Can completing the square apply to more higher degree?

I have some trouble about completing the square lessons.

As we know , minimum value of $\ \ 3x^{2} + 4x + 1$ is $-\frac{1}{3}$ when $x=-\frac{2}{3}$ by completing the square.

Then , how about minimum value of $\ \ x^5 +4x^4 + 3x^3 +2x^2 + x + 1$ when $(-2<x<2)$ ?

According to Wolfram Alpha , minimum value is $12 - 5\sqrt{5}$ when $x=-\frac{3}{2} + \frac{\sqrt{5}}{2}$

I try to put 5th degree to form $(x-a)^5$ or any nth degree but it doesn't work.

I'm just curious how completing the square can apply to higher deree equations to find their minimum or maximum value. Thank you for every help comments.

• sorry , I mean 2nd degree of x. – ABCDEFG user157844 Sep 15 '16 at 8:02
• $\ \ x^5 +4x^4 + 3x^3 +2x^2 + x + 1$ is irreducible over $\mathbb{Q}$, so you cannot factor it over $\mathbb{Q}$. And also over $\mathbb{R}$, not every quintic polynomial can be written as $(x-a)^5$. – Dietrich Burde Sep 15 '16 at 8:02
• What good will it do you if you were able to complete this to $(x-a)^5$? – Ivan Neretin Sep 15 '16 at 8:05
• When you complete the square for a second-degree polynomial, you essentially find its vertex. It happens that for quadratic polynomials the vertex is also the minimum or maximum point. Not so for odd degrees though. So in those cases you cannot find a minimum/maximum with this method, even if you restrict the interval like you did. – rubik Sep 15 '16 at 8:07
• I do not know what to say for $n=3$ or $4$, but in general for $5+$ it's definitely not possible. If you could, then you could solve equations of degree $5$ and higher using just radicals, which is definitely not possible, as proven by the Abel-Ruffini Theorem . I doubt there is known a simple proof of this, but you can check out this question to try to get an intuitive understanding as to why its impossible to solve quintic+ – Ovi Sep 15 '16 at 8:26

If you add and subtract $x^4+7x^3+8x^2+4x$ to your example, you get $$(x+1)^5 -x^4 -7x^3-8x^2-4x.$$ I've seen constest-type problems where such a trick works, because the thing you add and subtract is cleverly designed, but I don't see anything useful here.

If you look at the derivation of Cardano's formula for solving a general cubic, you can see that "completing the cube" is the major step. This site shows it:

https://www.math.ucdavis.edu/~kkreith/tutorials/sample.lesson/cardano.html

But as you can see, increasing the degree by one makes the problem much, much harder. So, in general, if we can solve or factor or otherwise manipulate a 5-th degree polynomial in to a useful form, we've just been very lucky.

Yes we have something same. But if we look to the topic like following.

$3x^2+4x+1$ has a complete square $3x^2+4x+\frac43$ corresponding to it in which their difference, $-\frac13$, is of degree maximum two degrees less than the original.

For any polynomial in one variable we have a unique multiple of a power of a $(x-a)$ in which the difference of the two is of degree maximum two degrees less than the degrees of them.

Before showing what I said, I try to give an algorithm to find that unique multiply of a complete power.

It is easy: If the initial polynomial is of the form $ax^n+bx^{n-1}+\dots+c$, then the unique multiply of a complete power is $$a(x+\frac{b}{na})^n,$$ which it would be expanded as $$a\left(x^n+n({ \frac{b}{na}})x^{n-1}+\dots+\frac{b^n}{n^na^n}\right)=a(x^n+\frac b a x^{n-1}+\dots+\frac {b^n}{n^na^n})=$$ $$ax^n+bx^{n-1}+\dots+\frac{b^n}{n^na^{n-1}}$$.

So by this: The multiple of complete power for $3x^2+4x+1$ is $3(x+\frac{4}{2\cdot 3})^2$, i.e. $3(x+\frac{2}{3})^2=3(x^2+\frac{2\cdot2}{3}+\frac{4}{9})=3x^2+4x+\frac43$.

And so for the quintic, $5x^5+4x^4+3x^3+2x^2+x+1$, the unique multiple of the complete power of a would be $5(x+\frac{4}{5\cdot5})^5$, i.e. $5(x+\frac4{25})^5$, which has an expanded form like: $5x^5+4x^4+\frac{32}{25}x^3+\frac{128}{625}x^2+\frac{256}{15625}x+\frac{1024}{1953125}$
which their difference is (of course of two degrees lesser):
$-\frac{43}{25}x^3 -\frac{1122}{625}x^2 -\frac{15369}{15625}x -\frac{1952101}{1953125}$.

I wish you got something of what you where looking for!

We would generally use the derivative for this. Set it to zero and find the extrema. Via this you can probably find some abstract factorisation that includes these extrema, but it wouod be hard to interpret I think. This is an interesting idea. I think I learnt that you cannot factorise up to order 5, but anything under that you can :) But these extrema won't all be the same value so why factorise it as $(x-a)^{5}$