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Ncert Solutions for Algebraic Expressions and Identities Class 8 CBSE Part 3



In this page we have NCERT book Solutions for Class 8th Maths:Algebraic Expressions and Identities for EXERCISE 3 . Hope you like them and do not forget to like , social share and comment at the end of the page.
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) a2– 9, 4a
 (v) pq + qr + rp, 0
Answer:
  1. 4p(q + r) =  4pq + 4pr
  2. ab(a - b) = a2b - ab2
  3. (a + b) (7a2b2) = 7a3b2 + 7a2b3
  4. (a2 - 9)(4a) = 4a3 - 36a2
  5. (pq + qr + rp) x 0 = 0
Question 2
 Find the product.
i) a2 x (2a22) x (4a26)
ii) (2xy/3) ×(-9x2y2/10)
      (iii)    (-10pq3/3) ×(6p3q/5)
      (iv)    ( x) x (x2) x (x3) x (x8)
Answer: We will use the below property extensively in above questions
am x an x ao = am+n+o
i) As you know
           So, we get
       a2 x (2a22) x (4a26) = 8a48
ii) (2xy/3) ×(-9x2y2/10)
=(-3x3y3/5)
iii) (-10pq3/3) ×(6p3q/5)
=(-4p4q4)
iv)  ( x) x (x2) x (x3) x (x8)
x14
Question 3
(a) Simplify 3x (4x – 5) + 3 and find its values for (i) x = 3 (ii) x =1/2
(b) Simplify a (a2+ 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
12x2 - 15x + 3 =108-45+3 = 66
(ii) putting x=1/2 in the equation we get

b)
 a(a2a+1)
=a3+a2+a
(i) putting a= 0 in the equation we get
03+02+0=0
(ii) putting a=1 in the equation we get
1+ 1+ 1 = 1 + 1 + 1 = 3
(iii) Putting a = -1 in the equation we get
-13+1-1 = -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: i) (p2 - pq) + (q2 - qr) + (r2 - pr)
= p2 + q2 + r2 - pq - qr - pr
ii) (2xz - 2x2 - 2xy) + (2yz - 2y2 - 2xy)
= 2xz - 4xy + 2yz - 2x2 - 2y2
iii) (40ln - 12lm + 8l2) - (3l2 - 12lm + 15ln)
= 40ln - 12lm + 8l2 - 3l2 - 12lm + 15ln
= 55ln - 24lm + 5l2
iv) = (-4ac + 4bc + 4c2) - (3a2 + 3ab + 3ac)
= -4ac + 4bc + 4c2 - 3a2 - 3ab - 3ac
= -7ac + 4bc + 4c2 - 3a2 - 3ab

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