Question 987565
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The roots of the polynomial  f(x) = {{{2x^3 - 5x^2 + 4x -1}}}  are  1  (of the multiplicity  2)  and   {{{1/2}}}.


The polynomial is factored as  {{{2x^3 - 5x^2 + 4x -1}}} = {{{(x-1)^2*(2x-1)}}}.  You can check it directly.


The plot of the polynomial is shown in &nbsp;<B>Figures 1</B> &nbsp;and &nbsp;<B>2</B>.


How to get these results? &nbsp;You can guess that &nbsp;1&nbsp; is the root. 


After this is done, &nbsp;you can make long division by dividing the given polynomial by the binomial &nbsp;(x-1). 

In this way you will be able to decrease the degree of the polynomial from &nbsp;3&nbsp; to &nbsp;2. 

Then you can find the roots of the obtained quadratic polynomial.

<TABLE> 
  <TR>
  <TD> 

{{{graph( 330, 400, -5.5, 5.5, -10.5, 10.5,
          2x^3 - 5x^2 + 4x -1
)}}}


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<B>Figure 1</B>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

 </TD>   
  <TD> 

{{{graph( 330, 400, -5.5, 5.5, -1.1, 1.1,
          2x^3 - 5x^2 + 4x -1
)}}}


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<B>Figure 2</B>
  </TD>
  </TR>
</TABLE>

Could you make the table of signs yourself?


Notice that since the polynomial contains the factor &nbsp;(x-1) &nbsp;in degree &nbsp;2, &nbsp;the original polynomial does not change the sign at the vicinity of the point &nbsp;x=1.