Question 1129875
{{{f(x)=-3x^5-x^3-4x^4-5}}}

Use the Rational Roots Test to Find All Possible Roots:


find every combination of ±{{{p/q}}} where{{{ p}}} is a factor of the constant and {{{q}}} is a factor of the leading coefficient

The constant term of this polynomial is {{{-5}}}, with factors {{{1}}} and {{{5}}}.

The leading coefficient is {{{-3}}}, with factors {{{1}}} and{{{ 3}}}.



Tests yields the following possible solutions:

±{{{1}}}, ±, {{{1/3}}}, ± {{{5}}}, ± {{{5/3}}}

test them:
{{{f(1)=-3*1^5-1^3-4*1^4-5=-3-1-4-5=-13}}}-> does not work
{{{f(-1)=-3*(-1)^5-(-1)^3-4*(-1)^4-5=3+1-4-5=-5}}}-> does not work
{{{f(1/3)=-3*(1/3)^5-(1/3)^3-4*(1/3)^4-5=-5.099}}}-> does not work
{{{f(-1/3)=-3*(-1/3)^5-(-1/3)^3-4*(-1/3)^4-5=-5}}}-> does not work
{{{f(5)=-3*(5)^5-(5)^3-4*(5)^4-5=-12005}}}-> does not work
{{{f(-5)=-3*(-5)^5-(-5)^3-4*(-5)^4-5=6995}}}-> does not work
{{{f(5/3)=-3*(5/3)^5-(5/3)^3-4*(5/3)^4-5=-2135/27}}}-> does not work
{{{f(-5/3)=-3*(-5/3)^5-(-5/3)^3-4*(-5/3)^4-5=595/81}}}-> does not work


This polynomial has no rational roots that can be found using Rational Root Test.


try to find roots using Newton method:


If {{{x[n] }}}is an approximation a solution of {{{f(x)=0}}} and if {{{f}}}&#8242;{{{(x[n])<>0}}}


the next approximation is given by, {{{x[n+1]=x[n]-f(x[n])}}}/{{{f}}}&#8242;{{{(x[n])}}}


{{{f(x)=-3x^5-x^3-4x^4-5}}}


{{{f}}}'{{{(x)= -x^2 (15 x^2 + 16 x + 3)}}}


{{{(-3x^5-x^3-4x^4-5)/(-x^2 (15 x^2 + 16 x + 3))=0}}}


{{{x}}} &#8776; {{{-1.46644}}}



{{{ graph( 600, 600, -10, 10, -10, 10, -3x^5-x^3-4x^4-5) }}}