Question 977664


<B>Solution 1</B>


The polynomial &nbsp;{{{y}}} = {{{x^3 - 2x^2 - 25x +50}}}&nbsp; has the real root &nbsp;{{{x}}} = {{{2}}}. 

Indeed, &nbsp;it is easy to check: &nbsp;{{{2^3 -2*2^2 - 25*2 + 50}}} = {{{0}}}.


Hence, &nbsp;this polynomial can be factored 


{{{x^3 - 2x^2 - 25 x +50}}} = {{{(x-2)}}}.{{{f(x)}}} 


with he linear factor &nbsp;{{{x-2}}}, &nbsp;where &nbsp;{{{f(x)}}}&nbsp; is a quadratic polynomial. 


To find &nbsp;{{{f(x)}}}, &nbsp;divide the polynomial &nbsp;{{{x^3 - 2x^2 - 25x +50}}}&nbsp; by &nbsp;{{{x-2}}} &nbsp;(long division). 

If you perform this division, &nbsp;you will get 


{{{x^3 - 2x^2 - 25x +50}}} = {{{(x-2)}}}.{{{(x^2 - 25)}}}.


Hence, &nbsp;two other roots of the given polynomial are the roots of the quadratic polynomial &nbsp;{{{x^2 - 25}}}, &nbsp;i.e. &nbsp;{{{x[1]}}} = {{{5}}}&nbsp; and &nbsp;{{{x[2]}}} = {{{-5}}}.


<B>Answer</B>. &nbsp;The roots of the given polynomial &nbsp;{{{x^3 - 2x^2 - 25x +50}}}&nbsp; are &nbsp;2, &nbsp;5&nbsp; and &nbsp;-5.



<B>Solution 2</B>


You can get factoring the polynomial &nbsp;{{{x^3 - 2 x^2 - 25 x +50}}}&nbsp; by grouping its terms:


{{{x^3 - 2x^2 - 25x +50}}} = {{{(x^3 - 2x^2)}}} - {{{(25x -50)}}} = {{{x^2(x-2)}}} - {{{25(x-2)}}} = {{{(x^2-25)*(x-2)}}}. 


It gives you the same roots.