Class 8 | NCERT Solution Maths Chapter 9 | Algebraic Expressions and Identities | Exercise 9.1
Exercise 9.1 Page No: 140
Q1. 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
Solution :
| Sl. No. | Expression | Term | Coefficient |
| i) | 5xyz 2 – 3zy | Term: 5xyz2 Term: -3zy | 5 -3 |
| ii) | 1 + x + x2 | Term: 1 Term: x Term: x2 | 1 1 1 |
| iii) | 4x 2 y 2 – 4x 2 y 2 z 2 + z 2 | 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 |
| we) | 0.3a – 0.6ab + 0.5b | Term : 0.3a -0.6ab 0.5b | 0.3 -0.6 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 + x4 , 7 + y + 5x, 2y – 3y2 , 2y – 3y2 + 4y3 , 5x – 4y + 3xy, 4z – 15z2 , ab + bc + cd + da, pqr, p2q + pq2 , 2p + 2q
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 listed three categories |
| 2y – 3y 2 | two terms | Binomial |
| 2y – 3y 2 + 4y 3 | 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 listed three categories |
| pqr | one term | Monomial |
| p 2 q + pq 2 | two terms | Binomial |
| 2p + 2q | two terms | Binomial |
| 7 + and + 5x | three terms | Trinomial |
3. 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) l 2 + m 2 , m 2 + n 2 , n 2 + l 2 , 2lm + 2mn + 2nl
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
= (2p 2 q 2 – 3pq + 4) + (5 + 7pq – 3p 2 q 2 )
= 2p 2 q 2 – 3p 2 q 2 – 3pq + 7pq + 4 + 5
= – p2q2 + 4pq + 9
iv) (l 2 + m 2 ) + (m 2 + n 2 ) + (n 2 + l 2 ) + (2lm + 2mn + 2nl)
= l 2 + l 2 + m 2 + m 2 + n 2 + n 2 + 2lm + 2mn + 2nl
= 2l2 + 2m2 + 2n2 + 2lm + 2mn + 2nl
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 – 11q + 5pq – 2pq2 + 5p2q
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
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