Use Newton's method to find the roots of 2x^3−9x^2+12x+15. How do I handle the cases when x>3 and x<3.

Use Newton's method to find the roots of
$2{x}^{3}-9{x}^{2}+12x+15$
for $x=3$, $x<-3$, and $x>3$.
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neobuzdanio
Newton's method is only guaranteed to converge to a solution (that satisfies ${f}^{\prime }\left(x\right)\ne 0$) if you start 'close enough'.
In this case the polynomial $p$ has exactly one real root, and since $p\left(-1\right)=-18,p\left(0\right)=5$, we see that it lies in $\left(-1,0\right)$.
Try starting at $-\frac{1}{2}$ for example.