NCERT Solutions For Class 8 Maths Chapter 9 Exercise -9.3

NCERT Solution For Class 8 Math Chapter – 9 Algebraic Expression Exercise – 9.3

Q1. Carry out the multiplication of the expression in each of the following pairs.

  • 4p , q + r
  • Ab , a – b
  • A + b , 7a2b2
  • A2 – 9, 4a
  • Pq + qr + rp , 0

Solution:-

 

  • 4p (q + r) = 4pq + 4pr
  • ab(a – b) = a2b – ab2
  • (a + b)(7a2b2) = 7a3b2 + 7a2b3
  • (a2 – 9)(4a) = 4a3 – 36a
  • (pq + qr + rp) x 0 = 0 (Anything multiplied by zero is zero)

 

Q2. Complete the table.

 

 

 

 

Solution:-

 

First expression Second Expression Product
(i) a B +
c +d
A( b + c + d)

= a x b + a x c + a x d

= ab + ac + ad

(ii) X + y – 5 5xy 5xy ( x + y -5)

= 5xy x x + 5xy x y – 5xy x 5

= 5x2y + 5xy2 – 25xy

(iii) p 6p2 – 7p + 5 P(6p2 -7p + 5)

= 6p3 – 7p2 + 5p

(iv) 4p2q2 P2 – q2 4p4q2 – 4p2 q4
(v) A + b + c abc A2bc + ab2c + abc2

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Q3. Find the product.

  • A2 x (2a22) x (4a26)
  • (2/3 xy) x (-9/10 x2y2)
  • (-10/3 pq3/) x (6/5p3q)
  • (x) x (x)2 x (x)3 x (x)4

Solution:-

 

  • A2 x (2a22) x (4a26) = ( 2 x 4) (a2 x a22 x a26) = 8 x a2 + 22 + 26 = 8a50
  • (2/3 x -9/10) ( x x x2 x y x y2)

= (-3/5 x3y3)

 

  • = (-10/3 x 6/5) (p x p3 x q3 x q )

= (-4p4q4)

 

  • = x1 + 2 + 3 + 4

x10

 

Q4. (A) Simplify 3x (4x – 5) + 3 and find its value 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

Solution:-

 

  • 3x(4x – 5) + 3

= 3x (4x) – 3x(5) + 3

= 12x2 – 154x + 3

 

  • Putting x = 3 in the equation we gets

2x2 – 15x + 3 = 12(32) – 15(3) + 3

= 108 – 45 + 6 = 66

 

  • Putting x = 1/2 in the equation we get

12x  – 15x + 3 = 12(1/2)2 – 15(1/2) + 3

= -3/2

 

  • A(a2 + a + 1) + 5

= a3 + a2 + a + 5

 

  • Putting a = 0 in the equation we get 03 + 02 + 0 + 5 = 5

 

  • Putting a = 1 in the equation we get

13 + 12 + 1 + 5 = 1 + 1 + 1 + 5 = 8

 

  • Putting a = -1 in the equation we get

(-1)3 + (-1)2 + (-1) + 5 = -1 + 1 – 1 + 5 = 4

 

Q5.  (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:-

 

 

  1. P(p – q) + q(q-r) + r(r-p)

= (p2 – pq) + (q2 – qr) + (r2 –  pr)

= p2 + q2 + r2 – pq – qr – pr

 

b)

=  2xz – 4xy + 2yz – 2x2 – 2y2

 

c)

= (40ln – 121lm + 8l2) – (3l2 – 121lm + 15ln)

= 25ln + 5l2

 

D)

= (-4ac + 4bc + 4c2) – (3a2 + 3ab + 3ac – (2ab – 2b2 + 2bc ))

= -7ac + 6bc + 4c2 – 3a2 – ab – 2b2

 

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NCERT Solution For Class 8 Math Chapter – 9 Algebraic Expression Exercise – 9.3 Q1. Carry out the multiplication of the…