Question 982451


To find zeroes of a polynomial 


{{{y}}} = {{{(x^2+2x-7)}}}.{{{(x^3+4x^2-21x)}}}


you need to find zeroes of the first factor  {{{(x^2+2x-7)}}}  and the the zeroes of the second factor  {{{(x^3+4x^2-21x)}}}. 


The first factor  {{{(x^2+2x-7)}}}  has the zeroes  {{{x[1]}}} = {{{-1 + 2sqrt(2)}}}  and  {{{x[2]}}} = {{{-1 - 2sqrt(2)}}}  (use the quadratic formula). 


The second factor is  {{{(x^3+4x^2-21x)}}} = {{{x}}}.{{{(x^2+4x-21)}}}. 


It has the zeroes  {{{x[3]}}} = {{{0}}},  {{{x[4]}}} = {{{4}}},  and  {{{x[5]}}} = {{{-7}}}. 


The first of these three zeroes is obvious  ({{{x[3]}}} = {{{0}}}).


The two others of these three zeroes you can obtain using the quadratic formula or the Viete's theorem 

(see the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/quadratic/lessons/Solving-quadratic-equations-without-quadratic-formula.lesson>Solving quadratic equations without quadratic formula</A>&nbsp; in this site).


<B>Answer</B>. &nbsp;The zeroes are &nbsp;{{{-1 + 2sqrt(2)}}}, &nbsp;{{{-1 - 2sqrt(2)}}}, &nbsp;{{{0}}}, &nbsp;{{{4}}}&nbsp; and &nbsp;{{{-7}}}.