What is expand ((2-y^2)^2)/8 ?

The solution to expand ((2-y^2)^2)/8 is (y^4)/8-(y^2)/2+1/2

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expand ((2-y^2)^2)/8

expand

\frac{( 2 - y ^{2} ) ^{2}}{8}

\frac{y ^{4}}{8} - \frac{y ^{2}}{2} + \frac{1}{2}

Solution steps

\frac{( 2 - y ^{2} ) ^{2}}{8}

(2 - y^{2})^{2} = 4 - 4 y^{2} + y^{4}

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

Apply the fraction rule:

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

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

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

Cancel

\frac{4}{8} : \frac{1}{2}

= \frac{1}{2} - \frac{4 y ^{2}}{8} + \frac{y ^{4}}{8}

Cancel

\frac{4 y ^{2}}{8} : \frac{y ^{2}}{2}

= \frac{1}{2} - \frac{y ^{2}}{2} + \frac{y ^{4}}{8}

Rewrite in standard form

= \frac{y ^{4}}{8} - \frac{y ^{2}}{2} + \frac{1}{2}

(3 x^{2} - \frac{1}{x})^{9}

- \frac{( 8 x )}{( x ^{2} - 4 ) ^{2}}

\frac{( s + 3 )}{s ( s + 2 ) ( s + 1 )}

- \frac{4}{s ( s ^{2} - 6 )}

\frac{50}{s ( s + 3 ) ( s + 6 )}

What is expand ((2-y^2)^2)/8 ?
The solution to expand ((2-y^2)^2)/8 is (y^4)/8-(y^2)/2+1/2