NCERT Solution for Class 8 Chapter – 9 Algebraic expression Exercise – 9.1
Q1. Identify the terms , their coefficient for each of the following expressions.
- 5xyz2 – 3zy
- 1 + x + x2
- 4x2y2 – 4x2y2z2 + z2
- 3 – pq + qr – p
- (x/2) + (y/2) – xy
- 3 a – 0.6ab + 0.5b
Solution:-
| SI . No | Expression | Term | Coefficient |
| 5xyz2 – 3zy | Term: 5xyz2
Term: -3zy |
5 – 3 |
| 1 + x + x2
4x2y2 – 4x2y2z2 + z2 |
Term: 1
Term: x Term: x2 Term: 4x2y2 Term: -4x2 y2z2 Term: z2 |
111
|
|
| 3 – pq + qr – p | Term: 3 – pq qr – p | 4-4 1 | |
| (x/2) + (y/2) – xy | Term: x/2 y/2 -xy | 3 – 1 1 – 1 | |
| 0.3a – 0.6ab + 0.5b | Term: 0.3a -0.6ab 0.5b | 1/2 1/2 – 1 | |
| 0.3 – 0.6 0.5 |
Q2. 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.
Bionomials, Contain only two terms.
Trinomials, Contain only three terms.
| x + y | Two term | Binomial |
| 1000 | One term | Monomial |
| X + x2 + x3 + x4 | Four term | Polynomial , and it does not fit in listed three categories |
| 2y – 3y2 | Two term | Binomial |
| 2y – 3y2 + 4y3 | Three term | Trinomial |
| 5x – 4y + 3xy | Three term | Trinomial |
| 4z – 15z2 | Two term | Binomial |
| Ab + bc + cd + da | Four term | Polynomial , and it does not fit in listed in three categories |
| pqr | One term | Monomial |
| P2q +pq2 | Two term | Binomial |
| 2p + 2q | Two term | Binomial |
| 7 + y + 5x | Three term | Trinomial
|
Q3. Add the following.
- Ab – bc, bc – ca, ca – ab
Solution:-
(ab – bc) + (bc – ca) + (ca – ab)
= 0
- A-b + ab , b – c + bc, c – a + ac
Solution:-
(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
- 2p2q2 – 3pq + 4 , 5 + 7pq – 3p2q2
Solution:-
2p2q2 – 3p2q2 – 3pq + 7pq + 4 + 5
-p2q2 + 4pq + 9
- (l2 + m2) + (m2 + n2) + (n2 + l2) + (2lm + 2mn + 2nl)
Solution:-
= 2l2 + 2m2 + 2n2 + 2lm + 2mn + 2nl
Q4. Subtract
- 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3
Solution:-
12a – 9ab + 5b – 3 – 4a + 7ab – 3b – 12
= 8a – 2ab + 2b – 15
- (5xy – 2yz – 2zx + 10xyz) – ( 3xy + 5yz – 7zx)
Solution:-
5xy – 2yz – 2zx + 10xyz – 3xy – 5yz + 7zx
= 2xy – 7yz + 5zx + 10xyz
( c ) (18 – 3p – 11p + 5pq – 2pq2 + 5p2q) – (4p2q – 3pq + 5pq2 – 8p + 7q – 10)
Solution:-
18 – 3p – 11p + 5pq – 2pq2 + 5p2q – 4p2q + 3pq – 5pq2 + 8p – 7q + 10
= 28 + 5p – 18q + 8pq – 7pq2 + p2q
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