What is expand ((2+x)/((10x^2-1)))(1000x^6-1) ?

The solution to expand ((2+x)/((10x^2-1)))(1000x^6-1) is 100x^5+10x^3+x+200x^4+20x^2+2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand ((2+x)/((10x^2-1)))(1000x^6-1)

expand

(\frac{2 + x}{( 10 x ^{2} - 1 )}) (1000 x^{6} - 1)

100 x^{5} + 10 x^{3} + x + 200 x^{4} + 20 x^{2} + 2

Solution steps

(\frac{2 + x}{10 x ^{2} - 1}) (1000 x^{6} - 1)

Apply rule:

(a) = a

(\frac{2 + x}{10 x ^{2} - 1}) = \frac{2 + x}{10 x ^{2} - 1}

= \frac{2 + x}{10 x ^{2} - 1} (1000 x^{6} - 1)

Apply the distributive law:

a (b - c) = ab - a c

\frac{2 + x}{10 x ^{2} - 1} (1000 x^{6} - 1) = \frac{2 + x}{10 x ^{2} - 1} \cdot 1000 x^{6} - \frac{2 + x}{10 x ^{2} - 1} \cdot 1

= \frac{2 + x}{10 x ^{2} - 1} \cdot 1000 x^{6} - \frac{2 + x}{10 x ^{2} - 1} \cdot 1

Simplify

\frac{2 + x}{10 x ^{2} - 1} \cdot 1000 x^{6} - \frac{2 + x}{10 x ^{2} - 1} \cdot 1 : 100 x^{5} + 10 x^{3} + x + 200 x^{4} + 20 x^{2} + 2

= 100 x^{5} + 10 x^{3} + x + 200 x^{4} + 20 x^{2} + 2

\frac{48 x ^{2}}{( x - 15 ) ( x + 5 ) ^{2}}

(sin (2 x) + 1)^{2}

4 (6 - x - 2 (3 - x - 5 = 9 (5 - 2 x)))

\frac{2 x - 6}{3 ( x ^{2} - 6 x + 5 ) ^{\frac{2}{3}}}

What is expand ((2+x)/((10x^2-1)))(1000x^6-1) ?
The solution to expand ((2+x)/((10x^2-1)))(1000x^6-1) is 100x^5+10x^3+x+200x^4+20x^2+2