What is expand ((4n+2)(2n+2)(4n+3))/9 ?

The solution to expand ((4n+2)(2n+2)(4n+3))/9 is (32n^3)/9+8n^2+(52n)/9+4/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand ((4n+2)(2n+2)(4n+3))/9

expand

\frac{( 4 n + 2 ) ( 2 n + 2 ) ( 4 n + 3 )}{9}

\frac{32 n ^{3}}{9} + 8 n^{2} + \frac{52 n}{9} + \frac{4}{3}

Solution steps

\frac{( 4 n + 2 ) ( 2 n + 2 ) ( 4 n + 3 )}{9}

Expand

(4 n + 2) (2 n + 2) (4 n + 3) : 32 n^{3} + 72 n^{2} + 52 n + 12

= \frac{32 n ^{3} + 72 n ^{2} + 52 n + 12}{9}

Apply the fraction rule:

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

\frac{32 n ^{3} + 72 n ^{2} + 52 n + 12}{9} = \frac{32 n ^{3}}{9} + \frac{72 n ^{2}}{9} + \frac{52 n}{9} + \frac{12}{9}

= \frac{32 n ^{3}}{9} + \frac{72 n ^{2}}{9} + \frac{52 n}{9} + \frac{12}{9}

Divide the numbers:

\frac{72}{9} = 8

= \frac{32 n ^{3}}{9} + 8 n^{2} + \frac{52 n}{9} + \frac{12}{9}

Cancel

\frac{12}{9} : \frac{4}{3}

= \frac{32 n ^{3}}{9} + 8 n^{2} + \frac{52 n}{9} + \frac{4}{3}

x^{(2)} ((x) d u + (u) d x)

2 e^{- x^{2} - y^{4}} y (- 2 y^{4} + 1)

(a - d) + (c - b)

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

\frac{q ( d - x ) ^{2}}{x ^{2}}

What is expand ((4n+2)(2n+2)(4n+3))/9 ?
The solution to expand ((4n+2)(2n+2)(4n+3))/9 is (32n^3)/9+8n^2+(52n)/9+4/3