For this case we have that the following function complies with the given conditions:
[tex] f (x) = 3x ^ 2 - 15x + 12
[/tex]
To prove it, let's find the roots of the polynomial:
[tex] 3x ^ 2 - 15x + 12 = 0
[/tex]
By doing common factor 3 we have:
[tex] 3 (x ^ 2 - 5x + 4) = 0
[/tex]
Factoring the second degree polynomial we have:
[tex] 3 (x-1) (x-4) = 0
[/tex]
Then, the solutions are:
Solution 1:
[tex] x-1 = 0
x = 1
[/tex]
Solution 2:
[tex] x-4 = 0
x = 4
[/tex]
Answer:
A second degree polynomial function f (x) that has a lead coefficient of 3 and roots 4 and 1 is:
[tex] f (x) = 3x ^ 2 - 15x + 12 [/tex]
