Find all the real zeros of the polynomial

Question

Find all the real zeros of the polynomial

P(x)=x⁴-4x³-6x²+4x+5

its real zeros are 

x1=

x2=

x3=

x4=

with x1≤x2≤x3≤x4

Answer 1

Possible degree-1 factors with rational coefficients are of the form (x+b) where b = ±1 or ±5

 

P(1) = 0, p(-1) = 0, p(5) = 0,  so 1, -1 and 5 are roots.   

 

dividing P(x) by (x-1)(x+1)(x-5) give (x+1),  so the root x = -1 occurs twice.

 

x1 = -1, x2= -1, x3 = 1, x4 = 5.

 

 

Answer 2

P( X ) = X^4 - 4X^3 - 6X^2 + 4X +5       X1 + X2 + X3 + X4 = 4

                                                             X1 + X2 + X3 +X4 = 5      ( 1)

                                                            

                                                            Given :

                                                              aX^4 + bX^3 + c X^2 + dX + e

                                                              

                                                                Roots: X1,X2, X3, X4

                                                               

                                                             X1 +X2 +x3 + X4 =-b/a

                                                              

                                                               X1 . X2 . X3 . X4 = c/a

 

 Roots   ±1 , ±5

   

     Test:

 

 P( -1) = ( -1)^4 - 4( -1) ^3 - 6 ( -1)^2 + 4 ( -1) +5

 

         = 1 + 4 - 6 -4 + 5 =0

 

      X1 = -1

 

 P( 1) = ( 1) ^4 - 4 ( 1)^3 - 6( 1)^2 + 4 ( 1) +5 =

        

             =  1 - 4 - 6 +4 + 5 =0

 

    From Equation1 :  X3 + X4 = 4 ,   X3 . X4 = -5

 

        then:

                      ( X3 +5) ( X4 - 1 ) = 0

 

                  X 3 = -5    X4 = 1

 

           X3 =-5 < X1= -1< X2, X4 = 1 

 

      X2 = +1