Question

Add:

$2,\frac{2}{3}x-\frac{5}{3}{x}^2+\frac{5}{2}{x}^3,\frac{-4}{3}+\frac{2x^2}{3}-\frac{x}{2},\frac{5x^3}{3}+3x^2+3x+\frac{6}{5}$.

Answer 1

So we are given,

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

$=2+(\frac{5}{2}{x}^3-\frac{5}{3}{x}^2+\frac{2}{3}x)+(\frac{2}{3}{x}^2-\frac{x}{2}-\frac{4}{3})+(\frac{5}{3}{x}^3+3x^2+3x+\frac{6}{5})$

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

$=\left(\frac{5\left(3\right)+5\left(2\right)}{3.2}\right)x^2+\left(\frac{-5\left(1\right)+2\left(1\right)+3\left(3\right)}{3}\right)x^2+\left(\frac{2\left(2\right)-1\left(3\right)+3\left(6\right)}{3.2}\right)x+\left(\frac{2\left(15\right)-4\left(5\right)+6\left(3\right)}{3.5}\right)$

$=\frac{25}{6}{x}^3+2x^2+\frac{19}{6}x+\frac{28}{15}$