The polynomial f(x) has degree 3. If f(-1) = 15, f(0) = 0, f(1) = 0, and f(2) = 12, then what are the x-intercepts of the graph of f?
+2666 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 \(\color{brown}\boxed{0,1,14}\)
+2666 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 \(\color{brown}\boxed{0,1,14}\)
