Chaim Ferguson

2022-10-13

How to solve ${r}^{3}-6{r}^{2}+11r-6=0$

n8ar1val

Beginner2022-10-14Added 12 answers

${r}^{3}-6{r}^{2}+11r-6=0\phantom{\rule{0ex}{0ex}}{r}^{3}-6({r}^{2}-2r+1)-r=0\phantom{\rule{0ex}{0ex}}r({r}^{2}-1)-6({r}^{2}-2r+1)=0\phantom{\rule{0ex}{0ex}}r(r-1)(r+1)-6(r-1{)}^{2}=0\phantom{\rule{0ex}{0ex}}(r-1)(r(r+1)-6(r-1))=0\phantom{\rule{0ex}{0ex}}(r-1)({r}^{2}+r-6r+6)=0\phantom{\rule{0ex}{0ex}}(r-1)({r}^{2}-5r+6)=0$

By solving the quadratic equation $({r}^{2}-5r+6)=0$, we get that ${r}_{1,2}=(5\pm 1)/2$, so finally:

$(r-1)(r-2)(r-3)=0$

By solving the quadratic equation $({r}^{2}-5r+6)=0$, we get that ${r}_{1,2}=(5\pm 1)/2$, so finally:

$(r-1)(r-2)(r-3)=0$

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