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One use of completing the square in a quadratic is to find its critical value. I thought whether this could also be possible with quartics; if nothing else, it would be a cool way to find extrema witouth taking a derivative and having to solve a cubic; so I tried to factor $x^4 + 4x^3 + 6x^2 + 6x + 2$ as $(x^2 + bx + c)^2+d$, and I arrived at an inconsistent system of equations.

Is there any deep reason why we can't solve quartics using the "completing the square" method?

P.S. If possible, please refrain from using advanced abstract algebra terminology.

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    As an aside, there is in fact a [general formula](https://en.wikipedia.org/wiki/Quartic_function#General_formula_for_roots) to solve for the roots of a degree four polynomial, but Galois Theory proves that [no general formula exists](https://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem) for *arbitrary* polynomials of higher degree. – 2017-02-24

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When we complete the square we split the quadratic into a product of two (identical) linear factors, and a constant term to correct for the error. If we were to do the same for a quartic, we should expect the correction term to be quadratic, and not linear.

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    Well for quadratics the correction factor is of degree zero (2 degrees less than the original quadratic), so perhaps the correction factor for a quartic is a quadratic. – 2017-02-24
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    Oops. Thanks for that. I've fixed that now. – 2017-02-24
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The direct analog for higher degrees is depressed polynomials.

For a polynomial $\;a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0\;$ the Tschirnhausen substitution $y = x + \frac{a_{n-1}}{n \, a_n}$ eliminates the term in the second highest power. In the case of a quadratic, that leaves just the square term, and a constant term.

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Quartics can be solved in a similar idea. As dxiv mentions, one can depress the quartic with the substitution $x=y-\frac b{4a}$ to get

$$ax^4+bx^3+cx^2+dx+e=ay^4+c'y^2+d'y+e'$$

For some new constants $c',d',e'$. We can then invite the "completing the square" step by setting this equal to zero:

$$ay^4+c'y^2+d'y+e'=0$$

And then subtracting a quadratic from both sides to get

$$ay^4+c''y^2+e''=ny^2-d'y+m$$

And solve for $c'',e'',n,m$ such that we get perfect squares:

$$(uy^2+u')^2=(vy+v')^2$$

Whereupon we can remove the squares to get

$$uy^2+u'=\pm(vy+v')$$

and complete the square again to solve for $y$, which solves for $x$.

Of course, since it's not possible to solve 5th degree polynomials and higher with radicals... this isn't going to work for any higher general polynomials.