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 $f$?
The polynomial will have the form
P(x) = Ax^3 + Bx^2 + Cx + D
And since (0, 0) is in the graph then
A(0)^3 + B(0)^2 + C(0) + D = 0 so...... D = 0
And we have these three equations
A(-1)^3 + B(-1)^2 + C (-1) = 15 → -A + B - C = 15 (1)
A(1)^3 + B(1)^2 + C(1) = 5 → A + B + C = -5 (2)
A(2)^3 + B(2)^2 + C(2) = 12 → 8A + 4B + 2C = 12 → 4A + 2B + C = 6 (3)
Adding (1) and (2) we have that 2B = 10 → B = 5
Using (2) and (3) and subbing for B we have
A + 5 + C = -5 → A + C = -10 → C = -10 - A (4)
4A + 10 + C = 6 → 4A + C = -4 → C = -4 - 4A (5)
Set (4) and (5) equal
-10 - A = -4 - 4A
3A = 6
A = 2
Using (4)
C = -10 - (2)
C = -12
So.....the polynomial is
P(x) = 2x^3 + 5x^2 -12x
The x intercepts can be found when P(x) = 0
2x^3 + 5x^2 - 12x = 0 factor
x (2x^2 + 5x^2 - 12) = 0
x ( 2x - 3) ( x + 4) = 0
Setting each factor to 0 and solving for x the x intercepts are -4, 0 and 3/2