SOLUTION: Find a polynomial function f of degree 3 with leading coefficient 1 whose graph has x- intercepts at (-1,0), (2,0), (4,0)

 
Two methods.  I'll do both:

======================================

Method 1:

Suppose the function is



then since it has x-intercept (-1,0) then when you substitute
-1 for x in that you get 0, so

 which equals 0, so

 or



and since it has x-intercept (2,0) then when you substitute
2 for x in that you get 0, so

 which equals 0, so

 or



and since it has x-intercept (4,0) then when you substitute
4 for x in that you get 0, so

 which equals 0, so

 or



So you have this system of 3 equations in 3 unknowns:



Solve that and get A=-5, B=2 and C=8

So substitute in



and get



-----------------

Method 2:

(-1,0), (2,0), (4,0)

If you had solved  you would have 
ended up with

    x = -1,       x = 2,       x = 4

before which you would have had

x + 1 =  0,   x - 2 = 0,   x - 4 = 0

before which you would have used the zero-factor
principle on this:



which you would have had after factoring, so before
factoring you would have had that multiplied out,
so we multiply it out









and since that was the solution to , then



Same answer either way.  Take your pick.

Edwin