What is expand ((2x-1)(4x+3))/8 ?

The solution to expand ((2x-1)(4x+3))/8 is x^2+x/4-3/8

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expand ((2x-1)(4x+3))/8

expand

\frac{( 2 x - 1 ) ( 4 x + 3 )}{8}

x^{2} + \frac{x}{4} - \frac{3}{8}

Solution steps

\frac{( 2 x - 1 ) ( 4 x + 3 )}{8}

Expand

(2 x - 1) (4 x + 3) : 8 x^{2} + 2 x - 3

= \frac{8 x ^{2} + 2 x - 3}{8}

Apply the fraction rule:

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

\frac{8 x ^{2} + 2 x - 3}{8} = \frac{8 x ^{2}}{8} + \frac{2 x}{8} - \frac{3}{8}

= \frac{8 x ^{2}}{8} + \frac{2 x}{8} - \frac{3}{8}

Divide the numbers:

\frac{8}{8} = 1

= x^{2} + \frac{2 x}{8} - \frac{3}{8}

Cancel

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

= x^{2} + \frac{x}{4} - \frac{3}{8}

(e^{3 t} - e^{2 t})^{2}

\frac{5}{2} cos (\frac{2 π}{365} (t - \frac{1067}{4})) + 12

\frac{x - 5}{( x - 2 ) ( x - 9 )}

\frac{10}{( s + 1 ) ( s + 3 ) ( s ^{2} + 2 s + 5 )}

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

What is expand ((2x-1)(4x+3))/8 ?
The solution to expand ((2x-1)(4x+3))/8 is x^2+x/4-3/8