# P(x)=3x^3-10x^2-5x has one root -2/3. What are the other roots?

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P(x)=3x^3-10x^2-5x.

The three roots of this equation can be found by equating 3x^3-10x^2-5x to 0.

=> 3x^3-10x^2-5x =0

=> x ( 3x^2 - 10x - 5) = 0

So one of the roots is 0

The other two roots are

x2 = [-b + sqrt (b^2 - 4ac)]/2a

=> x2 = [10 + sqrt (100 + 60)]/6

=> x2 = 10/6 + (sqrt 160) / 6

x3 = [-b - sqrt (b^2 - 4ac)]/2a

=> x3 = [10 - sqrt (100 + 60)]/6

=> x3 = 10/6 - (sqrt 160) / 6

**Therefore the roots are 0, 10/6 + (sqrt 160) / 6 and 10/6 - (sqrt 160) / 6.**

P(x) = 3x^3-10x^2-5x. To find the roots.

We shall factorise P(x) = 3x^3-10x^2-5x and equate each factor to zero to determine the roots.

We see that x could be factored. So x is a root.

P(x) = x(3x^2-10x-5).

We we find the roots of 3x^2-10x-5

We use the quadratic formula to find for what value of x the equation 3x^2-10x-5 = 0 is satisfied.

x1 = {10+sqrt(10^2-4*3(-5)}/2*3 = {10+sqrt160}/6 = (5+2sqrt10)/3.

x2 = {10-2sqrt10)/3.

Therefore the roots of P(x) are x= 0 , x1 = {5+2sqrt10}/3 and x2 {5-2sqrt10}/3.

The polynomial has 3 roots. If one root is known, the other 2 roots are x,y.

We'll use Viete's relations for finding the other 2 roots:

-2/3 + x + y = 10/3We'll add 2/3 both sides:

x + y = 10/3 + 2/3

x + y = 12/3

x + y = 4 (1)

(-2/3)*x*y = -2/3

We'll divide by -2/3 and we'll get:

xy = 1 (2)

With (1) and (2), we'll form the equation of the quadratic when knowing the sum and the product:

x^2 - Sx + P = 0

x^2 - 4x + 1 = 0

x1 = [4 + sqrt(16-4)]/2

x1 = (4 + 2sqrt3)/2

x1 = 2 + sqrt3

x2 = 2 - sqrt3 = y

**The other 2 roots of the equation are: {2 - sqrt3 ; 2 + sqrt3}.**