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Find all the real zeros of the polynomial

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2 Answers

 P( X)= X^3 - 5 X^2 - 15X +7      X1 + X2 + X3 = 5
 
                                                  X1 . X2 . X3 =7
 
 
       Rational roots:
 
       ±1 ,± 7
 
     P( 7) = 7^3 - 5 ( 7^2) -15*7 +7
    
              = 343 - 245 - 105  +7 = 0
 
            X3 = 7
 
                   X^2 + 2X -1                   
         X - 7 l X^3 -5 X^2 - 15X +7
                   X^3 ± 7X^2        
                           2X^2 - 15X
                           2 X^2 ± 14x
                                       -x +7
                                       - X+7 
                                          0
'     X^2 + 2X -1 =
 
       X = -2/2 ±√(8 ) /2        X1 = -1 - √2  , X2 = -1 +√2
 
       X1< X2 < X3
 

        -1 - √2 ‹ -1 + √2 < 7

If the polynomial has degree-1 factors with rational coefficients, then they are (x ± 1) or (x ± 7).
 
Substituting 
 
P(7) = 0, but the other values are not zeros.
 
Dividing P(x) by (x-7) gives x2 + 2x - 1 which has roots -1 ± √2
 
So the zeros are
 
x1 = -1 - √2
 
x2  = -1 + √2
 
x3  =  7