Prove that $3x^6+12x^5+9x^4-24x^3+9x^2-4x+3=0$ does not have any real root.

Question

I need to prove that $3x^6+12x^5+9x^4-24x^3+9x^2-4x+3=0$ does not have any real root. I tried analyzing the derivatives to see the maxima and minima but I can't compute them exactly, so I couldn't proceed further. Hints are appreciated.

Answer 1

One approach would be applying Sturm's Theorem to find the total number of real roots of the polynomial

$$p(x)=3x^6+12x^5+9x^4-24x^3+9x^2-4x+3$$

Sturm's theorem expresses the number of distinct real roots of $p$ located in an interval in terms of the number of changes of signs of the values of the Sturm sequence at the bounds of the interval. Applied to the interval of all the real numbers, it gives the total number of real roots of $p$.

The computation itself becomes slightly messy as you go further down the Sturm sequence. Although, it is still doable with the help of a Computer Algebraic System such as Mathematica. Suppose we wish to find the number of roots in some range for the polynomial

$$p(x)=3x^6+12x^5+9x^4-24x^3+9x^2-4x+3$$

Let

$$p_0(x)=3x^6+12x^5+9x^4-24x^3+9x^2-4x+3$$ $$p_1(x)=p'(x)=18 x^5 + 60 x^4 + 36 x^3 - 72 x^2 + 18 x - 4$$ The remainder of the Euclidean division of $p_0$ by $p_1$ is

$$-\frac{11x^4}{3}-16x^3+14x^2-\frac{16x}{3}+\frac{31}{9}$$

multiplying by $-1$ we obtain

$$p_2(x)=\frac{11x^4}{3}+16x^3-14x^2+\frac{16x}{3}-\frac{31}{9}$$

Next, the remainder of the Euclidean division of $p_1$ by $p_2$ is

$$\frac{22464x^3}{121}-\frac{20448x^2}{121}+\frac{7488x}{121}-\frac{2592}{121}$$

multiplying by $-1$ we obtain

$$p_3(x)=-\frac{22464x^3}{121}+\frac{20448x^2}{121}-\frac{7488x}{121}+\frac{2592}{121}$$

Next, the remainder of the Euclidean division of $p_2$ by $p_3$ is

$$\frac{43439x^2}{18252}-\frac{242x}{351}-\frac{7381}{6084}$$

multiplying by $-1$ we obtain

$$p_4(x)=-\frac{43439x^2}{18252}+\frac{242x}{351}+\frac{7381}{6084}$$

Next, the remainder of the Euclidean division of $p_3$ by $p_4$ is

$$-\frac{1920402432x}{15594601}+\frac{1249896960}{15594601}$$

multiplying by $-1$ we obtain

$$p_5(x)=\frac{1920402432x}{15594601}-\frac{1249896960}{15594601}$$

Next, the remainder of the Euclidean division of $p_4$ by $p_5$ is

$$\frac{8062408717}{12332584368}$$

multiplying by $-1$ we obtain

$$p_6(x)=-\frac{8062408717}{12332584368}$$

As this is a constant, this finishes the computation of the Sturm sequence. To find the number of real roots of $p_0$ one has to evaluate the sequences of the signs of these polynomials at $-\infty$ and $\infty$.

$$\text{For $-\infty$, the sequence of signs is: (+, −, +, +, −,−,−)}$$ $$\text{For $\infty$, the sequence of signs is: (+, +, +, −,−,+,−)}$$

Therefore

$$V(-\infty)-V(+\infty)=3-3=0$$

which shows that $p$ has no real roots.

Answer 2

I wanted to see if there was a way to prove this without heavy numerical calculation. I found such a proof, but it suggests, in two ways, that it is better to work with the original trigonometric form of the problem. For one thing, my proof requires a finicky division of the range of values of $x$ into subintervals on which the functions involved are monotonic. In the trigonometric version, an initial crude subdivision (it requires refinement) is suggested when one disposes of those values of $\theta$ for which the inequality $3 + \sin3\theta > \sin\theta + 2\sin2\theta$ (true for all $\theta$, incidentally, not just $\theta \in [0, \pi]$) is implied by the bounds on the value of the sine function. I didn't bother to complete that proof, because it is messy and uninteresting; but the proof of the polynomial version, given below, is trickier.

The second reason for being led back to the trigonometric version is that it was an initial unsimplified expression, in which $x = \tan\frac{\theta}2$, that suggested how to rewrite the polynomial expression for non-negative values of $x.$

First, though, we dispose of the simpler case of non-positive values of $x$: \begin{align*} p(x) & = 3x^6 + 12x^5 + 9x^4 - 24x^3 + 9x^2 - 4x + 3 \\ & > 3x^6 + 12x^5 + 9x^4 - 24x^3 \quad (x \leqslant 0) \\ & = 3x^3(x^3 + 4x^2 + 3x - 8) \\ & = 3x^3(x - 1)(x^2 + 5x + 8) \\ & = 3x^3(x - 1)\left[\left(x + \tfrac52\right)^2 + \tfrac74\right] \\ & \geqslant 0 \quad (x \leqslant 0), \end{align*} therefore: $$ p(x) > 0 \text{ for all } x \leqslant 0. $$

Now, no longer assuming that $x \leqslant 0$: \begin{align*} p(x) & = 3x^6 + 12x^5 + 9x^4 - 24x^3 + 9x^2 - 4x + 3 \\ & = 3q(x) + 4r(x), \text{ where:} \\ r(x) & = 3x^5 - x = x(3x^4 - 1), \\ q(x) & = (x^2 + 1)^3 - 24x^3 = (x^2 + 1)^3 - (2x)^3 \\ & = (x^2 - 2x + 1)[(x^2 + 1)^2 + 2x(x^2 + 1) + 4x^2] \\ & = (x - 1)^2[(x^2 + x + 1)^2 + 3x^2]. \end{align*} Differentiating: \begin{align*} r'(x) & = 15x^4 - 1, \\ q'(x) & = 6x(x - 1)u(x), \text{ where:} \\ u(x) & = x^3 + x^2 + 3x - 1, \\ u'(x) & = 3x^2 + 2x + 3 \\ & = 3\left(x + \tfrac13\right)^2 + \tfrac83 > 0. \end{align*}

Let $c$ be the unique real zero of $u(x)$. All we need to know is that $u(0) = -1$ and $u\left(\tfrac12\right) = \tfrac78$, whence $c$ is between $0$ and $\tfrac12$. (In fact, $c \bumpeq 0.2956$.) Also, define: \begin{gather*} a = 3^{-\tfrac14} \bumpeq 0.7598, \\ b = 15^{-\tfrac14} \bumpeq 0.5081. \end{gather*}

Choosing the arbitrary but convenient value of $\tfrac23$ to further subdivide the range of values of $x$, we have: $$ 0 < c < b < \tfrac23 < a < 1. $$

Now to prove that $p(x) > 0$ for all $x \geqslant 0$.

For a start, $q(x) \geqslant 0$ for all $x$, and $q(x) = 0$ iff $x = 1$. Since $r(x) \geqslant 0$ for $x \geqslant a$, we have: $$ p(x) > 0 \text{ for all } x \geqslant a. $$

This leaves us with the case $0 < x < a$.

We have $r(0) = r(a) = 0$, and $r$ is strictly decreasing in $[0, b]$, and strictly increasing in $[b, a]$. The minimum value of $r$ on the interval $[0, a]$ is, therefore: $$ r(b) = b(3b^4 - 1) = -\frac{4b}5 = -\frac{4}{5\sqrt[4]{15}} \bumpeq -0.4065. $$ On the other hand, $q(0) = 1$ and $q(1) = 0$, and $q$ is strictly increasing in $[0, c]$, and strictly decreasing in $[c, 1]$. The minimum value of $q$ on the interval $\left[0, \tfrac23\right]$ is, therefore: $$ q\left(\frac23\right) = \frac{469}{729} > \frac{459}{729} = \frac{17}{27}. % \bumpeq 0.6433. $$ Consequently, using the estimate $3^7 = 2187 < 3125 = 5^5$ $\therefore\ \sqrt[4]{15} > \tfrac95$: $$ p(x) = 3q(x) + 4r(x) > \frac{17}9 - \frac{16}{5\sqrt[4]{15}} > \frac{17}9 - \frac{16}9 = \frac19 \quad \left(0 \leqslant x \leqslant \frac23\right). $$

Although the remaining case $\tfrac23 < x < a$ is most easily handled by using a pocket calculator, I prefer to use the estimate $3^5 = 243 < 256 = 2^8$, which gives $3^{5/4} < 4$, or equivalently $a > \tfrac34$, whence $a^3 = \tfrac1{3a} < \tfrac49$, and: $$ q(a) = (a^2 + 1)^3 - 8a^3 = \left(1 + \frac1{\sqrt{3}}\right)^3 - 8a^3 > \frac{18 + 10\sqrt3}{9} - \frac{32}9 = \frac{10\sqrt3 - 14}{9}. $$ The minimum value of $q$ on the interval $\left[\tfrac23, a\right]$ is $q(a)$, for which we have just given a lower bound, and the minimum value of $r$ on the same interval is: $$ r\!\left(\frac23\right) = \frac23\cdot\left(\frac{16}{27} - 1\right) = -\frac{22}{81}, $$ therefore: $$ p(x) = 3q(x) + 4r(x) > % 3\times0.414970 - \tfrac{88}{81} \bumpeq 0.1585 \frac{10\sqrt3 - 14}{3} - \frac{88}{81} = \frac{270\sqrt3 - 466}{81} \quad \left(\frac23 \leqslant x \leqslant a\right). $$ The right hand side is positive, because $3\cdot270^2 = 218700 > 217156 = 466^2$. This completes the proof that $p(x) > 0$ for all $x$.