What is expand ((2y-3x+7)(2y+3x+1))/4 ?

The solution to expand ((2y-3x+7)(2y+3x+1))/4 is y^2+4y-(9x^2)/4+(9x)/2+7/4

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expand ((2y-3x+7)(2y+3x+1))/4

expand

\frac{( 2 y - 3 x + 7 ) ( 2 y + 3 x + 1 )}{4}

y^{2} + 4 y - \frac{9 x ^{2}}{4} + \frac{9 x}{2} + \frac{7}{4}

Solution steps

\frac{( 2 y - 3 x + 7 ) ( 2 y + 3 x + 1 )}{4}

Expand

(2 y - 3 x + 7) (2 y + 3 x + 1) : 4 y^{2} + 16 y - 9 x^{2} + 18 x + 7

= \frac{4 y ^{2} + 16 y - 9 x ^{2} + 18 x + 7}{4}

Apply the fraction rule:

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

\frac{4 y ^{2} + 16 y - 9 x ^{2} + 18 x + 7}{4} = \frac{4 y ^{2}}{4} + \frac{16 y}{4} - \frac{9 x ^{2}}{4} + \frac{18 x}{4} + \frac{7}{4}

= \frac{4 y ^{2}}{4} + \frac{16 y}{4} - \frac{9 x ^{2}}{4} + \frac{18 x}{4} + \frac{7}{4}

Divide the numbers:

\frac{4}{4} = 1

= y^{2} + \frac{16 y}{4} - \frac{9 x ^{2}}{4} + \frac{18 x}{4} + \frac{7}{4}

Divide the numbers:

\frac{16}{4} = 4

= y^{2} + 4 y - \frac{9 x ^{2}}{4} + \frac{18 x}{4} + \frac{7}{4}

Cancel

\frac{18 x}{4} : \frac{9 x}{2}

= y^{2} + 4 y - \frac{9 x ^{2}}{4} + \frac{9 x}{2} + \frac{7}{4}

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What is expand ((2y-3x+7)(2y+3x+1))/4 ?
The solution to expand ((2y-3x+7)(2y+3x+1))/4 is y^2+4y-(9x^2)/4+(9x)/2+7/4