# 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...

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#### 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 $x^2-3x+1$).