Factorise. (i) a4 - b4 (ii) p4 - 81 (iii) x4 - ( y + z)4
(iv) x4 - (x - z)4 (v) a4 - 2a2b2 + b4
Solution:
We will be using the following algebraic identities to solve the given questions
(a- b)2 = a2 - 2ab+ b2
a2 - b2 = (a - b)(a + b)
(i) a4 - b4
= (a2)2 - (b2)2
= (a2 - b2)(a2 + b2) [Since, a2 - b2 = (a - b)(a + b)]
= (a - b)(a + b)(a2 + b2 ) [Since, a2 - b2 = (a - b)(a + b)]
(ii) p4 - 81
= (p2)2 - (9)2
= (p2 - 9)(p2 + 9) [Since, a2 - b2 = (a - b)(a + b)]
= [(p)2 - (3)2](p2 + 9)
= (p - 3)(p + 3) (p2 + 9) [Since, a2 - b2 = (a - b)(a + b)]
(iii) x4 - (y + z)4
= (x2)2 - [(y + z)2]2
= [x2 - (y + z)2][x2 + (y + z)2] [Since, a2 - b2 = (a - b)(a + b)]
= [x - ( y + z)][x + ( y + z)][x2 + (y + z)2] [Since, a2 - b2 = (a - b)(a + b)]
= (x - y - z)(x + y + z)[x2 + (y + z)2]
(iv) x4 - (x - z)4
= (x2)2 - [( x - z)2]2
= [x2 - (x - z)2 ][ x2 + (x - z)2] [Since, a2 - b2 = (a - b)(a + b)]
= [x - (x - z)][x + (x - z )][x2 + ( x - z )2] [Since, a2 - b2 = (a - b)(a + b)]
= z(2x - z )[x2 + x2 - 2xz + z2] [Since, (a- b)2 = a2 - 2ab+ b2]
= z(2x - z )(2x2 - 2xz + z2)
(v) a4 - 2a2b2 + b4
= (a2)2 - 2 (a2)(b2) + (b2)2
= (a2 - b2)2 [Since, (a- b)2 = a2 - 2ab+ b2]
= [(a - b)(a + b)]2 [Since, a2 - b2 = (a - b)(a + b)]
= (a - b)2(a + b)2
☛ Check: NCERT Solutions for Class 8 Maths Chapter 14
Video Solution:
Factorise (i) a⁴ - b⁴ (ii) p⁴ - 81 (iii) x⁴ - ( y + z)⁴ (iv) x⁴ - (x - z)⁴ (v) a⁴ - 2a²b² + b⁴
Class 8 Maths NCERT Solutions Chapter 14 Exercise 14.2 Question 4
Summary:
Factorization of the following expressions (i) a4 - b4 (ii) p4 - 81 (iii) x4 - ( y + z)4 (iv) x4 - (x - z)4 (v) a4 - 2a2b2 + b4 are as follows (i) (a - b)(a + b)(a2 + b2 ) (ii) (p - 3)(p + 3) (p2 + 9) (iii) (x - y - z)(x + y + z)[x2 + (y + z)2] (iv) z(2x - z )(2x2 - 2xz + z2) (v) (a - b)2(a + b)2
☛ Related Questions:
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