Question 362572
Look at the factors of the leading coefficient, {{{1}}} and the constant coefficient {{{16}}}.
{{{q}}} : Factors of 1 : {1}
{{{p}}} : Factors of 16 : {1,2,4,8,16}
Possible zeros are of the form {{{p/q}}} 
{1/1,2/1,4/1,8/1,16/1} and their negative counterparts,
{-1,-2,-4,-8,-16}
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{{{graph(300,300,-10,10,-10,10,x^3+2x^2-8x-16)}}}
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Looks like {{{x=-2}}} is a possible candidate.
Verify using the equation.
{{{f(-2)=(-2)^3+2(-2)^2-8(-2)-16}}}
{{{f(-2)=-8+8+16-16}}}
{{{f(-2)=0}}}
The other two roots are therefore irrational.