Algebraic Expressions and Identities| NCERT Solutions| Class 8

NCERT Solutions of  Chapter: Algebraic Expressions and Identities. NCERT Solutions along with worksheets and notes for Class 8.

(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

Ans-

Polynomial do not fit in any given category Monomials Binomials Trinomials
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

Algebraic expressions and identities,A 3(iii) (iv),exercise 9.1,NCERT,Class 8

(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 + 5 pq2 – 8p + 7q -10 from 18 – 3p – 11q + 5pq – 2 pq2 + 5p2q

Ans-

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

12a – 9ab + 5b – 3

4a – 7ab + 3b + 12

(-)   (+)     (-)     (-)

8a – 2ab +2b – 15

Hence, subtraction of the given expressions is 8a – 2ab +2b – 15

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

5xy – 2yz – 2zx + 10xyz

3xy + 5yz – 7zx

(-)   (-)       (+)      (+)

2xy -7yz + 5zx +10xyz

Hence, subtraction of the given expressions is 2xy -7yz + 5zx +10xyz

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

18 – 3p – 11q + 5pq – 2pq2 + 5p2q

-10 – 8p +7q – 3pq + 5pq2 + 4p2q

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

28 + 5p – 18q + 8pq -7pq2 + p2q

Hence, subtraction of the given expressions is 28 + 5p – 18q + 8pq -7pq2 + p2q

 

Exercise 9.2

(1) Find the product of the following pairs of monomials.

(i) 4, 7p   (ii) – 4p, 7p  (iii) – 4p, 7pq  (iv) 4p3, -3p  (v) 4p, 0

Ans-

(i) 4, 7p

4 × 7p

=28p

Hence, the product of given expressions is 28p

(ii) – 4p, 7p

-4p × 7p

= – 28p2

Hence, the product of given expressions is – 28p2

(iii) – 4p , 7pq

-4p × 7pq

= -4 ×7 ×p × p × q

= -28p2q

Hence, the product of given expressions is -28p2q

(iv) 4p3, -3p

4p3 × -3p

= 4 × -3 × p3 × p

=-12p4

Hence, the product of given expressions is -12p4

(3) Complete the table of products.

First Monomial

_______________

Second Monomial

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

Ans-

First Monomial

____________

Second Monomial

2x -5y 3x2 -4xy 7x2y -9x2y2
2x 4x2 -10xy 6x3 -8x2y 14x3y -18x3y2
-5y -10xy 25y2 -15x2y 20xy2 -35x2y2 45x2y3
3x2 6x3 -15x2y 9x4 -12x3y 21x4y -27x4y2
-4xy -8x2y 20xy2 -12x3y 16x2y2 -28x3y2 36x3y3
7x2y 14x3y -35x2y2 21x4y -28x3y2 49x4y2 -63x4y3
-9x2y2 -18x3y2 45x2y3 -27x4y2 36x3y3 -63x4y3 81x4y4

Exercise 9.3

(1) Carry out the multiplication of the expressions in each of the following pairs:-

(i) 4p, q + r   (ii) ab, a-b   (iii) a + b, 7a2b2   (iv) a2 – 9, 4a  (v) pq + qr + rp, 0

Ans-

(i) 4p, q + r

4p × (q + r)

= (4p× q) + (4p× r) [Distributive Law]

= 4pq + 4pr

Hence, multiplication of given expressions is 4pq + 4pr

(ii) ab, a-b

ab × (a – b)

= (ab × a) – (ab × b)[Distributive Law]

= a2b – ab2

Hence, multiplication of given expressions is a2b – ab2

(iii) a + b, 7a2b2

(a + b) × 7a2b2

= (a × 7a2b2) + (b × 7a2b2) [Distributive Law]

= 7a3b2 + 7a2b3

Hence, multiplication of given expressions is 7a3b2 + 7a2b3

(iv) a2 – 9, 4a

(a2 – 9) × 4a

= (a2 × 4a) – (9 × 4a) [Distributive Law]

= 4a3 – 36a

Hence, multiplication of given expressions is 4a3 – 36a

(v) pq + qr + rp, 0

(pq + qr + rp)  × 0

= 0

Hence, multiplication of given expressions is 0.

 

(2) Complete the table.

First Expression Second Expression Product
(i) a b + c +d
(ii) x + y -5 5xy
(iii) p 6p2 – 7p +5
(iv) 4p2q2 P2 – q2
(v) a + b + c abc

Ans-

First Expression Second Expression Product
(i) a b + c +d a × (b + c + d)

= ab + ac + ad

(ii) x + y -5 5xy (x + y -5) × 5xy

=5x2y + 5xy2 – 25xy

(iii) p 6p2 – 7p +5 p × (6p2– 7p +5)

=6p3 – 7p2 + 5p

(iv) 4p2q2 P2 – q2 4p2q2 × (P2 – q2)

=4p4q2 – 4p2q4

(v) a + b + c abc (a + b + c) × abc

=a2bc + ab2c + abc2

(b) Simplify  a ( a2 + a + 1) +5 and find its value for (i) a = 0, (ii) a = 1 (iii) a = -1

(i) If a = 0

a ( a2 + a + 1) +5

= (a× a2 + a× a + a) + 5

= a3 + a2 +a + 5

Put a = 0

= 0 + 0 + 0 +5

=5

Hence, value of given expression if a = 0 is 5

(ii) If a = 1

a (a2 + a + 1) +5

= (a× a2 + a× a + a) + 5

= a3 + a2 +a + 5

Put a = 1

=(1)3 + (1)2 +1 + 5

= 1 + 1 +1 +5

=8

Hence, value of given expression if a = 1 is 8

(iii) If a = -1

a (a2 + a + 1) +5

= (a× a2 + a× a + a) + 5

= a3 + a2 +a + 5

Put a = -1

=(-1)3 + (-1)2 -1 + 5

= -1 + 1 -1 +5

=4

Hence, value of given expression if a = -1 is 4

 

5(a) Add p (p – q) , q ( q – r) and r (r – p)

(b) Add 2x (z – x – y) and 2y (z – y – x)

(c) Subtract: 3l (l – 4m + 5n) from 4l (10n – 3m + 2l)

(d) Subtract: 3a(a + b +c) – 2b(a – b + c) from 4c (- a + b + c)

(ii) (y – 8) and (3y – 4)

(y – 8) × (3y – 4)

= y (3y – 4) – 8 ( 3y – 4)

= y × 3y – 4y – 8 × 3y + 8 × 4

= 3y2 – 4y – 24y + 32

= 3y2– 28y + 32

Hence, the product of given expressions is 3y2– 28y + 32

(iii) (2.5l – 0.5m) and (2.5l + 0.5m)

(2.5l – 0.5m) × (2.5l + 0.5m)

= 2.5l ×(2.5l + 0.5m)- 0.5m (2.5l + 0.5m)

=2.5l × 2.5l + 2.5l × 0.5m – 0.5m × 2.5l – 0.5m × 0.5m

=6.25l2 + 1.25lm -1.25ml – 0.25m2

= 6.25l2– 0.25m2

Hence, the product of given expressions is 6.25l2– 0.25m2

(iv) (a + 3b) and (x + 5)

(a + 3b) × (x + 5)

=a(x + 5) + 3b (x +5)

=ax + 5a + 3bx + 15b

=ax +5a + 3bx + 15b

Hence, the product of given expressions is ax +5a + 3bx + 15b

(v) (2pq + 3q2) and (3pq – 2q2)

(2pq + 3q2) × (3pq – 2q2)

=2pq(3pq – 2q2) + 3q2(3pq – 2q2)

=2pq × 3pq – 2pq × 2q2 + 3q2 × 3pq – 3q2× 2q2

=6p2q2 – 4 pq3+ 9 pq3– 6q4

=6p2q2 + 5 pq3 – 6q4

Hence, the product of given expressions is 6p2q2 + 5 pq3 – 6q4

 

Helping Topics

Algebraic Expressions and Identities

Worksheet Class 8

 

 

 

 

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