<s>[INST] $x$ intercept problem How would I find the $x$ intercept of $x^5-x^3+2=0$?  I haven’t learned about things like synthetic division or any theorems, just algebraic manipulations.  
 [/INST] $$f(x)=x^5-x^3+2\implies f'(x)=5x^4-3x^2\implies f''(x)=20x^3-6x$$
The first derivative cancels twice at $x=0$ and at $x=\pm \sqrt{\frac{3}{5}}$.
$$f(0)=+2 $$
$$f\left(+\sqrt{\frac{3}{5}} \right)=2-\frac{6 \sqrt{\frac{3}{5}}}{25}\approx 1.8141 \qquad f''\left(+\sqrt{\frac{3}{5}} \right)=6 \sqrt{\frac{3}{5}}$$
$$f\left(-\sqrt{\frac{3}{5}} \right)=2+\frac{6 \sqrt{\frac{3}{5}}}{25}\approx 2.1859 \qquad f''\left(-\sqrt{\frac{3}{5}} \right)=-6 \sqrt{\frac{3}{5}}$$ So, ther is only one real root which is negative. By inspection, $f(-1)=2$, $f(-2)=-22$; so the root is close to $-1$.
To approximate the root, use Taylor series to get
$$f(x)=2+2 (x+1)-7 (x+1)^2+O\left((x+1)^3\right)$$ and the root to be considered is given by
$$x=\frac{1}{7} \left(\sqrt{15}-6\right)\approx -1.41043$$ Now, use Newton method which will update the guess according to
$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=\frac{-4 x_n^5+2 x_n^3+2}{3 x_n^2-5 x_n^4}$$ to get the following iterates
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & -1.410426192 \\
 1 & -1.354288685 \\
 2 & -1.347943175 \\
 3 & -1.347867906 \\
 4 & -1.347867896
\end{array}
\right)$$ which is the solution for ten significant figures.
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