What is expand ((2x-b+g)(2x-a+b))/4 ?

The solution to expand ((2x-b+g)(2x-a+b))/4 is x^2-(ax)/2+(ba)/4-(b^2)/4+(xg)/2-(ag)/4+(bg)/4

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expand ((2x-b+g)(2x-a+b))/4

expand

\frac{( 2 x - b + g ) ( 2 x - a + b )}{4}

x^{2} - \frac{a x}{2} + \frac{ba}{4} - \frac{b ^{2}}{4} + \frac{xg}{2} - \frac{a g}{4} + \frac{b g}{4}

Solution steps

\frac{( 2 x - b + g ) ( 2 x - a + b )}{4}

Expand

(2 x - b + g) (2 x - a + b) : 4 x^{2} - 2 a x + ba - b^{2} + 2 xg - a g + b g

= \frac{4 x ^{2} - 2 a x + ba - b ^{2} + 2 xg - a g + b g}{4}

Apply the fraction rule:

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

\frac{4 x ^{2} - 2 a x + ba - b ^{2} + 2 xg - a g + b g}{4} = \frac{4 x ^{2}}{4} - \frac{2 a x}{4} + \frac{ba}{4} - \frac{b ^{2}}{4} + \frac{2 xg}{4} - \frac{a g}{4} + \frac{b g}{4}

= \frac{4 x ^{2}}{4} - \frac{2 a x}{4} + \frac{ba}{4} - \frac{b ^{2}}{4} + \frac{2 xg}{4} - \frac{a g}{4} + \frac{b g}{4}

Divide the numbers:

\frac{4}{4} = 1

= x^{2} - \frac{2 a x}{4} + \frac{ba}{4} - \frac{b ^{2}}{4} + \frac{2 xg}{4} - \frac{a g}{4} + \frac{b g}{4}

Cancel

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

= x^{2} - \frac{a x}{2} + \frac{ba}{4} - \frac{b ^{2}}{4} + \frac{2 xg}{4} - \frac{a g}{4} + \frac{b g}{4}

Cancel

\frac{2 xg}{4} : \frac{xg}{2}

= x^{2} - \frac{a x}{2} + \frac{ba}{4} - \frac{b ^{2}}{4} + \frac{xg}{2} - \frac{a g}{4} + \frac{b g}{4}

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

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

(\frac{3 + z ^{2}}{2})^{6}

(x) (x - \frac{π}{2}) (x - π)

\frac{d}{d t} ((P (t)) P (t) (1 - P (t)))

What is expand ((2x-b+g)(2x-a+b))/4 ?
The solution to expand ((2x-b+g)(2x-a+b))/4 is x^2-(ax)/2+(ba)/4-(b^2)/4+(xg)/2-(ag)/4+(bg)/4