Since f(0) = 0,,,,then x = 0 is one root
So.....we have the form ax^3 + bx^2 + cx + d where d = 0
So we know that
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 (3)
Add (1) and (2) and we have that 2b = 10 ⇒ b = 5
Multiply (1) by 8 and add to (3) and we have that
12b - 6c = 132 sub 5 in for b and we have
12(5) - 6c = 132
60 - 6c = 132 subtract 60 from both sides
-6c = 72 divider both sides by -6
c = -12
And using (2) we can find a as
a + 5 - 12 = -5
a - 7 = -5 add 7 to both sides
a = 2
So.....the polynomial is 2x^3 + 5x^2 - 12x
Set to 0 and factor
x (2x^2 + 5x - 12) = 0
x (2x - 3) ( x + 4) = 0 set each factor to 0 and solve for x
x = 0 2x - 3 = 0 x + 4 = 0
x = 0 x = 3/2 x = -4 these are the x intercepts
Here is a graph : https://www.desmos.com/calculator/o040mctwqm