Question 1180205


{{{f(x)= x^3 - 11x^2 + 8x - 20}}}

Use Decartes' rule of signs...

So, the coefficients are {{{1}}},{{{-11}}},{{{8}}},{{{-20}}}

As can be seen, there are {{{3 }}}changes.

This means that there are {{{3 }}}or {{{1 }}}positive real roots.

To find the number of negative real roots, substitute {{{x}}} with {{{-x}}} in the given polynomial: 

{{{x^3 - 11x^2 + 8x - 20}}} becomes  {{{-x^3 - 11x^2 - 8x - 20}}}

The coefficients are {{{-1}}},{{{-11}}},{{{-8}}},{{{-20}}}.

As can be seen, there are{{{ 0}}} changes.

This means that there are{{{ 0}}} negative real roots.

Answer:

{{{3}}} or {{{1}}} positive real zeros 

{{{0}}} negative real zeros

{{{0}}} or {{{2}}}  imaginary zeros