What is factor 16n^4-1 ?

The solution to factor 16n^4-1 is (4n^2+1)(2n+1)(2n-1)

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factor 16n^4-1

factor

16 n^{4} - 1

(4 n^{2} + 1) (2 n + 1) (2 n - 1)

Solution steps

16 n^{4} - 1

Apply exponent rules

= (4 n^{2})^{2} - 1^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

(4 n^{2})^{2} - 1^{2} = (4 n^{2} + 1) (4 n^{2} - 1)

= (4 n^{2} + 1) (4 n^{2} - 1)

Factor

4 n^{2} - 1 : (2 n + 1) (2 n - 1)

= (4 n^{2} + 1) (2 n + 1) (2 n - 1)

x \times 3 x

20 lo g_{10} (0)

(102) (- - 90)

\frac{( k + 1 ) !}{k !}

1 2^{2} + 6^{2}