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

The solution to (1/4+(2x)/7)(1/4-(2x)/7) is ((7+8x)(7-8x))/(784)

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simplify (1/4+(2x)/7)(1/4-(2x)/7)

simplify

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

\frac{( 7 + 8 x ) ( 7 - 8 x )}{784}

Solution steps

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

Join

\frac{1}{4} + \frac{2 x}{7} : \frac{7 + 8 x}{28}

= \frac{7 + 8 x}{28} (\frac{1}{4} - \frac{2 x}{7})

Apply the fraction rule:

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

= \frac{( 7 + 8 x ) ( \frac{1}{4} - \frac{2 x}{7} )}{28}

Simplify

(7 + 8 x) (\frac{1}{4} - \frac{2 x}{7}) : \frac{( 7 + 8 x ) ( 7 - 8 x )}{28}

= \frac{\frac{( 7 + 8 x ) ( 7 - 8 x )}{28}}{28}

Apply the fraction rule:

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

= \frac{( 7 + 8 x ) ( 7 - 8 x )}{28 \cdot 28}

Multiply the numbers:

28 \cdot 28 = 784

= \frac{( 7 + 8 x ) ( 7 - 8 x )}{784}

I n (e^{2 x} + 1)^{2}

2 x e^{x} + e^{x} (x^{2} - 3)

cos (2 x) (2 cos (x) + 1)

\frac{3 ( - 3 + 33 )}{2} + 15

\frac{- y ^{2} + 2 y}{( y ^{2} - y + 1 ) ^{2}}

What is (1/4+(2x)/7)(1/4-(2x)/7) ?
The solution to (1/4+(2x)/7)(1/4-(2x)/7) is ((7+8x)(7-8x))/(784)