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My question is:

Find the square root -

$(x-1) (x^3 + 4) + (\frac{x}{2} + \frac{2}{x})^2$

The above is a polynomial.I would like to how to find its square root.

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Your'e looking for a polynomial $p(x)$ such that $(p(x))^2=(x-1) (x^3 + 4) + (\frac{x}{2} + \frac{2}{x})^2$, right? (which will not be a plynomial since it will have to include some negative powers of $x$) – Dennis Gulko Jun 9 '12 at 11:27
@Dennis Gulko: yes – mgh Jun 9 '12 at 11:28
The thing you are claiming is a polynomial is not a polynomial. – Chris Eagle Jun 9 '12 at 11:30

Not for all such expressions there exists a "root", but if it does, you can try this way: $$(x-1) (x^3 + 4) + (\frac{x}{2} + \frac{2}{x})^2=x^4-x^3+4x-4+\frac{x^2}{4}+2+\frac{4}{x^2}=\frac{1}{4x^2}\left(4x^6-4x^5+x^4+16x^3-8x^2+16\right)$$ There exists a $p(x)$ as you need if and only if there exists a polynomial $f(x)$ such that $(f(x))^2=4x^6-4x^5+x^4+16x^3-8x^2+16$. Since the polynomial on the RHS is of degree $6$, $f$ should be of degree $3$. Denote $f(x)=ax^3+bx^2+cx+d$. Square it and compare the coefficients to find $a,b,c,d$ if they exist.
Expanding, you get $$a^2x^6+2abx^5+(2ac+b^2)x^4+(2ad +2bc)x^3+(2bd+c^2)x^2+2cd x+d^2\\=4x^6-4x^5+x^4+16x^3-8x^2+16$$ So you have $a=\pm2$. Take $a=2$ (since $-f$ will be another root). $2ab=-4$, then $b=-1$.
Since $1=2ac+b^2=4c+1$, so $c=0$. Also, $16=2ad+2bc=4d+0$, hence $d=4$. It is left to check the remaining coefficients: $-8=2bd+c^2=-8+0$, $0=2cd$ and $16=d^2$.
Hence $f(x)=2x^3-x^2+4$. So $$p(x)=\pm\frac{2x^3-x^2+4}{2x}$$ satisfies your equation.

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Check it $5x^6$? i think it is $4x^6$ – Prasad G Jun 9 '12 at 11:58
+1 good work-thru. – draks ... Jun 9 '12 at 13:22

Hint $\ $ It follows simply using factorization by difference of squares. Namely if

$$\rm \dfrac{4x^2(x\!-\!1)(x^3\!+\!4) + (x^2\!+\!4)^2}{(2x)^2}$$ is a square then the numerator is a square $\rm\:f^2,\:$ so we have

$$\rm 4x^2(x\!-\!1)(x^3\!+\!4)\, =\, f^2\!-\!(x^2\!+\!4)^2 =\, (f\!-\!x^2\!-\!4)\,(f\!+\!x^2\!+\!4)$$

LHS has lead term $\rm 4x^6\:$ so $\rm\:f\:$ has lead term $\rm\:2x^3.\:$ LHS splits into such factors $\rm (2x^3\!-\!2x^2)(2x^3\!+\!8)$ which indeed differ by $\rm\,2\:\!(x^2\!+\!4).\:$ Thus $\rm\:f = 2x^3\!+\!8-(x^2\!+\!4) = 2x^3\!-\!x^2\!+\!4,\:$ up to sign.

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