What is expand ((x+1)(2x+1)(2x+3))/3 ?

The solution to expand ((x+1)(2x+1)(2x+3))/3 is (4x^3)/3+4x^2+(11x)/3+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand ((x+1)(2x+1)(2x+3))/3

expand

\frac{( x + 1 ) ( 2 x + 1 ) ( 2 x + 3 )}{3}

\frac{4 x ^{3}}{3} + 4 x^{2} + \frac{11 x}{3} + 1

Solution steps

\frac{( x + 1 ) ( 2 x + 1 ) ( 2 x + 3 )}{3}

Expand

(x + 1) (2 x + 1) (2 x + 3) : 4 x^{3} + 12 x^{2} + 11 x + 3

= \frac{4 x ^{3} + 12 x ^{2} + 11 x + 3}{3}

Apply the fraction rule:

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

\frac{4 x ^{3} + 12 x ^{2} + 11 x + 3}{3} = \frac{4 x ^{3}}{3} + \frac{12 x ^{2}}{3} + \frac{11 x}{3} + \frac{3}{3}

= \frac{4 x ^{3}}{3} + \frac{12 x ^{2}}{3} + \frac{11 x}{3} + \frac{3}{3}

Apply rule

\frac{a}{a} = 1

\frac{3}{3} = 1

= \frac{4 x ^{3}}{3} + \frac{12 x ^{2}}{3} + \frac{11 x}{3} + 1

Divide the numbers:

\frac{12}{3} = 4

= \frac{4 x ^{3}}{3} + 4 x^{2} + \frac{11 x}{3} + 1

\frac{( y + 3 ) ( x - 1 )}{( y - 2 ) ( x + 4 )}

(2 x + y = 5,) (- x + y = 2)

\frac{4 s - 2}{( 4 s ^{2} - 4 s + 5 ) ^{2}}

l imh \Rightarrow 0 (8 + h) 2 - 64 h

\frac{4 x + 1}{( x + 1 ) ^{2} ( x - 2 )}

What is expand ((x+1)(2x+1)(2x+3))/3 ?
The solution to expand ((x+1)(2x+1)(2x+3))/3 is (4x^3)/3+4x^2+(11x)/3+1