# Algorithm for finding the square root of a polynomial…

I'm going through Wallace Clarke Boyden's A First Book in Algebra, and there's a section on finding the square root of a perfect square polynomial, eg. $4x^2-12xy+9y^2=(2x-3y)^2$. He describes an algorithm for finding the square root of such a polynomial when it's not immediately apparent, but despite my best efforts, I find the language indecipherable. Can anyone clarify the process he's describing? The example I'm currently wrestling with is $x^6-2x^5+5x^4-6x^3+6x^2-4x+1$.

It's a lot of language to parse, but if anyone wants to take a stab at it, here's the original text:

To find the square root of a polynomial, arrange the terms with reference to the powers of some number; take the square root of the first term of the polynomial for the first term of the root, and subtract its square from the polynomial; divide the first term of the remainder by twice the root found for the next term of the root, and add the quotient to the trial divisor; multiply the complete divisor by the second term of the root, and subtract the product from the remainder. If there is still a remainder, consider the root already found as one term, and proceed as before.

I did some hunting online but didn't turn up anything useful. Is it possible this is an outdated method that's been abandoned for something cleaner?

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It's just saying start with the highest degree and work down. This may not be the fastest but will work or tell you that the thing is not really a square. So, begin with $x^3,$ since the square must be $x^6$ and we get one free choice, $\pm x^3.$ Next, $(x^3 + A x^2)^2 = x^6 + 2 A x^5 + \mbox{more}.$ So $2A = -2, A = -1.$

Alright, $(x^3 -x^2 + B x)^2 = x^6 - 2 x^5 + (2B+1)x^4 + \mbox{more}.$ So $2B+1 = 5$ and $B=2.$

Finally $(x^3 - x^2 + 2 x + C)^2 = x^6 - 2 x^5 + 5 x^4 +(2C -4)x^3 + \mbox{more}.$ So $2C-4 = -6$ and $C=-1.$

Then check $$(x^3 - x^2 + 2 x -1)^2 = x^6 - 2 x^5 +5 x^4 - 6 x^3 + 6 x^2 - 4 x + 1.$$ So it worked.

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Nice, thanks. I take it the $2A$, $2B+1$, $2C-4$ are the polynomial coefficients, like something akin to the binomial coefficients (but more terms, obviously)? –  ivan Mar 8 '13 at 5:31
@ivan, I deliberately solved one at a time for $A,B,C.$ If I were to take $(x^3 + A x^2 + B x + C )^2$ I would get mixtures of the letters $A,B,C$ in the outcome. My way it is not necessary to worry about those mixtures. –  Will Jagy Mar 8 '13 at 5:35
Oh cool! I actually get it now :) –  ivan Mar 8 '13 at 5:52