Math, asked by Faizangood, 5 months ago

Factorise ( 81a5 – 16a)

Answers

Answered by Anonymous

Step-by-step explanation:

$81a {}^{5} - 16a \\ = a(81a {}^{4} - 16) \\ = a(9 a {}^{2} ) {}^{2} - (4) {}^{2} \\ = a(9a {}^{2} + 4)(9a {}^{2} - 4) \\ = a(9a {}^{2} + 4)((3a) {}^{2} - (2) {}^{2} ) \\ = a(9a {}^{2} + 4)(3a + 2)(3a - 2)$

Faizangood: A polyhedron has 10 edges and 6 vertices. How many faces does this polyhedron have?...,... please

Anonymous: solve by Euler's formula . F+V-E = 2 here F = faces , V = vertices , E = edges . F+6-10 = 2 , F-4 = 2, F= 6

Faizangood: :) thank you so much

anindyaadhikari13: Correct answer.

Answered by anindyaadhikari13

Required Answer:-

Given To Factorise:

81a⁵ - 16a

Solution:

We have,

81a⁵ - 16a

= a(81a⁴ - 16)

= a[(9a²)² - (4)²]

= a(9a² + 4)(9a² - 4) [Using identity a² - b² = (a + b)(a - b)]

= a(9a² + 4)[(3a)² - (2)²] [Using identity a² - b² = (a + b)(a - b)]

= a(9a² + 4)(3a + 2)(3a - 2)

★ Hence, the factorised form of the polynomial is a(9a² + 4)(3a + 2)(3a - 2)

Answer:

Factorised form is - Hence, the factorised form of the polynomial is a(9a² + 4)(3a + 2)(3a - 2)

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