How do you calculate the root of a polynomial?

The best place to start is with the rational root theorem, which states that for (let "a" be any coefficient)

px^n + ax^(n - 1) + ... + ax + q

The rational roots will be made of Q/P, where Q is any factor of q and P is any factor of p

For instance, with x^3 - 2x^2 - 11x + 12, p = 1 and q = 12. Possible factors:

P => -1 , 1

Q => -12 , -6 , -4 , -3 , - 2 , -1 , 1 , 2 , 3 , 4 , 6 , 12

Since P is either -1 or 1, we can just say that Q/P reduces to Q

-12 , -6 , -4 , -3 , -2 , -1 , 1 , 2 , 3 , 4 , 6 , 12

Test each root. I'd start with 1, then -1, then 2, then -2, and so on, simply because smaller numbers are easier to work with and there's also a good chance that they'll be our roots anyways (because the other coefficients are so small, it's a big hint. Practice with other polynomials and you'll see what I mean)

x = 1

1^3 - 2 * 1^2 - 11 * 1 + 12 =>

1 - 2 - 11 + 12 =>

13 - 13 =>

0

x = 1 is a root, which means (x - 1) is a factor. We can try the other roots, or we can divide the polynomial by (x - 1) to reduce the power of the polynomial to a quadratic equation. Then we can use the quadratic formula to find the other 2 roots. That's the route I'll take simply because I don't like trial-and-error

(x^3 - 2x^2 - 11x + 12) / (x - 1)

x^3 / x = x^2

x^2 * (x - 1) = x^3 - x^2

x^3 - 2x^2 - x^3 + x^2 = -x^2

-x^2 / x = -x

-x * (x - 1) = -x^2 + x

-x^2 - 11x + x^2 - x = -12x

-12x / x = -12

-12 * (x - 1) = -12x + 12

-12x + 12 + 12x - 12 = 0

x^3 - 2x^2 - 11x + 12 = (x - 1) * (x^2 - x - 12)

x^2 - x - 12 = 0

x = (1 +/- sqrt(1 + 48)) / 2

x = (1 +/- 7) / 2

x = -6/2 , 8/2

x = -3 , 4

x = -3 , 1 , 4 are our roots

(x + 3) , (x - 1) , (x - 4) are our factors

x^3 - 2x^2 - 11x + 12 = (x−4)(x−1)(x+3)

So its roots are 4, 1 and -3

Hope this helps. Let me know if you need more help. I'm on Messenger or by email. Best of luck.

there is a formula for a polynomial like this :

x = cbrt [ q + sqrt ( q^2 + ( r - p)^3 )] + cbrt [ q - sqrt ( q^2 + ( r - p )^3 ) ] + p

p = - b / ( 3a)

q = p^3 + ( bc - 3ad ) / ( 6a^2 )

r = c / ( 3a )

polynomial is of the form ax^3 + bx^2 + cx + d = 0