THIS USER ASKED πŸ‘‡

Which second degree polynomial function f(x) has a lead coefficient of 3 and roots 4 and 1? f(x) = 3×2 + 5x + 4 f(x) = 3×2 + 15x + 12 f(x) = 3×2 – 5x + 4 f(x) = 3×2 – 15x + 12

THIS IS THE BEST ANSWER πŸ‘‡

In this case we need the following function to meet the given conditions:

f (x) = 3x ^ 2 - 15x + 12

To prove it, let’s find the roots of the polymy:

3x ^ 2 - 15x + 12 = 0

By doing a common factor of 3 we need to:

3 (x ^ 2 - 5x + 4) = 0

Our second stage polypolytic factor:

3 (x-1) (x-4) = 0

Then, the solutions are:

Solution 1:

x-1 = 0 \ x = 1

Solution 2:

x-4 = 0 \ x = 4

A second-degree polymorphic function f (x) having a coefficient of lead 3 and roots 4 and 1 is:

f (x) = 3x ^ 2 - 15x + 12