Factor using formulas?

Question

16a^2-1

Answer 1

Difference of two squares

16a² - 1 = 0

(4a - 1)(4a + 1)

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Answer 2

You see, anything of the form,
a^n - b^n , where n is positive integral can be expanded as,
(a-b)(a^(n-1) + a^(n-2).b +...+ b^(n-1))

Now consider when n=2, we obtain what is arguably the most useful formula in elementary algebra, the difference of two squares, for when n=2,
a^2 - b^2 = (a-b)(a+b) with all the terms in the middle diappearing,
so a^2 - b^2 = (a+b)(a-b)

Now for 16a^2 -1, this is obviously the same as [(4a)^2 -1^2]
= (4a-1)(4a+1)

Answer 3

a^2 - b^2 =(a + b)(a - b)

You can use the above formula to factorise your question;
16a^2 - 1 = (4a + 1)(4a -1)

Answer 4

16 a² - 1 = (4a - 1).(4a + 1)
Check
4a.(4a + 1) - 1.(4a + 1)
= 16a² + 4a - 4a - 1
= 16a² - 1

Answer 5

16a^2-1
= (4a+1) (4a-1)
= (4a+1) (2a^1/2+1) (2a^1/2-1)
......... proceed now using (2a^1/2-1)

Answer 6

isnt it just (4a+1)(4a-1) or are you asking for something else?

Answer 7

(4a)^2-(1)^2
(4a-1)(4a+1)