What is expand ((2x+12)^2)/4+((y-4)^2)/4 ?

The solution to expand ((2x+12)^2)/4+((y-4)^2)/4 is x^2+12x+(y^2)/4+40-2y

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expand ((2x+12)^2)/4+((y-4)^2)/4

expand

\frac{( 2 x + 12 ) ^{2}}{4} + \frac{( y - 4 ) ^{2}}{4}

x^{2} + 12 x + \frac{y ^{2}}{4} + 40 - 2 y

Solution steps

\frac{( 2 x + 12 ) ^{2}}{4} + \frac{( y - 4 ) ^{2}}{4}

Apply rule

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

= \frac{( 2 x + 12 ) ^{2} + ( y - 4 ) ^{2}}{4}

Expand

(2 x + 12)^{2} + (y - 4)^{2} : 4 x^{2} + 48 x + y^{2} + 160 - 8 y

= \frac{4 x ^{2} + 48 x + y ^{2} + 160 - 8 y}{4}

Apply the fraction rule:

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

\frac{4 x ^{2} + 48 x + y ^{2} + 160 - 8 y}{4} = \frac{4 x ^{2}}{4} + \frac{48 x}{4} + \frac{y ^{2}}{4} + \frac{160}{4} - \frac{8 y}{4}

= \frac{4 x ^{2}}{4} + \frac{48 x}{4} + \frac{y ^{2}}{4} + \frac{160}{4} - \frac{8 y}{4}

\frac{4 x ^{2}}{4} = x^{2}

\frac{48 x}{4} = 12 x

\frac{160}{4} = 40

\frac{8 y}{4} = 2 y

= x^{2} + 12 x + \frac{y ^{2}}{4} + 40 - 2 y

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\frac{25 ( s + 0.03 ) ( s + 1 )}{( s + 0.4 ) ( s ^{2} - 0.36 s + 0.16 )}

\frac{( x + 8 ) ( x - 6 )}{2}

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

What is expand ((2x+12)^2)/4+((y-4)^2)/4 ?
The solution to expand ((2x+12)^2)/4+((y-4)^2)/4 is x^2+12x+(y^2)/4+40-2y