Question

$rational\:roots\:x^3-7x+6$

solve

Answer 1

$\mathrm{Rational\:root\:test}:\quad x^3-7x+6\quad :\quad x=1,\:x=2,\:x=-3$

$\mathrm{For\:a\:polynomial\:equation\:with\:integer\:coefficients:\quad }a_nx^n+a_{n-1}x^{n-1}+\ldots +a_0$

$a_0=6,\:\quad a_n=1$

$\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:2,\:3,\:6,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1$

$\mathrm{The\:following\:rational\:numbers\:are\:candidate\:roots:}\quad \pm \frac{1,\:2,\:3,\:6}{1}$

$\mathrm{Validate\:the\:roots\:by\:plugging\:them\:into}\:x^3-7x+6=0:\quad x=1,\:x=2,\:x=-3$

$x=1,\:x=2,\:x=-3$

Answer 2

Explanation:

$\sf \: rational\:roots\:x^3-7x+6 \\ \\ \sf \to \: x( {x}^{2} - 7) + 6 \\ \\ \sf \to \: \: (x + 6)( {x}^{2} - 7) \\ \\ \sf \to \red{x \: = - 6 \: and \: x = \sqrt{7} }$