What is expand ((k+1)^2(k+2)^2)/4 ?

The solution to expand ((k+1)^2(k+2)^2)/4 is (k^4)/4+(3k^3)/2+(13k^2)/4+3k+1

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expand ((k+1)^2(k+2)^2)/4

expand

\frac{( k + 1 ) ^{2} ( k + 2 ) ^{2}}{4}

\frac{k ^{4}}{4} + \frac{3 k ^{3}}{2} + \frac{13 k ^{2}}{4} + 3 k + 1

Solution steps

\frac{( k + 1 ) ^{2} ( k + 2 ) ^{2}}{4}

(k + 1)^{2} = k^{2} + 2 k + 1

= \frac{( k ^{2} + 2 k + 1 ) ( k + 2 ) ^{2}}{4}

(k + 2)^{2} = k^{2} + 4 k + 4

= \frac{( k ^{2} + 2 k + 1 ) ( k ^{2} + 4 k + 4 )}{4}

Expand

(k^{2} + 2 k + 1) (k^{2} + 4 k + 4) : k^{4} + 6 k^{3} + 13 k^{2} + 12 k + 4

= \frac{k ^{4} + 6 k ^{3} + 13 k ^{2} + 12 k + 4}{4}

Apply the fraction rule:

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

\frac{k ^{4} + 6 k ^{3} + 13 k ^{2} + 12 k + 4}{4} = \frac{k ^{4}}{4} + \frac{6 k ^{3}}{4} + \frac{13 k ^{2}}{4} + \frac{12 k}{4} + \frac{4}{4}

= \frac{k ^{4}}{4} + \frac{6 k ^{3}}{4} + \frac{13 k ^{2}}{4} + \frac{12 k}{4} + \frac{4}{4}

Apply rule

\frac{a}{a} = 1

\frac{4}{4} = 1

= \frac{k ^{4}}{4} + \frac{6 k ^{3}}{4} + \frac{13 k ^{2}}{4} + \frac{12 k}{4} + 1

Cancel

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

= \frac{k ^{4}}{4} + \frac{3 k ^{3}}{2} + \frac{13 k ^{2}}{4} + \frac{12 k}{4} + 1

Divide the numbers:

\frac{12}{4} = 3

= \frac{k ^{4}}{4} + \frac{3 k ^{3}}{2} + \frac{13 k ^{2}}{4} + 3 k + 1

\frac{6}{2} (1 + 2)

\frac{2 x}{( x ^{2} + 1 ) ^{2}}

lo g_{10} (\frac{y z ^{3}}{x ^{5}})

δ (x - y)

(3 x - 5) (3 x + 2) (9 x^{2} - 9 x - 10)

What is expand ((k+1)^2(k+2)^2)/4 ?
The solution to expand ((k+1)^2(k+2)^2)/4 is (k^4)/4+(3k^3)/2+(13k^2)/4+3k+1