$x$ 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.  
 A: By the rational root theorem, if the function $f(x) = x^5 - x^3 + 2 = 0$ has a rational zero, then it must be among the rational numbers:
$1, -1, 2, -2$
However, we see that if we plug in any of these values of $x$ into our function, we will not get $0$. That is:
$f(1) = 2, \quad f(-1) = 2, \quad f(2) = 26, \quad f(-2) = -22$
Therefore, we may conclude that any value of $x$ which satisfies $x^5 - x^3 + 2 =0$ will not be a rational number (so, at best, we can only approximate the value of $x$ which does so). One of our options is to simply use a computer to approximate the solution for us. Otherwise, we would probably have to use Newton's Method to approximate this solution by hand (which is very tedious).
A: My answer is not totally correct but I hope it helps  a little :
Consider $f(x)=x^5-x^3+2$  . we have $f'(x)=5x^4-3x^2$ and $f'(x)=0$ if $x=0$ or $x=\pm \sqrt{\frac{3}{5}}$ and it is also possible to determine the sign of $f'(x)$ for all other $x$.
Note that $5x^4-3x^2>0$ if $5x^4>3x^2$ if $x^2>3/5$ . that is, $x >+ \sqrt{\frac{3}{5}}$ or $x <- \sqrt{\frac{3}{5}}$. Also $5x^4-3x^2<0$ if $- \sqrt{\frac{3}{5}}<x<+ \sqrt{\frac{3}{5}}$. 
Thus $f$ is increasing for $x <- \sqrt{\frac{3}{5}}$, decreasing between  $- \sqrt{\frac{3}{5}}$ and $+ \sqrt{\frac{3}{5}}$ and again increasing for $x>+ \sqrt{\frac{3}{5}}$.
For this $f$, finding $x$ for $f(x)=0$ is difficult. But IVT guranteed that such a   number  is between $-2$ and $-1$, since $f(-2)<0$ and $f(-1)>0$
with this information the graph look like this:(with the help of Desmos)  
A: $$f(x)=x^5-x^3+2\implies f'(x)=5x^4-3x^2\implies f''(x)=20x^3-6x$$
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.
