**NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.3 – Algebraic Expressions and Identities**

**NCERT Solutions for Class 8 Maths Chapter 9 Exercise 9.3 – Algebraic Expressions & Identities, has been designed by the NCERT to test the knowledge of the student on the topic – Multiplying a Monomial by a Polynomial**

- Multiplying a monomial by a binomial
- Multiplying a monomial by a trinomial

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

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

(ii) **ab****, ****a**** – ****b**

(iii) **a**** + ****b****, 7****a**^{2}**b**^{2}

(iv) **a**^{2}** − 9, 4****a**

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

**Solution:**

**(i)** (4p)×(q+r)=(4p×q)+(4p×r)=4pq+4pr

**(ii)** (ab)×(a-b)=(ab×a)+[ab×(- b)]=a²b-ab²

**(iii)** (a+b)×(7a²b²)=(a×7a²b²)+(b×7a²b²)=7a³b²+7a²b³

**(iv)** (a²-9)×(4a)=(a²×4a)+(- 9)×(4a)=4a³-36a

**(v)** (pq+qr+rp)×0=(pq × 0)+(qr×0)+(rp×0)=0

**Q.2 Complete the table**

**Solution:**

**Q.3 ****Find the product. **

**(i)**

**(ii)**

**(iii)**

**(iv)**

**Solution:**

**(i)**

**(ii) **

**(iii) **

**(iv) **

**Q.4**

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

**(b) Simplify a**** (****a**^{2}** + ****a**** + 1) + 5 and find its values for (i) ****a**** = 0, (ii) ****a**** = 1, (iii) ****a**** = − 1 **

**Solution:**

**(a)** 3x (4x − 5) + 3 = 12x^{2} − 15x + 3

(i) For x = 3,

12x^{2} − 15x + 3 = 12 (3)^{2} − 15(3) + 3

= 108 − 45 + 3

= 66

(ii) For x = ½

12x^{2} − 15x + 3 = 12 (1/2)^{2} − 15(1/2) + 3

= 3 – 15/2 + 3

= 6 – 15/2

= – 3/2

**(b)** a (a^{2} + a + 1) + 5 = a^{3} + a^{2} + a + 5

(i) For a = 0,

a^{3} + a^{2} + a + 5 = 0 + 0 + 0 + 5

= 5

(ii) For a = 1,

a^{3} + a^{2} + a + 5 = (1)^{3} + (1)^{2} + 1 + 5

= 1 + 1 + 1 + 5 = 8

(iii) For a = −1,

a^{3} + a^{2} + a + 5 = (−1)^{3} + (−1)^{2} + (−1) + 5

= − 1 + 1 − 1 + 5 = 4

** Q.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)

**Solution:**

**(a)** First expression = p (p − q) = p^{2} – pq

Second expression = q (q − r) = q^{2} – qr

Third expression = r (r − p) = r^{2} − pr

Adding three expressions, we get

Therefore, sum of the three given expressions is p^{2} + q^{2} + r^{2} − pq − qr – rp

**(b)** First expression = 2x (z − x − y) = 2xz − 2x^{2} − 2xy

Second expression = 2y (z − y − x) = 2yz − 2y^{2} − 2yx

Adding two expressions, we get

Therefore, sum of the two given expressions is − 2x^{2} − 2y^{2} − 4xy + 2yz + 2zx.

**(c)** First expression = 3l (l − 4m + 5n) = 3l^{2} − 12lm + 15ln

Second expression = 4l (10n − 3m + 2l) = 40ln − 12lm + 8l^{2
}Subtracting first expression from second, we get

Therefore, Subtracting 3l(l-4m+5n) from 4l(10n-3m+2l)

we get 25ln+5l²

**(d) **First expression = 3a (a + b + c) − 2b (a − b + c) = 3a^{2} +3ab + 3ac − 2ba + 2b^{2} − 2bc

= 3a^{2} + 2b^{2 }+ ab + 3ac − 2bc

Second expression = 4c (− a + b + c) = − 4ac + 4bc + 4c^{2
}Subtracting first expression from second, we get

Therefore, Subtract: 3a(a+b+c)-2b(a-b+c) from 4c(- a+b+c)

we get, -7ac+6bc+4c²-3a²-2b²-ab

**The next Exercise for** **NCERT Solutions for Class 8 Maths Chapter 9 Exercise 9.4 – Algebraic Expressions and Identities can be accessed by clicking here**

** Maths – NCERT Solutions Class 8**

**NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.3 – Algebraic Expressions and Identities**

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