Question

if two zeroes of x^3+x^2-5 x-5 are under root 5 and under root -5 then find its third zero
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Answer 1

let the third root be a.
roots are:  √5,  √-5 or  √5 i    which is a complex number.
we know that complex number roots appear in pairs.  The other root is a complex conjugate of the other root.  hence, the third root is :  - √5 i  or, -√-5

x³ + x² - 5 x - 5 = (x - √5) (x - i√5) (x + i √5 )
                      = (x - √5) (x² + 5)
                     = x³ - √5 x² + 5 x - 5√5          ---- (2)

  LHS and RHS  are not equal.  There is some discrepancy in the root  √-5 .  It is not a root of the given expression.

polynomial in  (2) has roots  √5 and √-5.
============================
Here we find the roots of the given equation:

let
x³ + x² - 5 x - 5 = (x - √5) (x² - a x + √5)
  find a from equating the coefficients.
     = (x - √5) (x² + (1+√5) x + √5)
 
      x² + (1+√5) x + √5 =                  ----- (1)
            roots are = [ - 1-√5 + - √(1+5+2√5-4√5)  ] / 2
                         = [ - 1 - √5  +- √(6-2√5) ] / 2

finding square root of  6 - 2 √5:

6 - 2 √5 = (a + b √5)² = a² + 5 b² + 2 a b √5
       a² + 5 b² = 6
       a b = - 1
       a² + 5/a² = 6
        a⁴ - 6 a² + 5 = 0      =>  a² = -5  or  -1
                 a = √5 i  or  i    or -√5 i  or - i
                 b = i /√5,  -i,  or  - i /√5    or  i

     a + b√5 = (√5+1) i ,  - (√5-1) i ,  -(√5+1) i  ,  (√5 - 1) i

So roots of polynomial in (1) are: - (1+√5) (1 + - i)  /2
                     or,     [ -1 -√5 + -  (√5 - 1) i ] /2

For the given equation: the roots are :
         √5,    - (1+√5) (1 + - i)  / 2,    [ -1 -√5 + -  (√5 - 1) i ] /2