Question

Trace the curve of \(\displaystyle{x}^{{3}}-{6}{x}^{{2}}+{11}{x}-{y}={6}\)

Answer 1

\(\displaystyle{x}^{{3}}-{6}{x}^{{2}}+{11}{x}-{y}={6}\)
\(\displaystyle{y}={x}^{{3}}-{6}{x}^{{2}}+{11}{x}-{6}\)
\(\displaystyle{p}{\left({x}\right)}={x}^{{3}}-{6}{x}^{{2}}+{11}{x}-{6}\)
\(\displaystyle{p}{\left({1}\right)}={1}-{6}+{11}-{6}={0}\)
(x-1) is a point of p(x)
Dividing p(x) by (x-2) we get
\(\displaystyle{q}{\left({x}\right)}={x}^{{2}}-{3}{x}-{21}{x}+{6}\)
\(\displaystyle{q}{\left({x}\right)}={x}{\left({x}-{3}\right)}-{2}{\left({x}-{3}\right)}\)
\(\displaystyle{q}{\left({x}\right)}={\left({x}-{2}\right)}{\left({x}-{3}\right)}\)
\(\displaystyle{p}{\left({x}\right)}={\left({x}-{1}\right)}{q}{\left({x}\right)}\)
\(\displaystyle{p}{\left({x}\right)}={\left({x}-{1}\right)}{\left({x}-{2}\right)}{\left({x}-{3}\right)}\)
Point of polynomial are x=1,2,3