Class 8 Maths Exercise 9.3 Solutions : Algebraic Expressions and Identities

In this page we have Class 8 Maths Exercise 9.3 Solutions for
Algebraic Expressions and Identities. This exercise has questions about multiplication of polynomials like mononial to binomial and mononial to Trinomial. Hope you like them and do not forget to like , social share
and comment at the end of the page.

Class 8 Maths Exercise 9.3 Solutions

Question 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) a^{2}- 9, 4a
(v) pq + qr + rp, 0 Answer:

4p(q + r) = 4pq + 4pr

ab(a - b) = a^{2}b - ab^{2}

(a + b) (7a^{2}b^{2}) = 7a^{3}b^{2} + 7a^{2}b^{3}

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

(pq + qr + rp) x 0 = 0

Question 2
Complete the table.
Answer:
(i) a(b + c + d) = ab + ac + ad

(ii) (x + y – 5) 5xy= 5x^{2}y + 5xy^{2} -25xy
(iii)p(6p^{2}– 7p + 5)=6p^{3} -7p^{2} +5p
(iv) 4p^{2}q^{2}(p^{2} - q^{2})=4p^{4}q^{2} -4p^{2}q^{4}
(v) (a + b + c)abc= a^{2}bc + ab^{2}c + abc^{2}

Question 2
Find the product.
(i) a^{2} x (2a^{22}) x (4a^{26})
(ii) (2xy/3) ×(-9x^{2}y^{2}/10)
(iii) (-10pq^{3}/3) ×(6p^{3}q/5)
(iv) x (x^{2}) x (x^{3}) x (x^{8}) Answer:We will use the below property extensively in above questions a^{m} x a^{n} x a^{o} = a^{m+n+o}
(i) As you know
So, we get a^{2} x (2a^{22}) x (4a^{26}) = 8a^{48}
(ii) (2xy/3) ×(-9x^{2}y^{2}/10) =(-3x^{3}y^{3}/5)
(iii) (-10pq^{3}/3) ×(6p^{3}q/5) =(-4p^{4}q^{4})
(iv)( x) x (x^{2}) x (x^{3}) x (x^{8})
= x^{14}

Question 3
(a) Simplify 3x (4x - 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 value for (i) a = 0, (ii) a = 1 (iii) a = - 1. Answer
(a)
(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
$12 \times \frac {1}{2} - \frac {15}{2} + 3 =3 - \frac {15}{2} -3=\frac {-3}{2}$
(b)
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

Question 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 - 4 m + 5 n) from 4l ( 10 n - 3 m + 2 l )
(d) Subtract: 3a (a + b + c ) - 2 b (a - b + c) from 4c ( - a + b + c ) Answer:
(a) (p^{2} - pq) + (q^{2} - qr) + (r^{2} - pr)
= p^{2} + q^{2} + r^{2} - pq - qr - pr
(b) (2xz - 2x^{2} - 2xy) + (2yz - 2y^{2} - 2xy)
= 2xz - 4xy + 2yz - 2x^{2} - 2y^{2}
(c) (40ln - 12lm + 8l^{2}) - (3l^{2} - 12lm + 15ln)
= 40ln - 12lm + 8l^{2} - 3l^{2} - 12lm + 15ln
= 55ln - 24lm + 5l^{2}
(d) = (-4ac + 4bc + 4c^{2}) - (3a^{2} + 3ab + 3ac)
= -4ac + 4bc + 4c^{2} - 3a^{2} - 3ab - 3ac
= -7ac + 4bc + 4c^{2} - 3a^{2} - 3ab

Summary

Class 8 Maths Exercise 9.3 Solutions has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail. You can download the solutions as PDF in the below Link also Download Class 8 Maths Exercise 9.3 as pdf

This chapter 9 has total 5 Exercise 9.1 ,9.2 ,9.3 ,9.4 and 9.5. This is the Third exercise in the chapter.You can explore previous exercise of this chapter by clicking the link below