NCERT Solutions For Class 8 Maths Chapter – 14 Factorisation Exercise – 14.3

Q1. Carry out the following divisions.

• 28x⁴/ 56x
• -36y³/ 9y²
• 66pq²r³/  11qr²
• 34x³y³z³/ 51xy²z³
• 12a⁸b⁸/ (-6a⁶b⁴)

Solution:-

 

• 28x⁴ = 2 x 2 x 7 x x x x x x x x

56x = 2 x 2 x 2 x 7 x x

 

28x⁴ / 56x = 2 x 2 x 7 x x x x  x x x x / 2 x 2 x 2 x 7 x x

= 1/2 x³

 

• -36y³ / 9y² = -2 x 2 x 3 x 3 x y x y x y/ 3 x 3 x y x y

= -4y

 

• 66pq²r³/ 11qr² = 2 x 3 x 11 x p x q x q x r x r x r / 11 x q x r x r = 6pqr

 

• 34x³y⁴z³/ 51xy²z³ = 2 x 17  x x x x x x x y x y x y x z x z x z/3 x 17  x x x y x y x z  x z x z

= 2/3 x²y

 

• 12a⁸b⁸/ (-6a⁶b⁴) = 2 x 2 x 3 x a⁸  x b⁸ / -2 x 3 x a⁶ x b⁴

= -2a²b⁴

 

Q2. Divide the given polynomial by the given monomial.

• (5x²– 6x) / 3x
• (3y⁸– 4y⁶ + 5y⁴) / y⁴
• 8(x³y²z²+ x²y³z² +x²y²z³)/  4x²y²z²
• (x³+ 2x² + 3x) / 2x
• (p³q⁶– p⁶q³) / p³q³

Solution:-

 

• 5x²– 6x = x(5x – 6)

(5x² – 6x) / 3x = x(5x – 6)/3x = 1/3  ( 5x – 6)

 

• 3y⁸– 4y⁴ + 5y⁴  = y⁴ (3y⁴ – 4y² + 5)

(3y⁸ – 4y⁶ + 5y⁴) + y⁴  = y⁴ (3y⁴ – 4y² + 5) / y⁴

= 3y⁴ – 4y² + 5

 

• 8( x³y²z²+ x²y³z² + x²y²z³) = 8x²y²z² ( x + y + z)

= 8(x³y²z² + x²y³z² + x²y²z³) / 4x²y²z³

= 2(x + y + z)

 

 

• x³+ 2x² + 3x = x(x² + 2x + 3)

(x³+ 2x² + 3x) / 2x = x(x³ + 2x² + 3)/2x = 1/2 (x²  + 2x + 3)

 

• P³q⁶– p⁶q³ = p³q³(q³ – p³)

(p³ q⁶ – p⁶ q³) / p³q³ = p³q³(q³– p³)/p³q³ = q³ – p³

 

 

Q3. Work out the following divisions.

• (10x – 25) / 5
• (10x – 25) / (2x – 5)
• 10y (6y + 21) /  5(2y + 7)
• 9x²y² (3z – 24) / 27xy(z – 8)
• 96abc (3a – 12) (5b – 30) /  144(a – 4) (b – 6)

Solution:-

 

• (10x – 25) / 5 = 5(2x – 5(/5 = 2x – 5
• (10x – 25) / (2x – 5) = 5(2x – 5)/ (2x – 5) = 5
• 10y(6y + 21) / 5(2y + 7) = 10y x 3(2y + 7)/ (2y + 7) = 6y
• 9x²y²(3z – 24) / 27xy(z – 8) = 9x²y³x 3(z – 8) / 27xy (z – 8) = xy
• 96abc(3a – 12)(5b – 30) / 144 (a- 4)/(b- 6) = 96 abc x 3(a – 4) x 5(b – 6)/144(a – 4) (b – 6) = 10abc

 

Q4. Divide as directed.

• 5(2x + 1)(3x + 5) / (2x + 1)
• 26xy (x + 5)(y – 4) / 13x (y – 4)
• 52pqr(p + q)(q +r)(r + p) / 104pq(q + r) (r + p)
• 20(y + 4) (y²+ 5y + 3) / 5(y + 4)
• X(x + 1) (x + 2) (x + 3) / x(x  + 1)

Solution:-

 

• 5(2x + 1) (3x + 5) / (2x + 1) = 5(2x + 1) (3x + 5)/(2x + 1)

= 5(3x + 5)

 

• 26xy (x + 5) (y – 4) / 13x ( y – 4) = 2 x 13 x xy (x + 5) (y – 4)/13x (y – 4)

= 2y (x + 5)

 

• 52pqr ( p + q) (q + r) (r +p) / 104 pq (q + r) (r+  p)

= 2 x 2 x 13 x p x q x r x (p + q) x (q + r) x (r + p)/ 2 x2 x 2 x 13 x p x q x (q + r) x ( r + p)

= 1/2 r(p + q)

 

• 20 (y + 4) (y²+ 5y + 3) = 2 x 2 x 5 x (y + 4) (y² + 5y + 3)

20 (y + 4) (y² + 5y + 3) + 5 ( y + 4) = 2 x 2 x 5 x ( y + 4) x (y² + 5y + 3)/ 5 x ( y + 4)

= 4(y² + 5y + 3)

 

 

• X(x +1) (x + 2) (x + 3) / x (x + 1) = x(x + 1) (x + 2) (x+ 3) / x(x + 1)

= (x + 2) (x + 3)

 

Q5. Factorise the  expressions and divide them as directed.

 

• (y² + 7y + 10) / (y + 5)
• (m² – 14m – 32 )/ (m + 2)
• 5p²– 25p + 20) / (p  – 1)
• 4yz (z² + 6z – 16)/  2y (z + 8)
• 5pq (p²– q²)/ 2p (p + q)
• 12xy (9x²– 16y²) / 4xy (3x + 4y)
• 39y³(50y² – 98 ) / 26y² ( 5y + 7)

Solution:-

 

• (y²+ 7y + 10)

= y² + 2y + 5y + 10

= y(y+2) + 5(y +2)

= (y + 2) (y + 5)

 

• (m²– 14m – 32)

= m² + 2m – 16m – 32

= (m – 16) (m + 2)

 

• (5p²– 25p + 20)

5(p² – 5p + 4)

= p² – 5p + 4

= p² – p – 4p + 4

= (p-1) (p-4)

 

 

(iv)

= z² + 6z – 16

= z² – 2z + 8z – 16

= (z – 2) (z + 8)

 

(v )

= 5pq(p²  – q²) / 2p( p + q) = 5pq (p – q)  (p + q) / 2p(p + q) = 5/2q (p – q)

 

(vi)

= 9x² – 16y²

= (3x)² – (4y)²

= (3x + 4y) (3x – 4y)

 

 

(vii )

= 50y² – 98

= 50y² – 98 = 2(25y² –  49 ) = 2((5y)² – 7²) = 2( 5y – 7) (5y + 7)

NCERT Solutions For Class 8 Science CHAPTER – 3