Question

"Question 2 Factorise the following expressions (i) 7x − 42 (ii) 6p − 12q (iii) 7a^2 + 14a (iv) −16z + 20z^3 (v) 20l^2m + 30 alm (vi) 5x^2y − 15xy^2 (vii) 10a^2 − 15b^2 + 20c^2 (viii) −4a^2 + 4ab − 4 ca (ix) x^2yz + xy^2z + xyz^2 (x) ax^2y + bxy^2 + cxyz

Class 8 Factorisation Page 220"

Answer 1

Factors:

In a product of two or more expressions each expression is called a factor of the product.

 

Factorization:

The process of writing a given expression as a product of two or more factors is called factorization.

Factors of a monomial:

The greatest common factor of 2 or more monomials is the largest common monomial. The largest common monomial is a product of the greatest common factor of the numerical Coefficients and the common variables with smallest powers.

 

Factorization when a common monomial factor occurs in each term:

An expression can be expressed as a product of the greatest common monomial and the quotient obtained by dividing the given expression by this greatest common monomial.

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Solution:

First, we need to factorize each term and then find common factors to factorize the expression

1) 7x-42
=  (7×x) – (7×6)
=7(x-6)

2) 6p-12q
=  (6×p) – (6×2q)
=6(p-2q)

3) 7a²+14a
=(7×a×a) +  (2×7×a)
=7a(a+2)

4) -16z+ 20z³

= -(2 x 2 x 2 x 2 x z) + (2 x 2 x 5 x z x z x z)

= (2 x 2x z)  [ - (2 x 2) +( 5 x z x z)]

=4z(-4 +5z²)

5) 20 l² m + 30 a l m
=(2×2×5×l×m×l)  + (2×3×5×a×l×m)
=2 x 5lm(2l+3a)

=2 x 5lm(2l+3a)

=10lm(2l+3a)

6) 5 x²y -15 xy²
=(5×x×x×y)  + (5×3×x×y×y)
=5xy(x-3y)

7) 10a² -15 b²+20c²
=(2×5×a×a)  - (5×3×b×b) + (5×2×2×c×c)

=5(2a²-3b²+4c²)

8) -4a² +4a b-4ca
=(-4×a×a)  +(4×b×a) - (4×c×a)

=-4a(a-b+c)

9)  x²yz + xy²z  +xyz²
=(x×x×y×z)  +(x×y×y×z) +(x×y×z×z)
=xyz(x+y+z)

10) ax²y + bxy²  +cxyz
=(x×x×y×a)  +(x×y×y×b) +(x×y×z×c)
=xy(ax+by+cz)

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Hope this will help you....

Answer 2

Answer:

(i) 7x − 42 = 7(x − 6)

(ii) 6p − 12q = 6(p − 2q)

(iii) 7a 2 + 14a = 7a(a + 2)

(iv) − 16z + 20 z 2 = 4z(5z 2 − 4)

(v) 20l 2m + 30alm = 2lm(10l + 15a)

(vi) 5x 2y − 15xy 2 = 5x(xy − 3y 2)

(vii)10a 2 − 15b 2 + 20c 2 = 5(2a 2 − 3b 2 + 4c 2)

(viii) − 4a 2 + 4ab − 4ca = − 4a(a − b + c)

(ix) x 2yz + xy 2z + xyz 2 = xyz(x + y + z)

(x) ax 2y + bxy 2 + cxyz = xyz(ax + by + cz