Given $P(x)=x^{4}-4x^{3}+12x^{2}-24x+24,$ then $P(x)=|P(x)|$ for all real $x$

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Solution 1

.In this question, you have to attempt to complete two squares, namely: $$ x^4-4x^3 + 12x^2-24x+24 = x^4 -4x^3+6x^2-4x+1 + 6x^2-20x+23 \\ = (x-1)^4 + 6x^2 -20x+23 $$

Now, as it turns out, we can complete the second square, and: $$ 6x^2-20x+23 = \frac{2}{3}(3x-5)^2 + \frac{19}{3} $$

Hence, we can rewrite the whole expression as: $$ x^4-4x^3 + 12x^2-24x+24 = (x-1)^4 + \frac{2}{3}(3x-5)^2 + \frac{19}{3} $$

It is a sum of positive expressions, hence is always positive, hence $P(x) = |P(x)|$.

Solution 2

Just for fun, observe that $$ P(x) = \begin{bmatrix} x^2 \\ x \\ 1 \end{bmatrix}^\intercal \begin{bmatrix} 1 && -2 && 0 \\ -2 && 12 && -12 \\ 0 && -12 && 24 \end{bmatrix} \begin{bmatrix} x^2 \\ x \\ 1 \end{bmatrix} = 24 - 24x + 12x^2 - 4x^3 + x^4 $$ and since the matrix in the middle is positive semi-definite (This can be determined from determinants of sub-matrices, etc.), it follows immediately $P(x) \geq 0$.

Solution 3

Here is an alternative solution. You can see that $P(x)$ is positive iff $P(x+c)$ is positive for all $x$ (think about why). So you can slightly simplify the polynomial by such substitution, for example

$$P(x+1) = x^4+6 x^2-8x+9$$

Now it's easy since you can check that $6x^2-8x+9$ corresponds to parabola with positive global minimum, hence all its values are positive. Since also $x^4$ is non-negative, you can combine it to see the whole polynomial is positive.

If you want to be a bit more explicit, you can rewrite the parabola on the right to get

$$P(x+1) = x^4 + 6\left(x-\frac{2}{3}\right)^2+\frac{19}{3}$$

Solution 4

Alternative solution : you can note that

$$ P(x)=2x^2+((x-2)^2)(x^2+6) $$

Solution 5

By completing the square for $x^4+12x^2\cdots$,

$$P(x)=(x^2+6)^2-4x^3-24x-12=(x^2+6)^2-4x(x^2+6)-12\\ =(x^2+6)(x^2-4x+6)-12.$$

The minimum value of the first factor is $6$ and that of the second is $2$.


Or completing the square for $x^2(x^2-4x\cdots)$,

$$P(x)=x^2(x-2)^2+8x^2-24x+24=x^2(x-2)^2+8(x^2-3x+3).$$

As you can easily check, the trinomial on the right has no real root.


Or completing the fourth power for $x^4-4x^3\cdots$,

$$P(x)=(x-1)^4+P(x)-(x^4-4x^3+6x^2-4x+1)\\=(x-1)^4+6x^2-20x+23$$ and the trinomial on the right has no real root.

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Updated on August 01, 2022

Comments

  • lmc
    lmc over 1 year

    $$P(x)=x^{4}-4x^{3}+12x^{2}-24x+24$$

    I'd like to prove that $P(x)=|P(x)|$. I don't know where to begin. What would be the first step?

    • DonAntonio
      DonAntonio about 7 years
      Well, the first and practically unique step would be, imo, to prove $\;P(x)\ge0\;\;,\;\;\forall\,,x\in\Bbb R\;$. What about the polynomial first derivative, minimum/maximum points and etc.?
    • lmc
      lmc about 7 years
      Oh, I see. As for the derivatives etc. I shouldn't use them here, since we haven't learned about them in class yet.
    • DonAntonio
      DonAntonio about 7 years
      @No Well, I didn't know that since you tagged your question "real analysis".
    • Tom Zych
      Tom Zych about 7 years
      I observe that $P(x)=x^4-P'(x)$, and a similar relation holds for all successive derivatives. I suspect this is meaningful, but I don't know what to make of it.
    • dxiv
      dxiv about 7 years
      @TomZych Neat observation, thanks for posting it. One possible way to use it is shown in my answer.
  • lmc
    lmc about 7 years
    This is brilliant, I wouldn't have ever thought of that.
  • Sarvesh Ravichandran Iyer
    Sarvesh Ravichandran Iyer about 7 years
    @Now_now_Draco_play_nicely To be fair, what gave me this hint is the first two terms of the expression, $x^4-4x^3$. It seemingly dropped a hint that I was to remove the fourth power from the expression. As luck would have it, the rest went through, otherwise I might have had trouble.
  • lmc
    lmc about 7 years
    How did you figure that out?
  • Steven Alexis Gregory
    Steven Alexis Gregory about 7 years
    Or $P(x) = (x-1)^4 + 6\left(x-\frac{5}{3}\right)^2+\sqrt{\frac{19}{3}}^2$
  • Ewan Delanoy
    Ewan Delanoy about 7 years
    @Now_now_Draco_play_nicely With a little bit of trial and error : first, I tried to factorize $P(x)$ and got nowhere. Then, I computed the minimum of $\frac{P(x)}{x^2}$ over $\mathbb R$ and found that it was equal to $2$. Then, I factorized $P(x)-2x^2$.
  • lmc
    lmc about 7 years
    Uh, that's way too advanced for me,
  • 6005
    6005 about 7 years
    A very cool answer! Thanks for posting.
  • Glen O
    Glen O about 7 years
    Incidentally, a nicer sum of squares is $(x^2-2x)^2+2(2x-3)^2+6$, avoiding any fractional values.
  • Sarvesh Ravichandran Iyer
    Sarvesh Ravichandran Iyer about 7 years
    @GlenO Thank you for the point out.
  • Tom Zych
    Tom Zych about 7 years
    Very nice! Short, sweet, and not hard to follow even for the mathematically rusty such as myself :)
  • dxiv
    dxiv about 7 years
    @TomZych Thanks. Yours was too good a hint to pass up ;-) Plus, a similar argument can prove more general propositions e.g. if $P(a) \gt0$ and $P(x)+P'(x) \gt 0$ for all $x \gt a$ then $P(x) > 0$ on $[a, \infty)$.
  • lmc
    lmc about 7 years
    @астонвіллаолофмэллбэрг I know I'm a bit late to reply, but didn't you get in the end that $P(x)$ is always positive, while you should get that it is always non-negative?
  • Sarvesh Ravichandran Iyer
    Sarvesh Ravichandran Iyer about 7 years
    @Now_now_Draco_play_nicely Good job, Sherlock, you have noticed that the minimum of this function is greater than $\frac{19}{3}$. Actually, if something is strictly positive, it is automatically non-negative, of course, In which case, $|P(x)| = P(x)$ would follow anyway. The absolute value function preserves non-negativity,therefore it preserves positivity as well.
  • Anurag Baundwal
    Anurag Baundwal about 5 years
    Good answer, and sorry I'm late, but how did you know that 1 would do the trick? Of all the values of $c$, why 1?
  • Sil
    Sil about 5 years
    Just tried couple of small integers. It is usually worth checking few simple substitutions just to see if it does not lead to more workable polynomial.
  • CiaPan
    CiaPan almost 4 years
    I still wonder what is wrong with this answer.
  • Rodrigo de Azevedo
    Rodrigo de Azevedo over 2 years
    You have done the obvious part and left the hard part as an exercise for the reader. Usually, it's the other way round.
  • CiaPan
    CiaPan over 2 years
    @RodrigodeAzevedo What a pity you didn't notify me with a @-link. I would have a chance to improve it.........
  • Rodrigo de Azevedo
    Rodrigo de Azevedo over 2 years
    It is your answer. You were notified.