Question

Multiply: $(\frac{1}{3}{x}^2-\frac{1}{2}x+5)$ by $(\frac{1}{2}{x}^2-\frac{1}{3}x+1)$

Answer 1

$\left(\frac{1}{3}{x}^2-\frac{1}{2}x+5\right)\left(\frac{1}{2}{x}^2-\frac{1}{3}x+1\right)$

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

$=\frac{1}{6}{x}^4-\frac{1}{9}{x}^3+\frac{1}{3}{x}^2-\frac{1}{4}{x}^3+\frac{1}{6}{x}^2-\frac{1}{2}x+\frac{5}{2}{x}^2-\frac{5}{3}x=5$

$=\frac{1}{6}{x}^4-\left(\frac{1}{9}+\frac{1}{4}\right)x^3+\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{5}\right)x^2-\left(\frac{1}{2}+\frac{5}{3}\right)x+5$

$=\frac{x^4}{6}-\frac{13}{36}{x}^3+\frac{18}{6}{x}^2-\frac{13}{6}x+5$