Factorise the expressions
(i) ax2 + bx (ii) 7p2 + 21q2 (iii) 2x3 + 2xy2 + 2xz2
(iv) am2 + bm2 + bn2 + an2 (v) (lm + l) + m + 1
(vi) y(y + z) + 9( y + z) (vii) 5y2 - 20 y - 8z + 2yz
(viii) 10ab + 4a + 5b + 2 (ix) 6xy - 4y + 6 - 9x
Solution:
For part (i), (ii), (iii), and (vi) - First we will find factors of each term then find out which factors are common in each term, and take out that common factor from expression.
For part (iv), (v), (vii), (viii), (ix) - There are 4 terms in each expression. First, we will make pair of two terms from which we can take out common factors and convert the expression of 4 terms into 2 terms expression then take out common factors from the remaining 2 terms.
Steps:
(i) ax2 + bx
= a × x × x + b × x
= x (ax + b)
(ii) 7p2 + 21q2
= 7 × p × p + 3 × 7 × q × q
= 7(p2 + 3q2)
(iii) 2x3 + 2xy2 + 2xz2
= 2x( x2 + y2 + z2 )
(iv) am2 + bm2 + bn2 + an2
= am2 + bm2 + an2 + bn2
= m2(a + b) + n2(a + b)
= (a + b)(m2 + n2 )
(v) (lm + l ) + m + 1
= lm + m + l + 1
= m(l + 1) + 1(l + 1)
= (l + 1)(m + 1)
(vi) y(y + z) + 9(y + z)
= (y + z)(y + 9)
(vii) 5y2 - 20 y - 8z + 2yz
= 5y2 - 20y + 2yz - 8z
= 5y(y - 4) + 2z(y - 4)
= (y - 4)(5y + 2z)
(viii) 10ab + 4a + 5b + 2
= 10ab + 5b + 4a + 2
= 5b(2a +1) + 2(2a +1)
= (2a +1)(5b + 2)
(ix) 6xy - 4 y + 6 - 9x
= 6xy - 9x - 4 y + 6
= 3x(2y - 3) - 2(2 y - 3)
= (2y - 3)(3x - 2)
☛ Check: NCERT Solutions for Class 8 Maths Chapter 14
Video Solution:
Factorise the expressions. (i) ax² + bx (ii) 7p² + 21q² (iii) 2x³ + 2xy² + 2xz² (iv) am² + bm² + bn² + an² (v) (lm + l) + m + 1 (vi) y(y + z) + 9( y + z) (vii) 5y² - 20 y - 8z + 2yz (viii) 10ab + 4a + 5b + 2 (ix) 6xy - 4 y + 6 - 9x
Class 8 Maths NCERT Solutions Chapter 14 Exercise 14.2 Question 3
Summary:
The expressions are factorized (i) ax2 + bx (ii) 7p2 + 21q2 (iii) 2x3 + 2xy2 + 2xz2 (iv) am2 + bm2 + bn2 + an2 (v) (lm + l) + m + 1 (vi) y(y + z) + 9( y + z) (vii) 5y2 - 20 y - 8z + 2yz (viii) 10ab + 4a + 5b + 2 (ix) 6xy - 4 y + 6 - 9x as follows (i) x(ax + b) (ii) 7(p2 + 3q2) (iii)2x( x2 + y2 + z2 ) (iv) (a + b)(m2 + n2 )(v) (l + 1)(m + 1) (vi) (y + z)(y + 9) (vii) (y - 4)(5y + 2z) (viii) (2a +1)(5b + 2) (ix) (2y - 3)(3x - 2)
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