SOLUTION: The polynomial $f(x)$ has degree 3. If $f(-1) = 15$, $f(0)= 0$, $f(1) = -5$, and $f(2) = 12$, then what are the $x$-intercepts of the graph of $f$? Please explain your work thoroug

Let  f(x) = ax^3 + bx^2 + cx + d  with unknown coefficients  a, b, c and d.


Since f(0) = 0, it implies d = 0   (to see it, simply substitute x= 0 into the polynomial).


So, you need to determine a, b and c.


f(-1) = 15  ====> -a +  b -  c = 15,     (1)

f(1)  = -5  ====>  a +  b +  c = -5,     (2)

f(2)  = 12  ====> 8a + 4b + 2c = 12.     (3)


Add equations (1) and (2). You will get   2b = 15 + (-5) = 10  ====>  b = 5.


Then equations (2) and (3) take the form

 a +   5 +  c = -5     (4)    (instead of (2))
8a + 4*5 + 2c = 12     (5)    (instead of (3))

or, which is the same

 a + c = -10,          (4')
4a + c =  -4,          (5')


Subtract (4') from (5') to get 3a = 6,   a = 2.

Then  from (4')  c = -12.


Thus the polynomial is  f(x) =  =  =  (factor ! ) =  = .


Thus the x-intercepts are , 0 (given !)  and  -4.