Show that the polynomial $P(x):=x^4-6x+6$ has no real roots .

Question

Show that the polynomial $$P(x):=x^4-6x+6$$ has no real roots.

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We need to solve this problem without using calculus. This is a problem from my son's olympiad textbook. Since the degree of the polynomial is $4$, our task seems difficult. I tried to factor the polynomial. For example, I was expecting an expression like $(x^2+1)(x^2+x+1)$. But unfortunately it didn't.

Rational root theorem obviously fails. Because, real roots don't exist.

I tried to rewrite the polynomial

$$x^4-6x+6=(x^2+ax+b)(x^2+cx+d)$$

So, it sufficies to show that $a^2-4b<0$ and $c^2-4d<0$.

But, I am not sure, is it good track or not.

Thanks for advance .

Answer 1

Write $$x^4-6x+6=(x^2-1)^2+2\left(x-\frac32\right)^2+\frac12$$ and then it is clear that $x^4-6x+6>0$ for all $x\in\mathbb{R}$.

Answer 2

Still another way :

If $x≤0$, then $x^4-6x≥0\thinspace .$ Therefore, $x>0\thinspace .$ Thus, using the AM-GM inequality you have :

$$x^3+\frac 2x+\frac 2x+\frac 2x=6≥4\sqrt [4]{8}$$

A contradiction .

Answer 3

We can use for example

$$x^4-6x+6 =(x^2-1)^2+2x^2-6x+5 >0$$

indeed $(x^2-1)^2\ge 0$ and

$$36-4\cdot 2\cdot 5 =-4<0$$

Answer 4

Hint: $\,P(x+1) = x^4 + 4 x^3 + 6 x^2 - 2 x + 1 = x^2(x+2)^2 + x^2 + (x-1)^2\,$.

Answer 5

This question attracted a huge echo, so let us write down a further decomposition of $f$ into a sum of squares: $$ x^4-6x+6=\left(x^2 -\frac 32\right)^2 + 3(x-1)^2 + \frac 34\ge \frac 34=0.75\ . $$ $\square$

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Note: In fact, using analytic tools, we can immediately catch the sign of $f'(x)=4x^3-6$, it is negative on $(-\infty, a)$, positive on $(a,\infty)$, here $a=(3/2)^{1/3}\approx 1.1447142425533\dots$, so $f$ decreases on the first interval, increases on the second one. So $a$ is the point where the best lower bound, the global minimal value $f(a)\approx 0.8487859085100\dots$ is taken.

Above, we obtained the lower bound $0.75$ which is good enough for the purpose of excluding real roots.

Answer 6

Another way, for $x<1$ we have $6-6x>0$ and for $x\ge1$

$$(x-1)^4+4(x-1)^3+6(x-a)^2+b=x^4+(8-12a)x+6a^2+b-3$$

which leads to $a=\frac76$ and $b=\frac 56$ that is

$$x^4-6x+6=(x-1)^4+4(x-1)^3+6\left(x-\frac76\right)^2+\frac56>0$$

Answer 7

To pursue your approach of factoring the quartic is a little troublesome because the exact expressions for the coefficients prove to be pretty terrifying (we catch the "bad end" of Ferrari's method).

We can conclude from the Rule of Signs that $ \ x^4 - 6x + 6 \ $ has either two or no positive and no negative real zeroes.* If we posit that there are two real and two complex-conjugate zeroes, we can begin with two quadratic factors $ \ (x^2 - ax + b)·(x^2 - 2cx + 6/b) \ \ , \ $ with $ \ c \ $ being the real part of the complex zeroes and $ \ \frac{6}{b} \ $ being the square of their modulus. The cubic term gives us $ \ a + 2c \ = \ 0 \ \ , \ $ so using this leads to equations for the quadratic and linear coefficients $$ b \ + \ \frac{6}{b} \ - \ a^2 \ \ = \ \ 0 \ \ \ , \ \ \ ab \ - \ \frac{6a}{b} \ \ = \ \ -6 \ \ . $$

$ ^{*} $ We can also see that $ \ x^4 - 6x \ = \ x·(x^3 - 6) \ $ has a zero at $ \ x \ = \ 0 \ $ and a "triple zero" at $ \ x \ = \ \sqrt[3]{6} \ \ , \ $ so "raising" the function curve by $ \ 6 \ $ units would produce a transformed curve with at most two positive real zeroes, or none if the minimum is $ \ y \ > \ -6 \ \ . \ $ (It turns out that it is, but only very slightly; this can be should be an inequality argument without calculus, but it is not simpler than the other (easier-to-follow) inequality-based answers already presented.)

Solving the non-linear $ \ a^2 \ = \ b \ + \ \frac{6}{b} \ \ , \ \ a·\left( \frac{6}{b} \ - \ b \right) \ = \ -6 \ \ $ simultaneously is what produces the rather complicated exact values for $ \ a \ $ and $ \ b \ \ . \ $

With a bit of computational help, we obtain a pair of solutions $ \ a \ \approx \ -2.35046 \ \ , \ \ b \ \approx \ 1.4857 \ \ ; \ \ a \ \approx \ +2.35046 \ \ , \ \ b \ \approx \ 4.0385 \ \ , \ $ corresponding to the factorization $ \ x^4 - 6x + 6 \ = \ (x^2 - 2.35046·x + 1.4857)·(x^2 + 2.35046·x + 4.0385) \ \ . $ Since $ \ \large{\frac{2.35046^2}{4} } \ \normalsize{\approx \ 1.38104 \ \ ,} \ $ neither of these factors has a non-negative discriminant. Both factors are irreducible over the real numbers, so the quartic polynomial has no real zeroes. (From the remark about how the complex zeroes are manifested in the coefficients of the quadratic factors, we could now "read off" what the four zeroes of this polynomial are.)

Answer 8

The discriminant of the depressed quartic $p(x)=x^4+dx+e$ is $\Delta=-27d^4+256e^3$. In our example, $d=-6$ and $e=6$, hence $\Delta=2^43^347>0.$ For quartics with real number coefficients, if the discriminant is positive, then the roots are either all real or all non-real. In this example, we can not have $4$ real roots due to Decartes's rule of signs. See also boojum's answer. We conclude that $P(x)>0$ for all $x\in\Bbb R$.