**I don"t desire the answer**) exactly how to obtain that instinct on factoring this polynomial? Or is over there a method?

**Remark**: Is there any kind of faster, non-rigorous way also, of recognize the square root?

Given that your polynomial$$x^4+2x^3-3x^2-4x+4$$has essence coefficients, and also the highest possible coefficient is one and the lowest four, if over there are any rational solutions, they have to be one of$$pm frac11,quad pm frac21,quad pmfrac41.$$Here, you need to run over various divisors of $4$ and $1$.

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You will certainly surely succeed v that in this case, since you will discover two double roots...

evaluate your polynomial in ~ $y-2$ and see what happens. This is to get rid of power 3 here. And also hopefully simplify things.

**edit**

let $f(x)=x^4+colorred2x^3-3x^2-4x+4$. Now evaluate in ~ $y-colorred2$, i.e. Plug $y-colorred2$ for $x$:

$$f(y-colorred2)=(y-colorred2)^4+2(y-colorred2)^3...=?$$

The example you offered is no a polynomial as there are negative powers associated (e.g. 1/x^2). I guess that you typical to uncover a polynomial which, if you take the square root, yields an additional polynomial.This is only possible if all the linear components are in even multiplicity, i.e. The polynomial has 2n worths for every root. If that is the case, you can take the square root as you described.

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**response come comment**

Ok as you execute not desire the answer, this is what I would do:- discover the rootsAs a hint, x = 1 is a root. Use factor division on the polynomial to mitigate its degree to a cubic equation and continue to find the roots.- inspect that all roots come in even frequencies, if not, us cannot take a "nice" square root.

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deserve to the square root of details quantities be turned into an additional square root plus one arbitrary continuous

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