What is (7t^2+1/4)(7t^2-1/4) ?

The solution to (7t^2+1/4)(7t^2-1/4) is ((28t^2+1)(28t^2-1))/(16)

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simplify (7t^2+1/4)(7t^2-1/4)

simplify

(7 t^{2} + \frac{1}{4}) (7 t^{2} - \frac{1}{4})

\frac{( 28 t ^{2} + 1 ) ( 28 t ^{2} - 1 )}{16}

Solution steps

(7 t^{2} + \frac{1}{4}) (7 t^{2} - \frac{1}{4})

Join

7 t^{2} + \frac{1}{4} : \frac{28 t ^{2} + 1}{4}

= \frac{28 t ^{2} + 1}{4} (7 t^{2} - \frac{1}{4})

Apply the fraction rule:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( 28 t ^{2} + 1 ) ( 7 t ^{2} - \frac{1}{4} )}{4}

Simplify

(28 t^{2} + 1) (7 t^{2} - \frac{1}{4}) : \frac{( 28 t ^{2} + 1 ) ( 28 t ^{2} - 1 )}{4}

= \frac{\frac{( 28 t ^{2} + 1 ) ( 28 t ^{2} - 1 )}{4}}{4}

Apply the fraction rule:

\frac{\frac{a}{b}}{c} = \frac{a}{b \cdot c}

= \frac{( 28 t ^{2} + 1 ) ( 28 t ^{2} - 1 )}{4 \cdot 4}

Multiply the numbers:

4 \cdot 4 = 16

= \frac{( 28 t ^{2} + 1 ) ( 28 t ^{2} - 1 )}{16}

(\frac{1 - cos ( 2 x )}{2})^{3}

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\frac{6 x}{( x ^{2} + 2 ) ^{2}}

\frac{( sec ( x ) ) ^{2} β - sin ( β ) tan ( β )}{cos ( x )}

- \frac{1}{s ( s + 2 )}

What is (7t^2+1/4)(7t^2-1/4) ?
The solution to (7t^2+1/4)(7t^2-1/4) is ((28t^2+1)(28t^2-1))/(16)