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

Solutions :

┬а ┬а Terms Coefficients
(i) 5xyz2 тИТ 3zy There are two terms in this expressions.
5xyz2 and 3zy
5 and тИТ 3
(ii) 1 + x + x2 There are two terms in this expressions.
1, x and x2
1, 1 and 1
(iii) 4x2y2 тИТ 4x2y2z2 + z2 There are three terms in this expressions.
4x2y2 , тИТ 4x2y2z2 and z2
4, тИТ 4 and 1
(vi) 3 тИТ pq + qr – rp There are four terms in this expressions.
3, тИТ pq, qr and тИТ rp
3, тИТ 1, 1 and тИТ 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 тИТ xy
\(\displaystyle \frac{1}{2}\) , \(\displaystyle \frac{1}{2}\) and тИТ 1
(vi) 0.3a тИТ 0.6ab + 0.5b There are three terms in this expressions.
0.3, тИТ 0.6ab and 0.5b
0.3, тИТ 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 тИТ 3y2, 2y тИТ 3y2 + 4y3, 5x тИТ 4y + 3xy, 4z тИТ 15z2, ab + bc + cd + da, pqr, p2q, 2p + 2q
Solutions :
Monomials :- 1000, pqr
Binomials :- x + y, 2y тИТ 3y2, 4z тИТ 15z2, p2q + pq2, 2p + 2q
Trinomials :- 7 + y + 5x, 2y тИТ 3y2 + 4y3, 5x тИТ 4y + 3xy
These polynomials do not fit in any of these three categories :- x + x2 + x3 + x4, ab + bc + cd + da

3. Add the following :

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

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

= ab тИТ ab + bc тИТ bc + ca тИТ ca
= 0

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

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

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

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

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

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

= 8a + 2b тИТ 2ab тИТ 16

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

remove the bracket –
= 5xy тИТ 2yz тИТ 2zx + 10xyz тИТ 3xy тИТ 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 тИТ 3xy тИТ 2yz тИТ 5yz + 7zx тИТ 2zx + 10xyz

= 2xy тИТ 7yz + 5zx + 10xyz

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

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

= 5p тИТ 18q + 8pq тИТ 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) тИТ 4p, 7p (iii) тИТ 4p, 7pq (iv) 4p3, тИТ 3p
(v) 4p, 0 ┬а ┬а ┬а

Solutions :┬а
(i) 4, 7p
= 4 ├Ч 7p
= 28p

(ii)┬а тИТ 4p, 7p
= тИТ 4p ├Ч 7p
= тИТ 28p2

(iii) тИТ 4p, 7pq
= тИТ 4p ├Ч 7pq
= тИТ 28p2q

(iv) 4p3, тИТ 3p
= 4p3 ├Ч тИТ 3p
= тИТ 12p4

(v) 4p, 0
= 4p ├Ч 0
= 0

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

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

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

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

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

3. Complete the table of products.

First monomial тЖТ
Second monomial тЖУ
2x тИТ 5y┬а 3x2 тИТ 4xy 7x2y тИТ 9x2y2
2x 4x2 ┬а ┬а ┬а ┬а ┬а
тИТ 5y ┬а ┬а тИТ 15x2y ┬а ┬а ┬а
3x2 ┬а ┬а ┬а ┬а ┬а ┬а
тИТ 4xy ┬а ┬а ┬а ┬а ┬а ┬а
7x2y ┬а ┬а ┬а ┬а ┬а ┬а
тИТ 9x2y2 ┬а ┬а ┬а ┬а ┬а ┬а

Solutions :┬а

First monomial тЖТ
Second monomial тЖУ
2x тИТ 5y┬а 3x2 тИТ 4xy 7x2y тИТ 9x2y2
2x 4x2 тИТ 10xy 6×3 тИТ 8x2y 14x3y тИТ 18x3y2
тИТ 5y тИТ 10xy 25y2 тИТ 15x2y 20xy2 тИТ 35x2y2 45x2y3
3x2 6x3 тИТ 15x2y 9x4 тИТ 12x3y 21x4y тИТ 27x4y2
тИТ 4xy тИТ 8x2y 20xy2 тИТ 12x3y 16x2y2 тИТ 28x3y2 36x3y3
7x2y 14x3y тИТ 35x2y2 21x4y тИТ 28x3y2 49x4y2 тИТ 63x4y3
тИТ 9x2y2 тИТ 18x3y2 45x2y2 тИТ 27x4y2 36x3y3 тИТ 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 :┬а
(i) 5a, 3a2, 7a4
volume of rectangular box = length ├Ч breadth ├Ч height
= 5a ├Ч 3a2 ├Ч 7a4
= 5 ├Ч 3 ├Ч 7 ├Ч a ├Ч a2 ├Ч a4
= 105a7

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

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

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

5. Obtain the product of

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

Solutions :
(i) xy, yz, zx
= xy ├Ч yz ├Ч zx┬а
= x2y2z2

(ii) a, тИТ a2, a3┬а
= a ├Ч (тИТ a2) ├Ч a3┬а
= тИТ a6

(iii) 2, 4y, 8y2, 16y3
= 2 ├Ч 4y ├Ч 8y2┬а├Ч 16y3
= 2 ├Ч 4 ├Ч 8 ├Ч 16 ├Ч y ├Ч y2 ├Ч y3
= 1024y3

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

(v) m, тИТ mn, mnp
= m ├Ч (тИТ mn) ├Ч mnp
= тИТ 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 тИТ b (iii) a + b, 7a2b2 (iv) a2 тИТ 9, 4a
(v) pq + qr + rp, 0 ┬а ┬а ┬а

Solutions :
(i) 4p, q + r┬а
= 4p ├Ч (q + r)
= 4p ├Ч q + 4p ├Ч r
= 4pq + 4pr

2. Complete the table.

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

Solutions :

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

(i) a ├Ч (b + c + d)
= ab + ac + ad

(ii) (x + y тИТ 5) ├Ч 5xy
= 5x2y + 5xy2 тИТ 25xy

(iii) p ├Ч (6p2 тИТ 7p + 5)
= 6p3 тИТ 7p2 + 5p

(iv) 4p2q2 ├Ч (p2 тИТ q2)
= 4p4q2 тИТ 4p2q4

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

3. Find the product.

(i) (a2) ├Ч (2a22) ├Ч (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 ├Ч x2 ├Ч x3 ├Ч x4

Solutions :
(i) (a2) ├Ч (2a22) ├Ч (4a26)
= 2 ├Ч 4 ├Ч a2 ├Ч a22 ├Ч 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 ├Ч x2 ├Ч x3 ├Ч x4
= x10

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

Solutions :
(a) (i) 3x (4x тИТ 5) + 3
= 3x ├Ч 4x тИТ 3x ├Ч 5 + 3
= 12x2 тИТ 15x + 3 ………………………….. (i)
putting x = 3 equ. in (i)
= 12 ├Ч (3)2 тИТ 9 ├Ч 5 + 3
= 12 ├Ч 9 тИТ 45 + 3
= 108 + 3 тИТ 45
= 111 тИТ 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 (p тИТ q), q (q тИТ r) and r (r тИТ p)┬а
(b) Add: 2x (z тИТ x тИТ y) and 2y (z тИТ y тИТ x)┬а
(c) Subtract: 3l (l тИТ 4m + 5n) from 4l (10n тИТ 3m + 2l )┬а
(d) Subtract: 3a (a + b + c ) тИТ 2b (a тИТ b + c) from 4c (тИТ a + b + c )┬а
Solutions :┬а
(a) Add: p (p тИТ q), q (q тИТ r) and r (r тИТ p)┬а
= p (p тИТ q) + q (q тИТ r) + r (r тИТ p)
= p2 тИТ pq┬а+ q2 тИТ qr┬а+ r2 тИТ rp
= p2 + q2 + r2 тИТ pq тИТ qr тИТ rp

(b) Add: 2x (z тИТ x тИТ y) and 2y (z тИТ y тИТ x)┬а
= 2x (z тИТ x тИТ y) + 2y (z тИТ y тИТ x)
= 2xz┬атИТ 2x2 тИТ 2xy┬а+ 2yz┬атИТ 2y2 тИТ 2xy
= тИТ 2x2 тИТ 2y2 тИТ 2xy┬атИТ 2xy┬а+ 2yz┬а+ 2xz
= тИТ 2x2 тИТ 2y2 тИТ 4xy + 2yz┬а+ 2xz

(c) Subtract: 3l (l тИТ 4m + 5n) from 4l (10n тИТ 3m + 2l )┬а
= 4l (10n тИТ 3m + 2l) тИТ 3l (l тИТ 4m + 5n)
= 4l ├Ч 10n тИТ 4l ├Ч 3m + 4l + 2l тИТ 3l ├Ч l тИТ 3l ├Ч (тИТ 4m) тИТ 3l ├Ч 5n
= 40ln┬атИТ 12lm┬а+ 8l2 тИТ 3l2 + 12lm┬атИТ 15ln
= 40ln┬атИТ 15ln┬атИТ 12lm┬а+ 12lm┬а+ 8l2 тИТ 3l2
= 25ln┬атИТ 0lm + 5l2
= 5l2 + 25ln

(d) Subtract: 3a (a + b + c ) тИТ 2b (a тИТ b + c) from 4c (тИТ a + b + c )┬а
= 4c (тИТ a + b┬а+ c) тИТ [3a (a + b┬а+ c) тИТ 2b (a тИТ b┬а+ c)]
= тИТ 4ac + 4bc + 4c2 тИТ [3a2 + 3ab┬а+ 3ac тИТ 2ab┬а+ 2a2 – 2bc]
= тИТ 4ac + 4bc + 4c2 тИТ 3a2 тИТ 3ab┬атИТ 3ac + 2ab тИТ 2b2 + 2bc
= тИТ 4ac тИТ 3ac + 4bc + 2bc тИТ 3ab┬а+ 2ab┬а+ 4c2 тИТ 3a2 тИТ 2b2
= тИТ 7ac + 6bc тИТ ab┬а+ 4c2 тИТ 3a2 тИТ 2b2
=┬а тИТ 3a2 тИТ 2b2 + 4c2 тИТ ab + 6bc тИТ 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 тИТ 3) (ii) (y тИТ 8) and (3y тИТ 4)
(iii) (2.5l тИТ 0.5m) and (2.5l + 0.5m)┬а (iv) (a + 3b) and (x + 5)
(v) (2pq + 3q2) and (3pq тИТ 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 тИТ 3)
= 2x (4x тИТ 3) + 5 (4x тИТ 3)
= 2x ├Ч 4x тИТ 2x ├Ч 3 + 5 ├Ч 4x тИТ 5 ├Ч 3
= 8x2 тИТ 6x + 20x тИТ 15
= 8x2 + 14x тИТ 15

(ii) (y – 8) and (3y тИТ 4)
= y (3y тИТ 4) тИТ 8 (3y тИТ 4)
= y ├Ч 3y тИТ y ├Ч 4 тИТ 8 ├Ч 3y тИТ 8 ├Ч (тИТ 4)
= 3y2 тИТ 4y тИТ 24y + 32
= 3y2 тИТ 28y + 32

(iii) (2.5l тИТ 0.5m) and (2.5l + 0.5m)
= 2.5l (2.5l + 0.5m) тИТ 0.5m (2.5l + 0.5m)
= 2.5l ├Ч 2.5l + 2.5l ├Ч 0.5m – 0.5m ├Ч 2.5l тИТ 0.5m ├Ч 0.5m
= 2.25l2 + 1.25lm тИТ 1.25lm тИТ 0.25m2
= 2.25l2 тИТ 0.25m2

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

(v) (2pq + 3q2) and (3pq тАУ 2q2)
= 2pq (3pq тИТ 2q2) + 3q2 (3pq тИТ 2q2)
= 2pq ├Ч 3pq тИТ 2q2 ├Ч 2pq + 3q2 ├Ч 3pq тИТ 3q2 ├Ч 2q2
= 2p2q2 тИТ 4pq3 + 9pq3 тИТ 6q4
= 6p2q2 + 5pq3 тИТ 6q4

2. Find the product.

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

Solutions :
(i) (5 тИТ 2x) (3 + x)
= 5 ├Ч (3 + x) тИТ 2x ├Ч (3 + x)
= 15 + 5x тИТ 6x тИТ 2x2
= 15 тИТ x тИТ 2x2

(ii) (x + 7y) (7x тИТ y)
= x ├Ч (7x тИТ y) + 7y ├Ч (7x тИТ y)
= 7x2 тИТ xy┬а+ 49xy тИТ 7y2
= 7x2 + 48xy тИТ 7y2

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

(iv) (p2 тИТ q2) (2p + q)
= p2 ├Ч (2p + q) тИТ q2 ├Ч (2p + q)
= 2p3 + p2q тИТ 2pq2 тИТ q3

3. Simplify.

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

Solutions :
(i) (x2 тАУ 5) (x + 5) + 25
= x2 ├Ч (x + 5) тИТ 5 ├Ч (x + 5) + 25
= x3 + 5x2 тИТ 5x тИТ 25 + 25
= x3 + 5x2 тИТ 5x

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

(iii) (t + s2) (t2 тИТ s)
= t ├Ч (t2 тИТ s) + s2 ├Ч (t2 тИТ s)
= t3 тИТ st + s2t2 тИТ s3

(iv) (a + b) (c тИТ d) + (a тИТ b) (c + d) + 2 (ac + bd)
= a ├Ч (c тИТ d) + b ├Ч (c тИТ d) + a ├Ч (c + d) тИТ b ├Ч (c + d) + 2ac + 2bd
= ac тИТ ad + bc тИТ bd + ac + ad тИТ bc тИТ bd + 2ac + 2bd
= ac + ac + 2ac + bc тИТ bc тИТ bd тИТ bd + 2bd тИТ ad + ad┬а
= 2ac + 2ac тИТ 2bd + 2bd
= 4ac

(v) (x + y) (2x + y) + (x + 2y) (x тАУ y)
= x ├Ч (2x + y) + y ├Ч (2x + y) + x ├Ч (x тИТ y) + 2y ├Ч (x тИТ y)
= 2x2 + xy + 2xy + y2 + x2 тИТ xy + 2xy тИТ 2y2
= 2x2 + x2 + xy + 2xy тИТ xy + 2xy + y2 тИТ 2y2
= 3x2 + 5xy тИТ xy тИТ y2
= 3x2 + 4xy тИТ y2

(vii) (1.5x тИТ 4y) (1.5x + 4y + 3) тИТ 4.5x + 12y
= 1.5x ├Ч (1.5x + 4y + 3) тИТ 4y ├Ч (1.5x + 4y + 3) тИТ 4.5x + 12y
= 2.25x2 + 6.0xy + 4.5x – 6.0xy тИТ 16y2 тИТ 12y тИТ 4.5x тИТ 12y
= 2.25x2 + 6.0xy тИТ 6.0xy + 4.5x тИТ 4.5x тИТ 16y2 тИТ 12y тИТ 12y
= 2.25x2 тИТ 16y2

(viii) (a + b + c) (a + b тИТ c)
= a ├Ч (a + b тИТ c) + b ├Ч (a + b тАУ c)┬а + c ├Ч (a + b тИТ c)
= a2 + ab тИТ ac + ab + b2 тИТ bc + ac + bc тИТ c2
= a2 + ab + ab тИТ ac + ac + b2 тИТ bc + bc тИТ c2
= a2 + 2ab + b2 тИТ c2
= a2 + b2 тИТ 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 тАУ 7) (2a тАУ 7) (iv)┬а
(v) (1.1m тАУ 0.4) (1.1m + 0.4)┬а (vi) (a2 + b2) (тИТ a2 + b2)
(vii) (6x тАУ 7) (6x + 7)┬а (viii) (тАУ a + c) (тАУ a + c)
(ix)┬а (x) (7a тАУ 9b) (7a тАУ 9b)

Solutions :
(i) (x + 3) (x + 3)
= (x + 3)2
[Use the Identity (a + b)2 = a2 + 2ab + b2 ]
= x2 + 2 ├Ч x ├Ч 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 ├Ч 2y ├Ч 5 + 52
= 4y2 + 20y + 25

(iii) (2a тАУ 7) (2a тАУ 7)
= (2a тАУ 7)2
[Use the Identity (a тИТ b)2 = a2 тИТ 2ab + b2 ]
= (2a)2 тИТ 2 ├Ч 2a ├Ч 7 + 72
= 4a2 тИТ 28a + 49

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

(vi) (a2 + b2) (тИТ a2 + b2)
= (b2 + a2) (b2 тИТ a2)
= (b2)2 тИТ (a2)2
[Use the Identity (a)2 тИТ (b)2 = (a + b) (a тИТ b) ]
= b4 тИТ a4

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

(viii) (тАУ a + c) (тАУ a + c)
= (c тИТ a) (c тИТ a)
= (c тИТ a)2
[Use the Identity (a тИТ b)2 = a2 тИТ 2ab + b2 ]
= c2 тИТ 2 ├Ч c ├Ч a + a2
= c2 тИТ 2ca + a2

(x) (7a тИТ 9b) (7a тАУ 9b)
= (7a тАУ 9b)2
[Use the Identity (a тИТ b)2 = a2 тИТ 2ab + b2 ]
= (7a)2 тИТ 2 ├Ч 7a ├Ч 9b + (9b)2
= 49a2 тИТ 126ab┬а+ 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)┬а
(iii) (4x тАУ 5) (4x тИТ 1) (iv) (4x + 5) (4x тАУ 1)
(v) (2x + 5y) (2x + 3y)┬а (vi) (2a2 + 9) (2a2 + 5)┬а
(vii) (xyz тАУ 4) (xyz тАУ 2) ┬а

Solutions :
(i) (x + 3) (x + 7)
by identity (x + a) (x + b) = x2 + (a + b) x + ab┬а
(x + 3) (x + 7) = x2 + (3 + 7) x + 3 ├Ч 7
= x2 + 10x + 21

(ii) (4x + 5) (4x + 1)
by identity (x + a) (x + b) = x2 + (a + b) x + ab┬а
(4x + 5) (4x + 1) = (4x)2 + (5 + 1) 4x + 5 ├Ч 1
= 16x2 + 6 ├Ч 4x + 5
= 16x2 + 24x + 5

(iii) (4x тАУ 5) (4x тАУ 1) = 16x тИТ 24x + 5
by identity (x тИТ a) (x тИТ b) = x2 тИТ (a + b) x + ab┬а
(4x тАУ 5) (4x тАУ 1) = (4x)2 тИТ (5 + 1) 4x + 5 ├Ч 1
= 16x2 тИТ 6 ├Ч 4x + 5
= 16x2 тИТ 24x + 5

(iv) (4x + 5) (4x тИТ 1) = 16x2┬а тИТ 4x + 20x тИТ 5
by identity (x + a) (x тИТ b) = x2 + (a тИТ b) x тИТ ab┬а
(4x + 5) (4x тАУ 1) = (4x)2 + (5 тИТ 1) 4x тИТ 5 ├Ч 1
= 16x2 + 4 ├Ч 4x тИТ 5
= 16x2 + 16x тИТ 5┬а

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

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

(vii) (xyz тАУ 4) (xyz┬атАУ 2)
by identity (x тИТ a) (x тИТ b) = x2 тИТ (a + b) x + ab┬а
(xyz тАУ 4) (xyz тАУ 2) = (xyz)2 тИТ (4 + 2) xyz┬а+ 4 ├Ч 2
= x2y2z2 тИТ 6xyz + 8

3. Find the following squares by using the identities.

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

Solutions :
(i) (b тИТ 7)2
by identity (a тИТ b)2 = a2 тИТ 2ab + b2┬а
(b тИТ 7)2 = (b)2 тИТ 2 ├Ч b ├Ч 7 + (7)2
= b2 тИТ 14b + 49

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

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

(v) (0.4p тИТ 0.5q)2
by identity (a тИТ b)2 = a2 тИТ 2ab + b2┬а
(0.4p тИТ 0.5q)2 = (0.4p)2 тИТ 2 ├Ч 0.4p ├Ч 0.5q + (0.5q)2
= 0.16p2 тИТ 0.4pq┬а+ 0.25q2

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

4. Simplify.

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

Solutions :
(i) (a2 тИТ b2)2
by identity (a тИТ b)2 = a2 тИТ 2ab + b2┬а
(a2 тИТ b2)2 = (a2)2 тИТ 2 ├Ч a2 ├Ч b2 + (b2)2
= a4 тИТ 2a2b2 + b4

(ii) (2x + 5)2 тИТ (2x тИТ 5)2
[by identity (a)2 тИТ (b)2 = (a + b) (a тИТ b) ]
(2x + 5)2 тИТ (2x тИТ 5)2 = (2x + 5 + 2x тИТ 5) {2x + 5 тИТ (2x тИТ 5)}
= 4x (2x + 5 тИТ 2x + 5)
= 4x ├Ч 10
= 40x

(iii) (7m – 8n)2 + (7m + 8n)2
by identity (a тИТ b)2 = a2 тИТ 2ab + b2 and (a + b)2 = a2 + 2ab + b2┬а
= [(7m)2 тИТ 2 ├Ч 7m ├Ч 8n + (8n)2] + [(7m)2 + 2 ├Ч 7m ├Ч 8n + (8n)2]

= 49m2 тИТ 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┬а
= [(4m)2 + 2 ├Ч 4m ├Ч 5n + (5n)2] + [(5m)2 + 2 ├Ч 5m ├Ч 4n + (4n)2]
= 16m2 + 40mn┬а+ 25n2 + 25m2 + 20mn┬а+ 16n2
= 16m2 + 25m2 + 80mn + 25n2 + 16n2
= 41m2 + 80mn┬а+ 16n2

(v) (2.5p тИТ 1.5q)2 тИТ (1.5p тИТ 2.5q)2
by identity (a тИТ b)2 = a2 тИТ 2ab + b2┬а
= [(2.5p)2 тИТ 2 ├Ч 2.5p ├Ч 1.5q + (1.5q)2] тИТ [(1.5p)2 – 2 ├Ч 1.5p ├Ч 2.5q + (2.5q)2]
= 6.25p2 тИТ 7.8pq┬а+ 2.25q2 тИТ [2.25p2 тИТ 7.8pq┬а+ 6.25q2]
= 6.25p2 тИТ 7.8pq + 2.25q2 тИТ 2.25p2 + 7.8pq┬атИТ 6.25q2
= 6.25p2 тИТ 2.25p2 + 2.25q2 тИТ 6.25q2
= 4p2 тИТ 4q`2

(vi) (ab + bc)2 тИТ 2ab2c
by identity (a + b)2 = a2 + 2ab + b2
= (ab)2 + 2 ├Ч ab ├Ч bc + (bc)2 тИТ 2ab2c
= a2b2 + 2ab2c┬а+ b2c2 тИТ 2ab2c
= a2b2 + b2c2 + 2ab2c┬атИТ 2ab2c
= a2b2┬а+ b2c2

(vii) (m2 тИТ n2m)2 + 2m3n2
by identity (a тИТ b)2 = a2 тИТ 2ab + b2 ┬а
= (m2)2 тИТ 2 ├Ч m2 ├Ч n2m┬а+ (n2m)2 + 2m3n2
= m4 тИТ 2m3n2 + n4m2 + 2m3n2
= m4 + n4m2

5. Show that.

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

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

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

(iv) (4pq + 3q)2 тИТ (4pq тИТ 3q)2 = 48pq2
L.H.S.
(4pq┬а+ 3q)2 тИТ (4pq┬атИТ 3q)2
by identity (a + b)2 = a2 + 2ab + b2 and┬аidentity (a тИТ b)2 = a2 тИТ 2ab + b2 рд╕реЗ –┬а
= (4pq)2 + 2 ├Ч 4pq ├Ч 3q + (3q)2 тИТ [{4pq)2 тИТ 2 ├Ч 4pq ├Ч 3q + (3q)2]
= 16p2q2 + 24pq2 + 9q2 тИТ [16p2q2 тИТ 24pq2 + 9q2]
= 16p2q2┬а+ 24pq2 + 9q2 тИТ 16p2q2 + 24pq2 тИТ 9q2
= 16p2q2 тИТ 16p2q2┬а+ 9q2 тИТ 9q2 + 24pq2 + 24pq2
= 48pq2
L.H.S. = R.H.S.

(v) (a тИТ b) (a + b) + (b тИТ c) (b + c) + (c тИТ a) (c + a) = 0
L.H.S.
(a тИТ b) (a + b) + (b тИТ c) (b + c) + (c тИТ a) (c + a)
(by identity (a)2 тИТ (b)2 = (a + b) (a тИТ b)┬а
= (a2 тИТ b2) + (b2 тИТ c2) + (c2 тИТ a2)
= a2 тИТ b2 + b2 тИТ c2 + c2 тИТ a2
= a2 тИТ a2 тИТ b2 + b2 тИТ c2 + 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 ├Ч 303 (vii) 78 ├Ч 82 (viii) 8.92
(ix) 10.5 ├Ч 9.5 ┬а ┬а ┬а

Solutions :

(i) 712
= (70 + 1)2
by identity (a + b)2 = a2 + 2ab + b2 ┬а
(70 + 1)2 = (70)2 + 2 ├Ч 70 ├Ч 1 + (1)2

= 4900 + 140 + 1
= 5041

(ii) 992
= (100 тИТ 1)2
by identity (a тИТ b)2 = a2 тИТ 2ab + b2┬а
(100 тИТ 1)2 = (100)2 тИТ 2 ├Ч 100 ├Ч 1 + (1)2
= 10000 тИТ 200 + 1
= 10001 тИТ 200
= 9801

(iii) 1022
= (100 + 2)2
by identity (a + b)2 = a2 + 2ab + b2┬а
(100 + 2)2 = (100)2 + 2 ├Ч 100 ├Ч 2 + (2)2
= 10000 + 400 + 4
= 10404

(iv) 9982
= (1000 тИТ 2)2
by identity (a тИТ b)2 = a2 тИТ 2ab + b2┬а
(1000 тИТ 2)2 = (1000)2 тИТ 2 ├Ч (1000) ├Ч 2 + (2)2
= 1000000 тИТ 4000 + 4
= 1000004 тИТ 4000
= 996004

(vi) 297 ├Ч 303
= (300 тИТ 3) (300 + 3)
by identity (a)2 тИТ (b)2 = (a + b) (a тИТ b)┬а
= (300)2 тИТ (3)2
= 90000 тИТ 9
= 89991

(vii) 78 ├Ч 82
= (80 тИТ 2) (80 + 2)
by identity (a)2 тИТ (b)2 = (a + b) (a тИТ b)┬а
= (80)2 тИТ (2)2
= 6400 тИТ 4
= 6396

(viii) 8.92
= (10.0 тИТ 1.1)2
by identity (a тИТ b)2 = a2 тИТ 2ab + b2┬а
(10.0 тИТ 1.1)2 = (10)2 тИТ 2 ├Ч 10 ├Ч 1.1 + (1.1)2

= 100 тИТ 22 + 1.21
= 100.21 тИТ 22
= 79.21

(ix) 10.5 ├Ч 9.5
= (10.0 + 0.5) (10.0 – 0.5)
by identity (a)2 тИТ (b)2 = (a + b) (a тИТ b)┬а
= (10)2 тИТ (0.5)2
= 100 тИТ 0.25
= 99.75

7. Using a2 тИТ b2 = (a + b) (a тИТ b), find┬а

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

Solutions :
(i) 512 тИТ 492
by identity a2 тИТ b2 = (a + b) (a тИТ b)┬а
512 тИТ 492 = (51 + 49) (51 тИТ 49)
= 100 ├Ч 2
= 200

(ii) (1.02)2 тИТ (0.98)2
by identity a2 тИТ b2 = (a + b) (a тИТ b)┬а
(1.02)2 тИТ (0.98)2 = (1.02 + 0.98) (1.02 тИТ 0.98)
= 2 ├Ч┬а0.04
= 0.08

(iii) 1532 тИТ 1472
by identity a2 тИТ b2 = (a + b) (a тИТ b)┬а
1532 тИТ 1472 = (153 + 147) (153 тИТ 147)

= 300 ├Ч 6
= 1800

(iv) 12.12 тИТ 7.92
by identity a2 тИТ b2 = (a + b) (a тИТ b)┬а
12.12 тИТ 7.92 = (12.1 + 7.9) (12.1 тИТ 7.9)

= 20 ├Ч 4.2
= 84

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

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

Solutions :
(i) 103 ├Ч 104
= (100 + 3) (100 + 4)
by identityрд╕рд░реНрд╡рд╕рдорд┐рдХрд╛ (x + a) (x + b) = x2 + (a + b) x + ab┬а
= (100)2 + (3 + 4) 100 + 3 ├Ч 4

= 10000 + 700 + 12
= 10712

(ii) 5.1 ├Ч 5.2
= (5.0 + 0.1) (5.0 + 0.2)
by identity (x + a) (x + b) = x2 + (a + b) x + ab┬а
= (5.0)2 + (0.1 + 0.2) 5.0 + 0.1 ├Ч 0.2

= 25 + 1.5 + 0.02
= 26.52

(iii) 103 ├Ч 98
= (100 + 3) (100 – 2)
by identity (x + a) (x + b) = x2 + (a + b) x + ab┬а
= (100)2 + {3 + (тИТ 2} 100 + 3 ├Ч (тИТ 2)

= 10000 + 1 ├Ч 100 + (тИТ 6)
= 10000 + 100 тИТ 6
= 10100 тИТ 6
= 10094

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

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