Mathematics
Factorise:
2x7 - 128x
Factorisation
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Answer
Given,
⇒ 2x7 - 128x
⇒ 2x(x6 - 64)
⇒ 2x[(x3)2 - (8)2]
⇒ 2x[(x)3 - (8)] [(x)3 + (8)]
⇒ 2x[x3 - 8] [x3 + 8]
⇒ 2x[x3 - 23] [x3 + 23]
By using the identity,
(a3 - b3) = (a - b)(a2 + ab + b2)
(a3 + b3) = (a + b)(a2 - ab + b2)
⇒ 2x[(x - 2)(x2 + 2 × x + (2)2)] [(x + 2)(x)2 - 2 × x + (2)2]
⇒ 2x(x - 2)(x2 + 2x + 4)(x + 2)(x2 - 2x + 4)
⇒ 2x(x - 2)(x + 2)(x2 + 2x + 4)(x2 - 2x + 4).
Hence, 2x7 - 128x = 2x(x - 2)(x + 2)(x2 + 2x + 4)(x2 - 2x + 4).
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