Question

Find a 3rd-degree polynomial with zeros of 0, 2, -3, that passes through the point (1,20).

a) f(x) = x^3 - 3x^2 - 2x + 6
b) f(x) = x^3 + 3x^2 - 2x - 6
c) f(x) = x^3 - x^2 - 2x + 6
d) f(x) = x^3 + x^2 - 2x - 6

Answer 1

To find a 3rd-degree polynomial with zeros of 0, 2, and -3 that passes through the point (1, 20), we can use the fact that if a polynomial has a zero at x = a, then it must have a factor of (x - a). We can find the value of a by substituting the coordinates of the given point into the equation and solving for a. The polynomial we are looking for is f(x) = 2.5x(x - 2)(x + 3).

Step-by-step explanation:

To find a 3rd-degree polynomial with zeros of 0, 2, and -3 that passes through the point (1, 20), we can use the fact that if a polynomial has a zero at x = a, then it must have a factor of (x - a).

Therefore, the polynomial we are looking for can be written as f(x) = a(x - 0)(x - 2)(x + 3), where a is a constant.

We can find the value of a by substituting the coordinates of the given point into the equation and solving for a.

When we substitute x = 1 and y = 20 into the equation, we get 20 = a(1 - 0)(1 - 2)(1 + 3). Simplifying this equation gives us a = 2.5.

Therefore, the polynomial we are looking for is f(x) = 2.5x(x - 2)(x + 3).