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

The solution to expand ((4x+12)^2)/4+((y-4)^2)/4 is 4x^2+24x+(y^2)/4+40-2y

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

expand

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

4 x^{2} + 24 x + \frac{y ^{2}}{4} + 40 - 2 y

Solution steps

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

Apply rule

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

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

Expand

(4 x + 12)^{2} + (y - 4)^{2} : 16 x^{2} + 96 x + y^{2} + 160 - 8 y

= \frac{16 x ^{2} + 96 x + y ^{2} + 160 - 8 y}{4}

Apply the fraction rule:

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

\frac{16 x ^{2} + 96 x + y ^{2} + 160 - 8 y}{4} = \frac{16 x ^{2}}{4} + \frac{96 x}{4} + \frac{y ^{2}}{4} + \frac{160}{4} - \frac{8 y}{4}

= \frac{16 x ^{2}}{4} + \frac{96 x}{4} + \frac{y ^{2}}{4} + \frac{160}{4} - \frac{8 y}{4}

\frac{16 x ^{2}}{4} = 4 x^{2}

\frac{96 x}{4} = 24 x

\frac{160}{4} = 40

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

= 4 x^{2} + 24 x + \frac{y ^{2}}{4} + 40 - 2 y

(t - a) (H (t - a) - H (t - b))

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

x^{2} y^{''} (x) + x (x + 2) y^{'} (x) - (x + 2) y

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

What is expand ((4x+12)^2)/4+((y-4)^2)/4 ?
The solution to expand ((4x+12)^2)/4+((y-4)^2)/4 is 4x^2+24x+(y^2)/4+40-2y