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.
 A: The systematic way to go about it, given that you know that there are two double roots, would be to find the common factors between the polynomial and its derivative using the Euclidean polynomial division, which gives the quadratic $\,\gcd(a^4 + 8 a^3 -64 a + 64, 4 a^3 + 24 a^2 - 64) = a^2 + 4 a - 8\,$.
A: You’re asking whether $\frac1{64}X^4 +\frac18X^3-X+1$ is a square.
I think it becomes more familiar-looking when you multiply by the square integer $64$ and get $X^4+8X^3 -64X+64$. If this is a square, it will be the square of a monic quadratic polynomial $X^2+\cdots$, and again if it’s a square, the next term in the quadratic will have to be $4X$. Still hoping that it’s a square, we try to decided on the constant term: must be $8$ or $-8$.
Now the constant term $8$ will be no good, gives all coefficients positive. So let’s try $(X^2+4X-8)^2$, andlo and behold, it expands out to your polynomial ! (Thus you want $\frac18$ times my quadratic)
A: In the last step it must be: $\frac{a^2}{4} + a \color{red}{-} 1 ) \color{red}{0}$. 
See the solution in the referenced format:
$\hspace{5cm}$
A: 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
$$ \left( \frac{a^2}{8} +  \frac{a}{2} -1\right)^2 = \frac{a^4}{64} +  \frac{a^3}{8}  -  a + 1. $$
