I have points (-1,0), (2,0) and (0, 8). How do I find its polynomial formula with a degree of 3?

Answer 1

You have two roots plus a third point, so
f(x) = (x+1)(x-2)(x-z) where z = the third root
f(x) = (x^2-x -2)(x-z)
But f(0) = 8, so 8 = 0^2-0 -2*0 +2z
2z=8
z=4
So f(x) = (x^2 -x-2)(x-4)= x^3 -5x^2 +2x +8

The previous answerers are wrong.

Answer 2

You can even do this with a polynomial of degree 2.

y = A(x-B)(x-C)

take B = -1 and C = 2 (from the first two points)
Then take x = 0 to find A.

8 = ABC = -2A -> A = -4

Polynomial is : y = 8 (x+1)(x-2) = - 4x² + 4x + 8

Answer 3

let the polynomial be y=ax^3+bx^2+cx+d
plugging in the values
equation (1) 0=-a+b-c+d
(2) 0=-8a+4b-2c+d
(3) 8=d
you need one more point to find a unique third degree polynomial

Answer 4

You can't. You need four points to find a polynomial of degree 3.