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

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

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

expand

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

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

Solution steps

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

Expand

(2 x + 1) (4 x - 1) : 8 x^{2} + 2 x - 1

= \frac{8 x ^{2} + 2 x - 1}{8}

Apply the fraction rule:

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

\frac{8 x ^{2} + 2 x - 1}{8} = \frac{8 x ^{2}}{8} + \frac{2 x}{8} - \frac{1}{8}

= \frac{8 x ^{2}}{8} + \frac{2 x}{8} - \frac{1}{8}

Divide the numbers:

\frac{8}{8} = 1

= x^{2} + \frac{2 x}{8} - \frac{1}{8}

Cancel

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

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

x (4 - 11)

π (16 x^{2} + 8 x + 1 = 20)

7 x - 2 l e f t (x - 3)

\frac{( 3 x - 1 ) ^{3}}{9}

5 sin (0.5 (x + 22.93119)) + 8

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