Find all the real zeros of the polynomial

Question

find all the real zeros of the polynomial

P(x)=x³-5x²-15x+7

its real zeros are 

x1=

x2=

x3=

with x1≤x2≤x3

Answer 1

 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

Answer 2

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