What is (1/(2x)+4)(1/(2x)-8) ?

The solution to (1/(2x)+4)(1/(2x)-8) is ((8x+1)(-16x+1))/(4x^2)

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simplify (1/(2x)+4)(1/(2x)-8)

simplify

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

\frac{( 8 x + 1 ) ( - 16 x + 1 )}{4 x ^{2}}

Solution steps

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

Join

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

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

Apply the fraction rule:

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

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

Simplify

(8 x + 1) (\frac{1}{2 x} - 8) : \frac{( 8 x + 1 ) ( - 16 x + 1 )}{2 x}

= \frac{\frac{( 8 x + 1 ) ( - 16 x + 1 )}{2 x}}{2 x}

Apply the fraction rule:

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

= \frac{( 8 x + 1 ) ( - 16 x + 1 )}{2 x \cdot 2 x}

Simplify

2 x \cdot 2 x : 4 x^{2}

= \frac{( 8 x + 1 ) ( - 16 x + 1 )}{4 x ^{2}}

2 π^{4} - 2 π^{4} cos (θ) + π^{4} sin (2 (1 - cos (θ)))

\frac{d ^{2}}{d ^{2} x} (x^{2} + y^{2} = a^{2})

(3, - \frac{π}{6}) 0 x

\frac{e ^{- 2 s}}{s ( s ^{2} + 2 s + 5 )}

a^{\frac{1}{3}} \cdot (a^{\frac{2}{3}} - a^{\frac{5}{3}})

What is (1/(2x)+4)(1/(2x)-8) ?
The solution to (1/(2x)+4)(1/(2x)-8) is ((8x+1)(-16x+1))/(4x^2)