How would I obtain the square root of this multinomial? I was doing some problems from a book I found on finding the square root of a polynomial expression. I came across this problem:
$$\frac{a^4}{64}+\frac{a^3}{8}-a+1$$
I utilised the method outlined here, and obtained the following result
$$\frac{a^4}{64}+\frac{a^3}{8}-a+1)\frac{a^2}{8} + \frac{a}{2}$$
$$\frac{a^2}{4} + \frac{a}{2} )\frac{a^3}{8} - a + 1$$
$$\frac{a^2}{4} + a + 1) -\frac{a^2}{4} - a + 1$$
$$\frac{a^2}{4} + a + 1 ) -2$$
I know that I didn't format it well, but, basically, when I used the method they suggested, I had a remainder at the end. I don't know whether I did something wrong or whether the polynomial is a perfect swuare.

We can start small and improve a bit at a time. We can begin with $\frac{a^2}{8}$ which does give the top term when squared. Next,
$$ \left( \frac{a^2}{8} + B a \right)^2 = \frac{a^4}{64} + B \frac{a^3}{4} + B^2 a^2. $$ To get the $a^3/8$ we take $B = 1/2.$ So far, we have
$$ \left( \frac{a^2}{8} +  \frac{a}{2} \right)^2 = \frac{a^4}{64} +  \frac{a^3}{8} + \frac{ a^2}{4}. $$ Not bad. Next,
$$ \left( \frac{a^2}{8} +  \frac{a}{2} +C\right)^2 = \frac{a^4}{64} +  \frac{a^3}{8} + \frac{ a^2}{4} + C\frac{a^2}{4} + C a + C^2. $$
To get rid of the $a^2$ term, we need only take $C = -1$ and get
$$ \left( \frac{a^2}{8} +  \frac{a}{2} -1\right)^2 = \frac{a^4}{64} +  \frac{a^3}{8}  -  a + 1. $$
