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

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We have x = 3 is a solution of x^3 - x^2 - 8x + 6 = 0. We need to find the other two roots.

As x = 3 is a root we can write the equation as

x^3 - x^2 - 8x + 6 = 0

=> x^3 - 3x^2 + 2x^2 - 6x - 2x + 6 = 0

=> x^2(x - 3) + 2x(x - 3) - 2(x - 3) = 0

=> (x - 3)(x^2 + 2x - 2) = 0

The roots of x^2 + 2x - 2 = 0 are

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

=> -1 + sqrt 3

x2 = -1 - sqrt 3

**The other roots of the equation are -1 + sqrt 3 and -1 - sqrt 3**

If x = 3 is the solution of the equtaion, then the polynomial x^3-x^2-8x+6 is divisible by x - 3.

We'll write the rule of division:

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

We'll remove the brackets from the right side:

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

We'll combine like terms form the right side:

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

We'll compare the expressions from both sides and we'll get:

a = 1

b - 3a = -1

b - 3 = -1

b = 3 - 1

b = 2

-3c = 6

c = 6/-3

c = -2

The quotient ax^2 + bx + c = x^2 + 2x - 2.

Now, we'll determine the other 2 solutions of the equation (the equation has 3 solutions and one of them is x = 3).

x^2 + 2x - 2 = 0

We'll apply the quadratic formula:

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

x1 = (-2+2sqrt3)/2

x1 = -1 + sqrt3

x2 = -1 - sqrt3

All the solutions of the equation are: { -1 - sqrt3 ; -1 + sqrt3 ; 3}.