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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

expand

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

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

Solution steps

\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{- x ^{2} - 3}{( x ^{2} - 3 ) ^{2}}

\frac{π}{2} (x - 3)

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

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

(y - z^{2}) d y

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