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

The solution to expand ((2x+3)^2-x^4)/2 is -(x^4)/2+2x^2+6x+9/2

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expand ((2x+3)^2-x^4)/2

expand

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

- \frac{x ^{4}}{2} + 2 x^{2} + 6 x + \frac{9}{2}

Solution steps

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

Expand

(2 x + 3)^{2} - x^{4} : 4 x^{2} + 12 x + 9 - x^{4}

= \frac{4 x ^{2} + 12 x + 9 - x ^{4}}{2}

Apply the fraction rule:

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

\frac{4 x ^{2} + 12 x + 9 - x ^{4}}{2} = \frac{4 x ^{2}}{2} + \frac{12 x}{2} + \frac{9}{2} - \frac{x ^{4}}{2}

= \frac{4 x ^{2}}{2} + \frac{12 x}{2} + \frac{9}{2} - \frac{x ^{4}}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2 x^{2} + \frac{12 x}{2} + \frac{9}{2} - \frac{x ^{4}}{2}

Divide the numbers:

\frac{12}{2} = 6

= 2 x^{2} + 6 x + \frac{9}{2} - \frac{x ^{4}}{2}

Rewrite in standard form

= - \frac{x ^{4}}{2} + 2 x^{2} + 6 x + \frac{9}{2}

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e^{(x^{2} - 3) (x + 4)}

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\frac{3 x ( x ^{2} - 5 )}{x ^{2} - 1}

\frac{40 ( 4 cos ( 40 t ) - sin ( 40 t ) )}{17}

What is expand ((2x+3)^2-x^4)/2 ?
The solution to expand ((2x+3)^2-x^4)/2 is -(x^4)/2+2x^2+6x+9/2