NCERT Solutions For Class 8 Maths Chapter 9

NCERT Solutions Class 8 Maths Algebraic Expressions and Identities

NCERT Solutions for class 8 Maths chapter 9 Algebraic expression & Identities is crucial for the students of class 8 to excel in their examination. These solution help students to frame a better understanding of the topic. To ease the fear of maths, we at BYJU’S provide NCERT Solution for Class 8 Maths Chapter 9 Algebraic Expressions. Student can download the NCERT Solution for class 8 Maths Chapter 9 pdf or can view it online by following the link. These solution are provided in detailed manner, where one can find step-by-step solution to all the questions of NCERT Solutions for class 8 maths chapter 9.

NCERT Solutions Class 8 Maths Chapter 9 Exercises

Exercise 9.1

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

(I) 5abc2 – 3cb

Terms :  5abc2  


Coefficients: 5, -3

(II) 1+a+a2

Terms: 1, a, a2

Coefficients: 1, 1, 1

(III) 4x2y– 4 x2y2z+  z2

Terms: 4x2y2   ,  -4 x2y2z2    ,  Z2

Coefficient: 4,  -4,  1

(IV) 3 – xy + yz – zx

Terms:   3:  -xy,  yz,  -zx

Coefficient:  3:  -1,  1,  -1

(V) a2+b2ab

Terms: a2,  b2,  -ab

Coefficient: 12,  12,  -1


Terms: 0.3x,  -0.6xy,  0.5y

Coefficient: 0.3,  -0.6,  0.5


Q.2. Check whether the following polynomials are monomials, binomials or trinomials. Find out which polynomials do not fit any of these three categories?

1)  x+y,

2)  1000,

3)  x+x2+x3+x4,

4)  7+y+5x,

5)  2y3y2,

6)  2y3y2+4y3,

7)  5x-4y+3xy,

8)  4z15z2,

9)  ab+bc+cd+da,

10)  pqr,

11)  p2q+pq2,

12)  2p+2q,


Monomials: 1000,   pqr

Binomials: x+y,   2y3y2,  4z15z2,   p2q+pq2,    2p+2q

Trinomials: 7+y+5x,    2y3y2+4y3,      5x-4y+3xy

Polynomials that do not fit any of these categories are :

x+x2+x3+x4, ab+bc+cd+da


Q.3.Add the following :

Note: The given expressions written in separate rows, with like terms one below the other and then the addition of these expressions are done.

(I)ab – bc,    bc – ca,    ca – ab


+   bc-ca

+  -ab+ca

=           0

(II) x – y+xy,       y-z+yz,        z-x+xz

x – y + xy

+      y -z+yz

+       -x+z +xz

=        xy+yz+xz

(III) 2a2b23ab+45+7ab3a2b2


+    3a2b2+7ab+5



(IV) a2+b2b2+c2,c2+a2,2ab+2bc+2ca


+     b2+c2

+     c2+a2

+     2ab+2bc+2ca

=      2a2+2b2+2c2+2ab+2bc+2ca


Q.4. (i)Substract 4x-7xy+3y+12 from 12x-9xy+5y-3


12x – 9xy + 5y –  3

4x – 7xy + 3y + 12

(-)     (+)    (-)    (-)

8x – 2xy + 2y – 15

(ii)Substract 3xy +5yz -7zx from 5xy-2yz-2zx+10xyz

5xy – 2yz -2zx +10xyz

3xy + 5yz -7zx

(-)      (-)      (+)

2xy-7yz + 5zx  +10xyz

(iii) Substract4p2q3pq+5pq28p+7q10from183p+11q+5pq2pq2+5p2q




(+)          (+)   (-)  (+)   (-)            (-)



Exercise: 9.2

Q.1.For the following pairs of monomials find the product.

(I)5, 6a

Answer: 5×6×a=30a

(II)-5a, 6 a

Answer: 5a×6a×=5×a×6×a=(5×6)×(a×a)=30a2

(III) )-5a, 6 ab

Answer: 5a×6ab×=5×a×6×a×b=(5×6)×(a×a×b)=30a2b

(IV) ) 5a3,- 4 a

Answer: 5a3×4a=5×(4)×a×a×a×a=20a4

(V)5a, 0

Answer: 5a×0=5×a×0=0


Q.2.calculate the area of rectangles.Where the pairs of monomials  are lengths and breadths respectively.

NOTE: area of rectangle =length×breadth

  • (a, b)

Area= a×b=ab

  • (10a, 5b)

Area = 10a×5b=10×5×a×b=50ab

  • (20p2,5q2)

Area = 20p2×5q2=20×5×p2×q2=100p2q2

  • (4a,3a2)

Area = 4a×3a2=4×3×a×a2=12a3

  • (4ab,3bc)

Area= 4ab×3bc=4×3×a×b×b×c=12ab2c


Q.3.Complete the table of product.

First monomial

Second monomial



-5y 3x2 -4xy 7x2y 9x2y2
2x 4x2
-5y 15x2y



First monomial

Second monomial



-5y 3x2 -4xy 7x2y 9x2y2
2x 4x2 10xy 6x3 8x2y 14x3y 18x3y2
-5y   10xy 15x2y 15x2y 20xy2 35x2y2 45x2y3
3x2 6x3 15x2y 9x4 12x3y 21x4y 27x4y2
-4xy 8x2y 20x2y 12x3y 16x2y2 28x3y2 36x3y3



Q.4.Rectangular boxes with the length ,breadth , and height are given respectively. Find the volume.

(I) 5x,3x2,7x4

Answer: Volume=5x×3x2×7x4=5×3×7×x×x2×x4=105x7

(II)2p, 4q, 8r

Answer: Volume=2p×4q×8r=2×4×8×p×q×r=64pqr

(III) ab,2a2b,2ab2

Answer:  Volume=ab×2a2b×2ab2=2×2×ab×a2b×ab2=4a4b4

(IV)p, 2q, 3r

Answer:  Volume=p×2q×3r=2×3×p×q×r=6pqr


Q.5.Find the Product of the following:

(I)ab, bc, ca

Answer: ab×bc×ca= a2b2c2

(II) x,x2,x3

Answer: x×(x2)×x3=x6

(III) 2,4a,8a2,16a3

Answer: 2×4a×8a2×16a3=1024a6

(IV)x, 2y, 3z, 6xyz

Answer: x×2y×3z×6xyz=36x2y2z2

(V)m, -mn, mnp

Answer: m×mn×mnp=m3n2p