# Given that x=3 is a solution of x^3-x^2-8x+6=0 find all the other solutions of this equation.

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Since an equation could be written as a product of linear factors, when knowing it's roots, we'll write the given equation as:

x^3-x^2-8x+6=(x-3)(ax^2 + bx + c)

We'll remove the brackets from the right side and we'll get:

x^3-x^2-8x+6 = ax^3 + bx^2 + cx - 3ax^2 - 3bx - 3c

We'll combine like terms from the right side and we'll get:

x^3-x^2-8x+6 = ax^3 + x^2(b - 3a) + x(c-3b) - 3c

For the expressions from both sides to be equal, we'll have the followings equalities:

**a = 1**

b - 3a = -1 => b - 3 = -1 => **b = 2**

c - 3b = -8 => c - 3*2 = -8 => c = -8 + 6 => **c = -2**

-3c = 6

c = -2

The equation ax^2 + bx + c = x^2 + 2x - 2

To determine all the solutions of the equation, we'll put x^2 + 2x - 2 = 0.

x^2 + 2x - 2 = 0

We'll apply the quadratic formula:

x1 = [-2+sqrt(4+8)]/2

x1 = (-2+2sqrt3)/2

We'll simplify and we'll get:

x1 = -1 + sqrt3

x2 = -1 - sqrt3

**The solutions of the equation x^3-x^2-8x+6 = 0 are: { -1 - sqrt3 ; -1 + sqrt3 ; 3}.**

Given x= 3 is a root of x^3 -x^2 -8x+6 = 0.

So by remainder theorem x-3 is factor of x^3-x^2-8x+6.

Therefore we divide x^3-x^2-8x-6 by x-3.

x-3)x^3-x^2-8x+6( x^2

x^3 -3x^2

------------------------

x-3 ) 2x^2-8x +6(x^2+2x

2x^2 -6x

------------------------

x-3) -2x +6(x^2+2x-2

-2x +6

-----------------------.

0.

Therfore x^2+2x-2 is also a factor of x^3-x^2-8x+6.

Therefore the zeros of x^2+2x-2 are by quadratic formula,

x2 = {-2 +sqrt(2^2-4(-2))}/2

x2 = (-12+2sqrt2)/2

x2 = (-1+sqrt2) and

x3 = -(1+sqrt2) are the other two roots.

Therefore (-1+sqrt2) and -(1+sqrt2) are the other roots.