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

The solution to expand ((2x+3)(3x-1))/2 is 3x^2+(7x)/2-3/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

expand

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

3 x^{2} + \frac{7 x}{2} - \frac{3}{2}

Solution steps

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

Expand

(2 x + 3) (3 x - 1) : 6 x^{2} + 7 x - 3

= \frac{6 x ^{2} + 7 x - 3}{2}

Apply the fraction rule:

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

\frac{6 x ^{2} + 7 x - 3}{2} = \frac{6 x ^{2}}{2} + \frac{7 x}{2} - \frac{3}{2}

= \frac{6 x ^{2}}{2} + \frac{7 x}{2} - \frac{3}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3 x^{2} + \frac{7 x}{2} - \frac{3}{2}

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

\frac{5}{( 2 x - 4 ) ^{2}}

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

\frac{1}{2} (e^{t} + e^{- t})

(A e^{- k x} + B e^{k x})^{2}

What is expand ((2x+3)(3x-1))/2 ?
The solution to expand ((2x+3)(3x-1))/2 is 3x^2+(7x)/2-3/2