How to find the Square Root of a Polynomial

Question

$4x^4 + 4x^3 - 11x^2 -6x + 9$

How do you find the square root of this polynomial? I really don't understand. Please provide an easy-to-understand explanation. Thanks.

Answer 1

A square root of $4x^4+4x^3-11x^2-6x+9$ would be a polynomial $P(x)$ with the property that $$P(x) \cdot P(x) = 4x^4+4x^3-11x^2-6x+9.$$ We want to find $P(x)$ with this property, or prove that there is no such. $P(x)$ must have degree 2; since otherwise its square would not have degree 4. (For example, if $P(x)$ contained an $x^3$ term, then $P(x)\cdot P(x)$ would contain an $x^6$ term, which we don't want.) If there is such a polynomial, we can write it as $$P(x) = ax^2 + bx + c$$ for some numbers $a,b,c$.

Now we have $$(ax^2 + bx+ c)^2 = 4x^4+4x^3-11x^2-6x+9.$$

Now you can expand the left side and equate the coefficients of the two sides, obtaining five equations in $a,b,$ and $c$, which are not hard to solve.

Does that help?

Answer 2

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 $c^2=9$.

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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 $\gcd(f,f')$: $$\begin{align}\gcd(f,f')&= \gcd(4x^4 + 4x^3 - 11x^2 -6x + 9,16x^3+12x^2-22x-6)\\&=\gcd(4x^4 + 4x^3 - 11x^2 -6x + 9,8x^3+6x^2-11x-3)\\ &=\gcd(8x^3+6x^2-11x-3,2x^2+x-3)\\ &=\gcd(2x^2+x-3,0)\\ &=2x^2+x-3\end{align}$$

Answer 3

If you know that the polynomial is a perfect square, then the square root algorithm works. For example

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$$\sqrt{4x^4 + 4x^3 - 11x^2 -6x + 9}$$

\begin{array}{lcccccccccccccc} &&2x^2 && +x && -3\\ &&---&---&---&---&---\\ &|& 4x^4 & 4x^3 & -11x^2 & -6x & +9\\ 2x^2 && 4x^4\\ &&---&---&---\\ &&& +4x^3 & -11x^2 \\ 4x^2+x &|& &+4x^3 &+x^2\\ &&&---&---&---&---\\ &&&& -12x^2 &-6x &+9\\ 4x^2+ 2x -3 &|&&&-12x^2 &-6x &+9\\ &&&&---&---&---\\ &&&&&&0\\ \end{array}

\begin{array}{ll} \text{STEP}\;1. &\text{Compute the square root of the leading term (4x^4) and put it,}\\ & \text{(2x^2), in the two places shown.} \\ \text{STEP}\;2. &\text{Subtract and then append the next two terms ($4x^3-11x^2$).}\\ \text{STEP}\;3. &\text{Double the currently displayed quotient $(2x^2 \to 4x^2)$,} \\ & \text{then add the new term, $x$, so that $x(4x^2 + x)$ will "cancel out" the} \\ & \text{$4x^3$ in $4x^3 -11x^2$.}\\ \text{STEP}\;4. &\text{Subtract and then append the next two terms ($-6x+9$).}\\ \text{STEP}\;5. &\text{Double the currently displayed quotient} \\ & \text{ $(2x^2 + x \to 4x^2 + 2x)$, then add a new term, $x$,} \\ & \text{then add a new term, $-3$, so that $-3(4x^2 + 2x - 3)$ will "cancel out"} \\ & \text{ the $-12x^2$ in $-12x^2 -6x + 9$.} \end{array}

Answer 4

Adjusting the coefficient of $x^3,$

$$4x^4+4x^3+\cdots+x^2=(2x^2)^2+2\cdot2x^2\cdot x+x^2$$

Adjusting the coefficient of $x^2$ and the constants $$4x^4+4x^3-11x^2+\cdots+3^2=(2x^2)^2+2\cdot2x^2\cdot x+x^2-2\cdot2x^2\cdot3+\cdots+3^2$$

Finally,

$$4x^4+4x^3-11x^2-6x+3^2=(2x^2)^2+2\cdot2x^2\cdot x+x^2-2\cdot2x^2\cdot3+3^2-2\cdot x\cdot3=(2x^2+x-3)^2$$