$x$ 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.  

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)  
