What is factor 16u^4-1 ?

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

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factor 16u^4-1

factor

16 u^{4} - 1

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

Solution steps

16 u^{4} - 1

Apply exponent rules

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

Apply Difference of Two Squares Formula:

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

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

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

Factor

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

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

x^{(- \frac{1}{2})}

(- 12)^{2}

6 w - 7 + 12 w - 3 z

\frac{- \frac{3}{2}}{3}

0.5 x + x