f(x) is non negative for all real x

anemone

MHB POTW Director
Staff member
Hi MHB,

I have encountered an interesting math problem which I couldn't solve.

Problem:
Find all values of such that $x^6-6x^5+12x^4+ax^3+12x^2-6x+1$ is non-negative for all real .

1st attempt:

I first let $f(x)=x^6-6x^5+12x^4+ax^3+12x^2-6x+1$

Then if it has repeated roots (of multiplicity of 2 or 4), then the function of f will always greater than zero and we are done.

This leads to a few cases to be considered:

1. $f(x)=(x-p)^4(x-q)^2$
2. $f(x)=(x-p)^2(x-q)^2(x-r)^2$
3. $f(x)=(x-p)^2(\text{always positive})$
4. $f(x)=(x-p)^4(\text{always positive})$

But I notice even if I compare the coefficients of the $x^n$ terms, I can get nothing useful from this attempt and hence, the possible values of a remain unknown.

2nd attempt:

I noticed $\dfrac{x^6+1}{2} \ge x^3$, $\dfrac{6x^5+6x}{2} \ge 6x^3$ and hence $-\dfrac{6x^5+6x}{2} \le -6x^3$, $\dfrac{12x^4+12x^2}{2} \ge 12x^3$.

So, by putting them together, $x^6-6x^5+12x^4+ax^3+12x^2-6x+1 \ge2x^3-12x^3+24x^3+ax^3>=26x^3+ax^3$

If we want $f(x)\ge 0$, then no conclusion can be drawn from this silly attempt.

3rd attempt:

I noticed f(x) could be rewritten as $x^6-6x^5+12x^4+ax^3+12x^2-6x+1=x^4(x^2-6x+12)+ax^3+12x^2-6x+1$

That is, $x^2-6x+12$ and $12x^2-6x+1$ are both quadratic functions with the nice changing of the coefficients, but I don't know what that indicates...perhaps I'm not thinking enough but I will and will add to this thread if I think of something useful from this.

Could someone please point me with the correct direction so that I can get this problem solved?

MarkFL

Staff member
I would try an analysis of the discriminant, which is:

$$\displaystyle \Delta=729(a+14)(a-38)\left((a+12)(a+16) \right)^2$$

From Wikipedia:

For a polynomial of degree $n$ with real coefficients, we have:

• $\Delta>0$: for some integer $k$ such that $$\displaystyle 0\le k\le\frac{n}{4}$$, there are $2k$ pairs of complex conjugate roots and $n-4k$ real roots, all different;
• $\Delta<0$: for some integer $k$ such that $$\displaystyle 0\le k\le\frac{n-2}{4}$$, there are $2k+1$ pairs of complex conjugate roots and $n-4k-2$ real roots, all different;
• $\Delta=0$: at least 2 roots coincide, which may be either real or not real (in this case their complex conjugate also coincide).

Opalg

MHB Oldtimer
Staff member
$x^6-6x^5+12x^4+ax^3+12x^2-6x+1 = (x^3 -3x^2 -3x +1)^2 + 9x^2(x\pm1)^2 + \:?x^3.$ If the coefficient of $x^3$ in that last expression is not zero then the polynomial must take negative values. There are only two values of $a$ that will achieve that.

anemone

MHB POTW Director
Staff member
I would try an analysis of the discriminant, which is:

$$\displaystyle \Delta=729(a+14)(a-38)\left((a+12)(a+16) \right)^2$$

From Wikipedia:

For a polynomial of degree $n$ with real coefficients, we have:

• $\Delta>0$: for some integer $k$ such that $$\displaystyle 0\le k\le\frac{n}{4}$$, there are $2k$ pairs of complex conjugate roots and $n-4k$ real roots, all different;
• $\Delta<0$: for some integer $k$ such that $$\displaystyle 0\le k\le\frac{n-2}{4}$$, there are $2k+1$ pairs of complex conjugate roots and $n-4k-2$ real roots, all different;
• $\Delta=0$: at least 2 roots coincide, which may be either real or not real (in this case their complex conjugate also coincide).

It's good to know this theorem exists. For the first case where $\Delta>0$: for some integer $k$ such that $$\displaystyle 0\le k\le\frac{6}{4}=0 \le k \le \frac{3}{2}$$, there are $2(\frac{3}{2})=3$ pairs of complex conjugate roots and no real root.

So, we have$$\displaystyle \Delta=729(a+14)(a-38)\left((a+12)(a+16) \right)^2 >0$$ and this gives the solution set $a<-14$ or $a>38$.

But when I checked the answer with wolfram, that is, I let $a=37$, what I get is the given function is still greater than zero for all real x... 1. http://www.wolframalpha.com/input/?i=graph+y%3D+x^6%E2%8%926x^5%2B12x^4%2B37x^3%2B12^2-6x%2B1

2. solve x^6?6x^5+12x^4+37x^3+12^2-6x+1=0 - Wolfram|Alpha

I also have no idea how did you come up with the formula for the discriminant ($$\displaystyle \Delta=729(a+14)(a-38)\left((a+12)(a+16) \right)^2$$) but I will look it up online. $x^6-6x^5+12x^4+ax^3+12x^2-6x+1 = (x^3 -3x^2 -3x +1)^2 + 9x^2(x\pm1)^2 + \:?x^3.$ If the coefficient of $x^3$ in that last expression is not zero then the polynomial must take negative values. There are only two values of $a$ that will achieve that.
Hi Opalg,

Since

$(x^3 -3x^2 -3x +1)^2 + 9x^2(x+1)^2 =(x+1)^2((x^2 -4x +1)^2 + 9x^2)=x^6-6x^5+12x^4+38x^3+12x^2-6x+1$

So we have

$x^6-6x^5+12x^4+ax^3+12x^2-6x+1 = (x+1)^2((x^2 -4x +1)^2 + 9x^2) + (a-38)x^3$

Now, in order to get $x^6-6x^5+12x^4+ax^3+12x^2-6x+1>0$, that means we must set $(x+1)^2((x^2 -4x +1)^2 + 9x^2) + (a-38)x^3>0$.

So, $a=38$ must be one of the value of $a$ that meets the requirement of the problem.

But I'm thinking we could also have to consider the possibility like when $x=-1$, then $a=37, 36, \cdots$ works too. That means we have a lot of $a$ values that will meet the requirement of the problem, but not just two $a$ values. $(x+1)^2((x^2 -4x +1)^2 + 9x^2) + (a-38)x^3=0+ (a-38)(-1)= (37-38)(-1)=1>0$

I guess I must have been missing something very important here...

I don't know how $(x^3 -3x^2 -3x +1)^2 + 9x^2(x-1)^2$ is going to help in this case... Last edited:

Opalg

MHB Oldtimer
Staff member
I don't know how $(x^3 -3x^2 -3x +1)^2 + 9x^2(x-1)^2$ is going to help in this case... You are quite right! My previous attempt was more or less complete nonsense. There was just one thing right about it, namely the polynomial $p(x) = (x^3 -3x^2 -3x +1)^2 + 9x^2(x+1)^2 = x^6 -6x^5 + 12x^4 + 38x^3 + 12x^2 - 6x+ 1$ is a sum of squares and is therefore always non-negative. Therefore $38$ is a possible value for $a$.

It is also true that $q(x) = (x^2-3x+1)^2(1+x^2) = x^6 -6x^5 + 12x^4 - 12x^3 + 12x^2 - 6x+ 1$ is always positive. Therefore $-12$ is a possible value for $a$.

Now let $0\leqslant\lambda\leqslant1$, and form the convex combination $\lambda q(x) + (1-\lambda)p(x)$. That is equal to $x^6 -6x^5 + 12x^4 + (38-50\lambda)x^3 + 12x^2 - 6x+ 1$. Also, it is a convex combination of two positive functions and is therefore positive. As $\lambda$ goes from $0$ to $1$ the coefficient of $x^3$ goes from $38$ to $-12$. So $a$ can take every value in the interval $[-12,38]$.

It seems that these are the only possible values for $a$, but I don't yet see how to prove that.

Opalg

MHB Oldtimer
Staff member
Now I see how to complete the proof. If $f(x) = x^6-6x^5+12x^4+ax^3+12x^2-6x+1$ then $f(-1) = 38-a$, which is negative if $a>38$. So we must have $a\leqslant 38$. With a bit more calculation (in fact, make that a lot more) you can check that $f\bigl(\frac12(3+\sqrt5)\bigr) = (a+12)\bigl(\frac12(3+\sqrt5)\bigr)^3$. That will be negative if $a<-12$, so we must have $a\geqslant-12$.

To help you on your way with that calculation, the powers of $\frac12(3+\sqrt5)$ are $$\bigl(\tfrac12(3+\sqrt5)\bigr)^2 = \tfrac12(7+3\sqrt5),$$ $$\bigl(\tfrac12(3+\sqrt5)\bigr)^3 = \tfrac12(18+8\sqrt5),$$ $$\bigl(\tfrac12(3+\sqrt5)\bigr)^4 = \tfrac12(47+21\sqrt5),$$ $$\bigl(\tfrac12(3+\sqrt5)\bigr)^5 = \tfrac12(123+55\sqrt5),$$ $$\bigl(\tfrac12(3+\sqrt5)\bigr)^6 = \tfrac12(322+144\sqrt5).$$

anemone

MHB POTW Director
Staff member
Thanks again Opalg for your replies. I hate to keep bothering you but when I let $a$ to lie outside the interval of which you found, for example $a=39$ and also $a=70$, I noticed that the original function of $x^6-6x^5+12x^4+ax^3+12x^2-6x+1$ is always greater than zero as well...  MarkFL

Staff member
Check your terms that are supposed to contain $x^2$. The variable is missing and you have $12^2$ instead. anemone

MHB POTW Director
Staff member
Check your terms that are supposed to contain $x^2$. The variable is missing and you have $12^2$ instead. Oh my...I'm sorry! This is probably the most silly blunder that I have ever made this year!

Now, everything looks fine and good to me.

Thanks to both of you and Opalg for the guidance on how to tackle the problem correctly.

Opalg

MHB Oldtimer
Staff member
Oh my...I'm sorry! This is probably the most silly blunder that I have ever made this year!  We all do that sort of thing more often than we like to admit.

While I am about it, I should point out that my argument for showing that $a\geqslant-12$ was unnecessarily complicated. I had already pointed out that $f(x) = x^6 -6x^5 + 12x^4 + ax^3 + 12x^2 - 6x+ 1 = (x^2-3x+1)^2(1+x^2) + (a+12)x^3$. If you put $x=\frac12(3+\sqrt5)$ in that equation then you immediately get $f\bigl(\frac12(3+\sqrt5)\bigr) = (a+12)\bigl(\frac12(3+\sqrt5)\bigr)^3$ (because $\frac12(3+\sqrt5)$ is a root of $x^2-3x+1$).