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 a^{2 + 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)

 

• = x^{1 + 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

12x²    – 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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