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- Algebraic Expressions : Solution of NCERT Exercise 9.2 and 9.3

**Exercise 2**

**1. Find the product of the following pairs of monomials.**

**(i) 4, 7p**

**Answer:** *4* x *7 p* = *28p*

**(ii) – 4p, 7p**

**Answer:** *- 4p* x *7p* = *-28p*^{2}

**(iii) – 4p, 7pq**

**Answer**: *- 4p* x *7pq* = *-28p ^{2}*q

**(iv) 4p3, – 3p**

**Answer: ***4p ^{3}q* x

**(v) 4p, 0**

**Answer:** *4p* x *0* = *0*

**2. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.**

**(p, q); (10m, 5n); (20x ^{2} 5y^{2}; (4x, 3x^{2}); (3mn, 4np)**

**Answer:** Area = Length x breadth

(i) *p* x *q* = *pq*

(ii)*10m* x *5n* = *50mn*

(iii) *20x*^{2} x *5y*^{2} = *100x ^{2}*y

(iv) *4x* x *3x*^{2} = *12x*^{3}

(v)*3mn* x *4np* = *12mn ^{2}*p

**3. Complete the following table of products:**

**4. Obtain the volume of rectangular boxes with the following length, breadth and height respectively.**

**(i) 5a, 3a^{2}, 7a^{8} (ii) 2p, 4q, 8r (iii) xy, 2x^{2}y, 2xy^{2}(iv) a, 2b, 3c**

**Answer:** Volume = length x breadth x height

(i) *5a* x *3a*^{2} x *7a*^{8} = *105a ^{11}*

(ii) *2p* x *4q* x *8r* = *64pqr*

(iii) *xy* x *2x ^{2}*y x

(iv) a x 2b x 3c = 6abc

**5. Obtain the product of**

**(i) xy, yz, zx (ii) a, – a ^{2} a^{3}(iii) 2, 4y, 8y^{2} 16y^{3}(iv) a, 2b, 3c, 6abc (v) m, – mn, mnp**

**Answer:** (i) *x ^{2}*y

*(ii) –a ^{5}*

*(iii) 1024y ^{6}*

*(iv) 36a ^{2}*b

*(v) –m ^{3}*n

**Exercise 3**

**1. Carry out the multiplication of the expressions in each of the following pairs.**

**(i) 4p, q + r**

**Answer:** *4p(q + r) = 4pq + 4pr*

**(ii) ab, a – b**

**Answer:** *ab(a - b) = a ^{2}b - ab*

**(iii) a + b, 7aÂ²bÂ²**

**Answer:** *(a + b) (7a ^{2}*b

**(iv) a ^{2}– 9, 4a**

**Answer:** *(a ^{2} - 9)(4a) = 4a^{3} - 36a*

**(v) pq + qr + rp, 0**

**Answer:** *(pq + qr + rp)* x *0 = 0*

**2. Find the product.**

**(i) a^{2} x (**

**Answer:** As you know

*a*^{m} x *a*^{n} x *a*^{o} = **a ^{m+n+o}**

So, we get

*a*^{2} x (*2a ^{22}*) x (

(iv) * x* x

= **x ^{14}**

3. (a) Simplify 3x (4x – 5) + 3 and find its values for (i) x = 3 (ii) x =1/2

(i) putting x=3 in the equation we get

12x^{2} - 15x + 3

=108-45+3 = 66

(ii) putting x=1/2 in the equation we get

**(b) Simplify ****a (a ^{2}+ a + 1) + 5**

**Answer:** *a(a ^{2}a+1)*

*=a ^{3}+a^{2}+a*

(i) putting a= 0 in the equation we get

*0 ^{3}+0^{2}+0=0*

(ii) putting a=1 in the equation we get

*1 ^{3 }+ 1^{2 }+ 1 = 1 + 1 + 1 = 3*

(iii) putting a = -1 in the equation we get

*-1 ^{3}+1^{2 }-1 = -1 + 1 + 1 = 1*

**5. (a) Add: p ( p – q), q ( q – r) and r ( r – p)**

**Answer:** *(p ^{2} - pq) + (q^{2} - qr) + (r^{2} - pr)*

*= p ^{2} + q^{2} + r^{2} - pq - qr - pr*

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

**Answer:** *(2xz - 2x ^{2} - 2xy) + (2yz - 2y^{2} - 2xy)*

*= 2xz - 4xy + 2yz - 2x ^{2} - 2y^{2} *

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

**Answer:** *(40ln - 12lm + 8l ^{2}) - (3l^{2} - 12lm + 15ln)*

*= 40ln - 12lm + 8l ^{2} - 3l^{2} - 12lm + 15ln*

*= 55ln - 24lm + 5l*^{2}

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

*= (-4ac + 4bc + 4c ^{2}) - (3a^{2} + 3ab + 3ac)*

*= -4ac + 4bc + 4c ^{2} - 3a^{2} - 3ab - 3ac*

*= -7ac + 4bc + 4c ^{2} - 3a^{2} - 3ab*

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