.02
= 26.52

(iii) 103 × 98
= (100 + 3) (100 – 2)
by identity (x + a) (x + b) = x2 + (a + b) x + ab
= (100)2 + {3 + (− 2} 100 + 3 × (− 2)

= 10000 + 1 × 100 + (− 6)
= 10000 + 100 − 6
= 10100 − 6
= 10094

(iv) 9.7 × 9.8
= (9.0 + 0.7) (9.0 + 0.8)
by identity (x + a) (x + b) = x2 + (a + b) x + ab
= (9)2 + (0.7 + 0.8) 9 + 0.7 × 0.8
= 81 + 1.5 × 9 + 0.56
= 81 + 13.5 + 0.56
= 95.06

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