Question 176377
If the coefficients of a polynomial are integers, it is natural to look for roots which are also integers. Any such root must divide the constant term.
:
Given{{{ x^3 - 4x^2 + 2x + 1 =0.}}} 
:
our constant term is 1 ...so 1 and -1 are the only possibles
:
lets try 1:{{{(1)^3-4(1)^2+2(1)+1=0}}}so 1 is a factor
:
lets try -1:{{{(-1)^3-4(-1)^2+2(-1)+1=-6}}}so -1 is not a factor
:
This relationship is always true: If a polynomial has rational roots, then those roots will be fractions of the form (plus-or-minus) (factor of the constant term) / (factor of the leading coefficient). However, not all fractions of this form are necessarily zeroes of the polynomial. Indeed, it may happen that none of the fractions so formed is actually a zero of the polynomial.


since 1 is our constant and 1 is our leading coefficient the only possible rational roots are 1 divided by 1 or -1 which is the same as those terms we have already tested.
:<pre>
 
   |
  1|1 -4  2   1
   |___1_-3__-1___________________
    1 -3 -1   0
:</pre>
so when we factor out the root 1 we get
:
{{{x^2-3x-1=0}}} 
:and using the quadratic formula we get 2 irrational roots from this depressed equation

{{{3+sqrt(13)/2}}} and {{{3-sqrt(13)/2}}}

 *[invoke solve_quadratic_equation 1, -3, -1]