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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.

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  • $\begingroup$ Have you seen the (complicated) method for finding square roots of numbers(with remainder)? This is nothing but a mimic of that method with polynomials. If you like I can provide you with an answer that shows how the two methods match up. If you do not know the method I mentioned above then the answer below is massively helpful. $\endgroup$ – астон вілла олоф мэллбэрг May 8 '18 at 3:53
  • $\begingroup$ Wonder if Newton's method would work. $\endgroup$ – marty cohen May 8 '18 at 4:04
  • $\begingroup$ Just pointing out, but you can convert the link into a PDF which might help those who have old laptops. $\endgroup$ – Frank W. May 8 '18 at 4:06
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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:

$\hspace{5cm}$enter image description here

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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 $$ \left( \frac{a^2}{8} + \frac{a}{2} -1\right)^2 = \frac{a^4}{64} + \frac{a^3}{8} - a + 1. $$

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  • $\begingroup$ Shucks darn, you wrote faster than I could. $\endgroup$ – Lubin May 8 '18 at 3:56
  • $\begingroup$ @Lubin sometimes I take my time. math.stackexchange.com/questions/2767201/… This fellow was making something out of concrete and had a poor idea of what the (six) shapes ought to be. I was afraid he would hurt himself, so I spent many hours on finding a way that he could construct a correct model out of cardboard before attempting the concrete again. $\endgroup$ – Will Jagy May 8 '18 at 14:42
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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 $\,\gcd(a^4 + 8 a^3 -64 a + 64, 4 a^3 + 24 a^2 - 64) = a^2 + 4 a - 8\,$.

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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)

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