NCERT Solution For Class 8 Math Chapter – 14 Factorisation  Exercise – 14.1

Q1. Find the common factors of the given terms.

• 12x , 36
• 22x
• 14pq , 28p²q²
• 2x , 3x², 4
• 6 abc , 24ab²  , 12a²b
• 16 x³, -4x² , 32x
• 10pq , 20qr , 30 rp
• 3x²y³, 10x³y² , 6x²y²z

Solution:-

 

• Factors  of 12x and 36

2x  = 2 x 2 x 3 x x

36 = 2 x 2 x 3 x 3

Common factors of 12x and 36 are 2,3,2

And 2 x 2 x 3 = 12

 

• Factors of 2y and 22xy

2y =  2xy

22xy = 2 x 11 x x x y

Common factors of 2y and 22xy are 2,y

And , 2xy = 2y

 

• Factors of 14pq and 28p²q

14pq = 2 x 7 x pq

28p²q = 2 x 2 x 7 x p x p x q

Common factors of 14pq and 28p²q are 2.7,p,q

And 2 x 7 x p x q = 14pq

 

• Factors of 2x , 3x²and 4

2x = 2  x x

3x² = 3 x x x

4 = 2 x 2

Common  factors of 2x, 3x² and 4 is 1.

 

• Factors of 6abc , 24ab²and 12a²b

6abc = 2 x 3 x a x b x c

24ab² = 2 x 2 x 2 x 3 x a x b x c

12a²b = 2x 2 x 3 x a x a x b

Common factors of 6 abc , 24ab² and 12a²b are 2,3 , a , b  and 2 x 3 x a x b = 6ab

 

• Factors of  16x³, -4x²  and 32x

16x³ =  2 x 2 x 2 x 2 x x x  x x x

-4x² = -1  x 2 x 2 x x x x

32x = 2 x 2 x 2 x 2 x 2 x  x

Common factor of 16x³ ,  -4x² and 32x are 2,2,x and 2 x 2 x x = 4x

 

• Factors of 10 pq, 20qr and 30rp

10pq = 2 x 5 x q x p

20qr = 2 x  x 5 x q x r

30rp = 2 x 3 x 5 x r x p

Common factors of 10 pq ,, 20qr and 30rp are 2,5  and, 2 x 5  = 10

 

• Factors of 3x²y³, 10x³y² and 6x²y²z

3x²y³ = 3 x  x x x x y x y x y

10x³y² = 2 x 5 x x x x x x x y x y

6x²y²z = 3 x 2 x x x x x y x y x z

Common factors of 3x²y³ , 10x³y² and 6x²y²z are x² , y² and x²  x y²  = x²y²

 

Q2. Factorise the following expressions

• 7x – 42
• 6p – 12q
• 7a²+ 14a
• -16z + 20z³
• 20l²m  + 30alm
• 5x²y – 15xy²
• 10a²– 15b² + 20c² 
• -4a² + 4ab – 4ca
• X²yz + bxy² + cxyz
• ax²y + bxy²+ cxyz

Solution:-

 

• 7x = 7 x x

42 = 2 x 3 x 7

7x – 42 = (7 x x) – (2 x 3 x 7) = 7(x – 6)

 

• 6p = 2 x 3 x p

12q = 2 x 3 x 2 x q

6p – 12q = (2 x 3 x p) – (2 x 2 x 3 x q)

= 2 x 3 [ p – (2 x q)]

= 6 (p – 2q)

 

• 7a²= 7 x a x a

14a = 7 x 2 x a

7a² + 14 a = (7 x a x a) + (2 x 7 x a)

= 7 x a [ a +2] =  7a ( a +2)

 

• 16z = 2 x 2 x 2 x 2 x z

20z³ = 2x 2x 5 x z x z x z

-16z + 20z³ = -(2×2 x 2 x 2 x z) + (2 x 2 x 5 x z x z x z)

= 4z(-4 + 5z²)

 

• 20l²m = 2 x 2 x 5 x l x l x m

30alm = 2 x 3 x 5 x a x l x m

20l²m + 30alm = (2 x 2 x 5 x l x l x m) + (2 x 3 x 5 x a x l x m)

= 10lm (2l + 3a)

 

• 5x²y = 5 x x x x x y

15xy² = 3 x 5 x x x y x y

5x²y – 15xy² = (5 x x x x x y) – (3 x 5 x x x y x y)

= 5xy (x – 3y)

 

• 10a²– 15b² + 20c

10a² = 2 x 5 x a x a

-15b² = -1 x 3 x 5 x bx b

20c² = 2 x 2 x 5 x c x c

 

10a² – 15b² + 20c² = 5(2a² – 3b² + 4c²)

 

• -4a²+ 4ab – 4ca

4a² = -1 x 2 x 2 x a x a

4ab = 2 x 2 x a x b

4ca = -1 x 2 x 2 x c x a

-4a² + 4ab – 4ca = 4a(-a + b – c)

 

• X²yz + xy²z + xyz²

X²yz = x x x x y x z

Xy²z = x x y x y x z

Xyz² = x x y x z x z

 

X²yz + xy²z + xyz² = xyz( x + y + z)

 

 

• Ax²y + bxy²+ cxyz

Ax²y = a x x x x x y

Bxy² = b x x x y x y

Cxyz = c x x y x z

 

Ax²y + bxy² + cxyz = xy (ax + by + cz)

 

Q3. Factorise.

• x²+ xy + 8x + 8y
• 15xy – 6x + 5y – 2
• ax + bx – ay – by
• 15pq + 15 + 9q + 25p
• Z –  7 + 7xy – xyz

Solution:-

 

• X²+ xy + 8x + 8y =  x x x + x x y + 8  x x + 8 x y

= x(x+ y) + 8(x+ y)

= (x + y) (x + 8)

 

• 15xy – 6x + 5y  – 2 = 3 x 5 x x x y – 3 x 2 x x +  5xy – 2

= (5y  2) (3x + 1)

 

• Ax + bx – ay – by = a x x + b  x x- a x y -b x y

= x(a + b) -y (a + b)

= (a + b) (x – y)

 

• 15pq + 15 + 9q + 25p = 15pq + 9q + 25p + 15

= (5p +  3) (3q + 5)

 

• Z – 7 + 7xy – xyz = z – x x y x z – 7 + 7 x x x y

= (1 – xy) (z – 7)