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NCERT · Class 8 · Maths

Chapter 9: Algebraic Expressions and Identities

Practice each question, then open the answer or worked solution.

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Exercise 9.1

Textbook page 140 · 4 questions
Q1Written answer

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 – p (v) (x/2) + (y/2) – xy (vi) 0.3a – 0.6ab + 0.5b

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Worked solution
Sl. No. Expression Term Coefficient
i) 5xyz2 – 3zy Term: 5xyz2

Term: -3zy

 5 -3
ii) 1 + x + x2 Term: 1
Term: x
Term: x2		 1 1 1
iii) 4x2y2 – 4x2y2z2 + z2 Term: 4x2y2
Term: -4 x2y2z2
Term :  z2		 4 -4 1
iv) 3 – pq + qr – p Term : 3 -pq qr -p 3 -1 1 -1
v) (x/2) + (y/2) – xy Term : x/2 Y/2 -xy ½ 1/2 -1
vi) 0.3a – 0.6ab + 0.5b Term : 0.3a -0.6ab 0.5b 0.3 -0.6 0.5
Q2Written answer

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 + x4 , 7 + y + 5x, 2y – 3y2 , 2y – 3y2 + 4y3 , 5x – 4y + 3xy, 4z – 15z2 , ab + bc + cd + da, pqr, p2q + pq2 , 2p + 2q

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Worked solution

Let us first define the classifications of these 3 polynomials:

Monomials contain only one term.

Binomials contain only two terms.

Trinomials contain only three terms.

x + y two terms Binomial
1000 one term Monomial
x + x2 + x3 + x4 four terms Polynomial, and it does not fit in the listed three categories
2y – 3y2 two terms Binomial
2y – 3y2 + 4y3 three terms Trinomial
5x – 4y + 3xy three terms Trinomial
4z – 15z2 two terms Binomial
ab + bc + cd + da four terms Polynomial, and it does not fit in the listed three categories
pqr one term Monomial
p2q + pq2 two terms Binomial
2p + 2q two terms Binomial
7 + y + 5x three terms Trinomial
Q3Written answer

Add the following.

(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

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Worked solution

i) (ab – bc) + (bc – ca) + (ca-ab)

= 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

= a – a +b – b +c – c + ab + bc + ca

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

= ab + bc + ca

iii) 2p2q2 – 3pq + 4, 5 + 7pq – 3p2q2

= (2p2q2 – 3pq + 4) + (5 + 7pq – 3p2q2)

= 2p2q2 – 3p2q2 – 3pq + 7pq + 4 + 5

= – p2q2 + 4pq + 9

iv)(l2 + m2) + (m2 + n2) + (n2 + l2) + (2lm + 2mn + 2nl)

= l2 + l2 + m2 + m2 + n2 + n2 + 2lm + 2mn + 2nl

= 2l2 + 2m2 + 2n2 + 2lm + 2mn + 2nl

Q4Written answer

(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 – 11q + 5pq – 2pq2 + 5p2q

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Worked solution

(a) (12a – 9ab + 5b – 3) – (4a – 7ab + 3b + 12)

= 12a – 9ab + 5b – 3 – 4a + 7ab – 3b – 12

= 12a – 4a -9ab + 7ab +5b – 3b -3 -12

= 8a – 2ab + 2b – 15

b) (5xy – 2yz – 2zx + 10xyz) – (3xy + 5yz – 7zx)

= 5xy – 2yz – 2zx + 10xyz – 3xy – 5yz + 7zx

=5xy – 3xy – 2yz – 5yz – 2zx + 7zx + 10xyz

= 2xy – 7yz + 5zx + 10xyz

c) (18 – 3p – 11q + 5pq – 2pq2 + 5p2q) – (4p2q – 3pq + 5pq2 – 8p + 7q – 10)

= 18 – 3p – 11q + 5pq – 2pq2 + 5p2q – 4p2q + 3pq – 5pq2 + 8p – 7q + 10

=18+10 -3p+8p -11q – 7q + 5 pq+ 3pq- 2pq^2 – 5pq^2 + 5 p^2 q – 4p^2 q

= 28 + 5p – 18q + 8pq – 7pq2 + p2q