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.

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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