Question 696202
The rational roots of {{{P(x)=2x^3+2x^2-19x+20}}} (if they exist) will be of the form
{{{p/q}}} and {{{-p/q}}} where {{{p}}} is a factor of the independent term {{{20}}} and {{{q}}} is a factor of the leading coefficient {{{2}}}.
Factors of {{{20}}} : {{{1}}}, {{{2}}}, {{{4}}}, {{{5}}}, {{{10}}}, {{{20}}}. Factors of {{{2}}} : {{{1}}}, {{{2}}}.
Possible rational roots: {{{-20}}}, {{{-10}}}, {{{-5}}}, {{{-4}}}, {{{-2}}}, {{{-1}}}, {{{-5/2}}}, {{{-1/2}}}, {{{1/2}}}, {{{5/2}}}, {{{1}}}, {{{2}}}, {{{4}}}, {{{5}}}, {{{10}}}, and {{{20}}}.
I tried them, and found that {{{P(-4)=0}}}, and that
{{{2x^3+2x^2-19x+20-(x+4)(2x^2-6x+5))}}}
The roots would be {{{highlight(x=-4)}}},
and the solutions to {{{2x^2-6x+5=0}}}.
However, {{{2x^2-6x+5=0}}} has no real solution.
If you are studying complex numbers, you could find another two complex roots.