How to find the Square Root of a Polynomial $4x^4 + 4x^3 - 11x^2 -6x + 9$
How do you find the square root of this polynomial? I really don't understand. Please provide an easy-to-understand explanation. Thanks.

If this is the square of some polynomial $ax^2+bx+c$, we find
$$\begin{align}4x^4 + 4x^3 - 11x^2 -6x + 9&= (ax^2+bx+c)^2\\&=a^2x^4+2abx^3+(b^2+2ac)x^2+2bcx+c^2\end{align}$$
By equating coefficients, we find $a^2=4$, hence (as we may play with the sign) wlog. $a=2$. Next, $4=2ab$ gives us $b=1$. Then $-6=2bc$ gives us $c=-3$. We check that all wiorks out, i.e. that $-11=b^2+2ac$ and $c^2=9$.

Alternatively, you can make use of the fact that the derivative of $g(x)^2$ is $2g(x)g'(x)$. Hence if $f(x)=g(x)^2$, then $g$ is a common divisor of $f$ and $f'$. Use the euclidean algorithm to find $\gcd(f,f')$:
$$\begin{align}\gcd(f,f')&= \gcd(4x^4 + 4x^3 - 11x^2 -6x + 9,16x^3+12x^2-22x-6)\\&=\gcd(4x^4 + 4x^3 - 11x^2 -6x + 9,8x^3+6x^2-11x-3)\\
&=\gcd(8x^3+6x^2-11x-3,2x^2+x-3)\\
&=\gcd(2x^2+x-3,0)\\
&=2x^2+x-3\end{align}$$
