polynomial

Let the function be in the form $f(x) = ax^3 + bx ^2 + cx + d$

 

From $f(0) = 0$, we know that $d = 0$

 

From the remaining 3, we can form the system: 

 

$-a + b -c = 15$                           (i)

$a + b + c = 0$                                 (ii)

$8a + 4b + 2c = 12$                        (iii)

 

Adding (i) and (ii) gives us $2b = 15$, meaning $b = 7.5$

 

Substituting this into (iii) gives us: $8a + 30 + 2c = 12$, which means $8a + 2c = - 18$. 

 

Substituting this into (ii) gives us: $a + c = -7.5$. 

 

Solving this new system, we find that $a = -0.5$ and $c = -7$. 

 

This means that the polynomial is $f(x)=-0.5x^3 + 7.5x^2 -7x$

 

Now, recall the sum of the roots, by Vieta's, is $-{b \over a} = 15$. 

 

Because the polynomial is of degree 3, we know that there must be 3 real roots. 

 

But, recall that we know 2 of the roots, (0 and 1), so the remaining root is $15 - 1 = 14$. 

 

Thus, the x-intercepts are $\color{brown}\boxed{0,1,14}$