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

The solution to expand ((n^2-3n-2)^2)/4 is (n^4)/4-(3n^3)/2+(5n^2)/4+3n+1

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expand ((n^2-3n-2)^2)/4

expand

\frac{( n ^{2} - 3 n - 2 ) ^{2}}{4}

\frac{n ^{4}}{4} - \frac{3 n ^{3}}{2} + \frac{5 n ^{2}}{4} + 3 n + 1

Solution steps

\frac{( n ^{2} - 3 n - 2 ) ^{2}}{4}

(n^{2} - 3 n - 2)^{2} = n^{4} - 6 n^{3} + 5 n^{2} + 12 n + 4

= \frac{n ^{4} - 6 n ^{3} + 5 n ^{2} + 12 n + 4}{4}

Apply the fraction rule:

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

\frac{n ^{4} - 6 n ^{3} + 5 n ^{2} + 12 n + 4}{4} = \frac{n ^{4}}{4} - \frac{6 n ^{3}}{4} + \frac{5 n ^{2}}{4} + \frac{12 n}{4} + \frac{4}{4}

= \frac{n ^{4}}{4} - \frac{6 n ^{3}}{4} + \frac{5 n ^{2}}{4} + \frac{12 n}{4} + \frac{4}{4}

Apply rule

\frac{a}{a} = 1

\frac{4}{4} = 1

= \frac{n ^{4}}{4} - \frac{6 n ^{3}}{4} + \frac{5 n ^{2}}{4} + \frac{12 n}{4} + 1

Cancel

\frac{6 n ^{3}}{4} : \frac{3 n ^{3}}{2}

= \frac{n ^{4}}{4} - \frac{3 n ^{3}}{2} + \frac{5 n ^{2}}{4} + \frac{12 n}{4} + 1

Divide the numbers:

\frac{12}{4} = 3

= \frac{n ^{4}}{4} - \frac{3 n ^{3}}{2} + \frac{5 n ^{2}}{4} + 3 n + 1

\frac{n ( 3 n + 5 )}{3}

(x) (1 - x^{2})

\frac{1 - 2 ( x + h ) - 1 - 2 x}{h}

5 \cdot (2 x + 3) x + 54

(4 u + v) (4 u - v)

What is expand ((n^2-3n-2)^2)/4 ?
The solution to expand ((n^2-3n-2)^2)/4 is (n^4)/4-(3n^3)/2+(5n^2)/4+3n+1