What is expand 1/2 x^4-5/3 x^3-1/2 x^2+6x+1 ?

The solution to expand 1/2 x^4-5/3 x^3-1/2 x^2+6x+1 is (x^4)/2-(5x^3)/3-(x^2)/2+6x+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand 1/2 x^4-5/3 x^3-1/2 x^2+6x+1

expand

\frac{1}{2} x^{4} - \frac{5}{3} x^{3} - \frac{1}{2} x^{2} + 6 x + 1

\frac{x ^{4}}{2} - \frac{5 x ^{3}}{3} - \frac{x ^{2}}{2} + 6 x + 1

Solution steps

\frac{1}{2} x^{4} - \frac{5}{3} x^{3} - \frac{1}{2} x^{2} + 6 x + 1

\frac{1}{2} x^{4} = \frac{x ^{4}}{2}

\frac{5}{3} x^{3} = \frac{5 x ^{3}}{3}

\frac{1}{2} x^{2} = \frac{x ^{2}}{2}

= \frac{x ^{4}}{2} - \frac{5 x ^{3}}{3} - \frac{x ^{2}}{2} + 6 x + 1

Combine the fractions

\frac{x ^{4}}{2} - \frac{x ^{2}}{2} : \frac{x ^{4} - x ^{2}}{2}

= \frac{x ^{4} - x ^{2}}{2} - \frac{5 x ^{3}}{3} + 6 x + 1

Apply the fraction rule:

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

\frac{x ^{4} - x ^{2}}{2} = \frac{x ^{4}}{2} - \frac{x ^{2}}{2}

= \frac{x ^{4}}{2} - \frac{x ^{2}}{2} - \frac{5 x ^{3}}{3} + 6 x + 1

Rewrite in standard form

= \frac{x ^{4}}{2} - \frac{5 x ^{3}}{3} - \frac{x ^{2}}{2} + 6 x + 1

0.12 + \frac{0.12}{3.5 s} + 0.105 s

4 (2 +, + (2 + 5 i) (6 - 2 i))

\frac{x ( x - 3 )}{x + 4}

\frac{( 20 - t ) ( - t + 10 )}{400}

2 (3 x + 6 y = 33)

What is expand 1/2 x^4-5/3 x^3-1/2 x^2+6x+1 ?
The solution to expand 1/2 x^4-5/3 x^3-1/2 x^2+6x+1 is (x^4)/2-(5x^3)/3-(x^2)/2+6x+1