Question 78251
<pre>
Problem: Find all the real zeros of the polynomial. Use the quadratic formula if necessary.

P(x)= x^3 + 4x^2 + 3x -2


To get the possible zeros of the polynomials p/q can help us, where p= 2
it is the last term. q = 1, this is the coefficient of the first term

We will factor p and q
p = 1*2 and -1*-2
  = 1, 2, -1, -2

q = 1*1 and -1*-1
  = 1, -1


possible zeros p/q are 1, -1, 2, -2

Test using synthetic division which among them are the real zeros of the polynomial

1   |   1    4     3    -2
             1     5     8
    _______________________
        1    5     8    6    --- > not a real zero


-1  |   1   4    3    -2
           -1   -3     0
    _____________________
        1   3    0    -2    ==> Not a real zero


2  |   1    4    3    -2
            2    12   30
   _____________________
       1    6    15   28   ====> not a real zero


-2    |  1   4   3   -2
            -2  -4    2
      _________________
         1   2  -1   0    ===> a Real zero

the factor now is (x + 2) (x^2 + 2x - 1)
                              |
                               --> we can use quadratic formula to get the factor for this where a = 1, b = 2 and c = -2

             {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

             {{{x = (-2 +- sqrt( 2^2-4*1*(-1) ))/(2*1) }}}     
               
             {{{x = (-2 +- sqrt( 4 +4 ))/(2) }}}     

             {{{x = (-2 +- sqrt(8))/(2) }}}

             {{{x = (-2 +- 2 (sqrt(2)))/(2) }}}          

             {{{x = -1 +- 1 (sqrt(2))}}}  

             {{{x = -1 +- sqrt(2)}}}  

There are 3 real zeros of the polynomial

   x= -2, 

   {{{x = -1 + sqrt(2)}}}, 

   {{{x = -1 - sqrt(2)}}}  




Rmromero