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

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

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

expand

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

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

Solution steps

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

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

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

Expand

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

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

Apply the fraction rule:

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

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

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

Apply rule

\frac{a}{a} = 1

\frac{4}{4} = 1

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

Cancel

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

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

Divide the numbers:

\frac{4}{4} = 1

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

(C x + D) (x^{2} + 2 x + 1)

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10 (\frac{x ^{2}}{36100})

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