Advertisements
Advertisements
प्रश्न
Factorise : 16p4 – 1
Advertisements
उत्तर
16p4 – 1 = 24p4 – 1
= (22)2(p2)2 – 12
= (22p2)2 – 12
Comparing with a2 – b2 (a + b)(a – b) where a = 22p2 and b = 1
∴ (22p2)2 – 12 = (22p2 + 1)(22p2 – 1)
= (4p2 + 1)(4p2 – 1)
∴ 16p4 – 1 = (4p2 + 1)(4p2 – 1)
= (4p2 + 1)(22p2 – 12)
= (4p2 + 1)[(2p)2 – 12]
= (4p2 + 1)(2p + 1)(2p – 1) ...[∵ using a2 – b2 = (a + b)(a – b)]
∴ 16p4 – 1 = (4p2 + 1)(2p + 1)(2p – 1)
APPEARS IN
संबंधित प्रश्न
Factorise the following expressions
m2 + m – 72
Using the identity (a + b)(a – b) = a2 – b2, find the following product
(6x + 7y)(6x – 7y)
Factorise the following algebraic expression by using the identity a2 – b2 = (a + b)(a – b)
z2 – 16
Simplify using identities
(3p + q)(3p – q)
Using suitable identities, evaluate the following.
(9.7)2 – (0.3)2
Factorise the following using the identity a2 – b2 = (a + b)(a – b).
y4 – 625
Factorise the following using the identity a2 – b2 = (a + b)(a – b).
p5 – 16p
The sum of first n natural numbers is given by the expression `n^2/2 + n/2`. Factorise this expression.
Verify the following:
(m + n)(m2 – mn + n2) = m3 + n3
Verify the following:
(a2 – b2)(a2 + b2) + (b2 – c2)(b2 + c2) + (c2 – a2) + (c2 + a2) = 0
