c'mon, it's just a quadratic:
x(2x^2+5x-12) = 0
x(2x-3)(x+4) = 0
...
I already asked this question,I just wanted to clarify the answer.
Problem: The polynomial f(x) has degree 3. If f(-1)=15, f(0)=0, f(1)=-5, f(2)=12, then what are the x-intercepts of the graph of f?
Steve: clearly, f(0) = 0
so,
f(x) = ax^3+bx^2+cx
= x(ax^2+bx+c)
-1(a-b+c) = 15
1(a+b+c) = -5
2(4a+2b+c) = 12
solve those and you wind up with
f(x) = 2x^3+5x^2-12x
Now just solve for the roots of f(x)
How do I solve for the roots?
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