Factorise the following expressions
(i) 7x - 42  (ii) 6p -12q  (iii) 7a²+14a  (iv) -16z + 20z³
(v) 20l²m + 30alm  (vi) 5x²y -15xy²  (vii) 10a² -15b² + 20c²
(viii) -4a² + 4ab - 4ca  (ix) x²yz + xy²z + xyz²
(x) ax²y + bxy² + cxyz

Solution:

Factorization is a method of finding factors for any mathematical object, be it a number, a polynomial or any algebraic expression. Thus, factorization of an algebraic expression refers to finding out the factors of the given algebraic expression.

(i) 7x - 42

7x = 7 × x

42 = 2 × 3 × 7

The common factor is 7.

Therefore, 7x - 42 = (7 × x) - (2 × 3 × 7 )

= 7(x - 6)

(ii) 6p -12q

6p = 2 × 3 × p

12q = 2 × 2 × 3 × q

The common factors are 2 and 3.

Therefore, 6p -12q = (2 × 3 × p) - (2 × 2 × 3 × q)

= 2 × 3(p - 2 × q)

= 6(p - 2q)

(iii) 7a²+14a

7a²= 7 × a × a

14a = 2 × 7 × a

The common factors are 7 and a .

Therefore, 7a² + 14a = (7 × a × a) + (2 × 7 × a)

= 7 × a(a + 2)

= 7a(a + 2)

(iv) -16z + 20z³

16z = -1 × 2 × 2 × 2 × 2 × z

20z³ = 2 × 2 × 5 × z × z × z

The common factors are 2, 2, and z.

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

= (2 × 2 × z)[-(2 × 2) + (5 × z × z)]

= 4z(-4 + 5z²)

(v) 20l²m + 30alm

20l²m= 2 × 2 × 5 × l × l × m

30alm = 2 × 3 × 5 × a × l × m

The common factors are 2, 5, l and m.

Therefore, 20l²m + 30alm = (2 × 2 × 5 × l × l × m) + (2 × 3 × 5 × a × l × m)

= (2 × 5 × l × m)[(2 × l) + (3 × a)]

= 10lm(2l + 3a)

(vi) 5x²y -15xy²

5x²y = 5 × x × x × y

15xy² = 3 × 5 × x × y × y

The common factors are 5, x, and y.

Therefore, 5x²y - 15xy² = (5 × x × x × y) - (3 × 5 × x × y × y)

= 5 × x × y[x - (3 × y)]

= 5xy(x - 3y)

(vii) 10a² -15b² + 20c²

10a² = 2 × 5 × a × a

15b² = 3 × 5 × b × b

20c² = 2 × 2 × 5 × c × c

The common factor is 5.

Therefore, 10a² -15b² + 20c² = (2 × 5 × a × a) - (3 × 5 × b × b) + (2 × 2 × 5 × c × c)

= 5[(2 × a × a) - (3 × b × b) + (2 × 2 × c × c)]

= 5(2a² - 3b² + 4c² )

(viii) -4a² + 4ab - 4ca

4a² = 2 × 2 × a × a

4ab = 2 × 2 × a × b

4ca = 2 × 2 × c × a

The common factors are 2, 2, and a .

Therefore, - 4a² + 4ab - 4ca = -(2 × 2 × a × a) + (2 × 2 × a × b) - (2 × 2 × c × a)

= 2 × 2 × a (-a + b - c)

= 4a(-a + b - c)

(ix) x²yz + xy²z + xyz²

x²yz = x × x × y × z

xy²z= x × y × y × z

xyz² = x × y × z × z

The common factors are x, y, and z.

Therefore, x²yz + xy²z + xyz² = (x × x × y × z) + (x × y × y × z) + (x × y × z × z)

= x × y × z (x + y + z)

= xyz(x + y + z)

(x) ax²y + bxy² + cxyz

ax²y= a × x × x × y

bxy² = b × x × y × y

cxyz= c × x × y × z

The common factors are x and y.

Therefore, ax²y + bxy² + cxyz = (a × x × x × y) + (b × x × y × y) + (c × x × y × z)

= (x × y)[(a × x) + (b × y) + (c × z)]

= xy(ax + by + cz)

☛ Check: NCERT Solutions for Class 8 Maths Chapter 14

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Video Solution:

Factorise the following expressions (i) 7x - 42 (ii) 6p-12q (iii) 7a²+14a (iv) -16z + 20z³ (v) 20l²m + 30alm (vi) 5x²y -15xy² (vii) 10a² -15b² + 20c² (viii) -4a² + 4ab - 4ca (ix) x²yz + xy²z + xyz² (x) ax²y + bxy² + cxyz

Class 8 Maths NCERT Solutions Chapter 14 Exercise 14.1 Question 2

Summary:

The following expressions (i) 7x - 42 (ii) 6p-12q (iii) 7a²+14a (iv) -16z + 20z³ (v) 20l²m + 30alm (vi) 5x²y -15xy² (vii) 10a² -15b² + 20c² (viii) -4a² + 4ab - 4ca (ix) x²yz + xy²z + xyz² (x) ax²y + bxy² + cxyz are factorised as: (i) 7(x - 6) (ii) 6(p - 2q) (iii) 7a(a + 2) (iv) 4z(-4 + 5z²) (v) 10lm(2l + 3a) (vi) 5xy(x - 3y) (vii) 5(2a² - 3b² + 4c²)(viii) 4a(-a + b - c) (ix) xyz(x + y + z) (x) xy(ax + by + cz)

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