# Show that $2x^6+12x^5+30x^4+60x^3+80x^2+30x+45=0$ has no real roots

I tried solving the above question but was unable to prove it. I used Descartes rule of sign, factorisation techniques, and many other things but could not figure out the solution.

• @Hendrix: mh, as the leading coefficient is positive and the polynomial is said to have no real roots, this is not a great discovery. – Yves Daoust Aug 3 '19 at 14:25
• @Hendrix That's a rather pointless statement, since that is precisely what needs to be shown. – TheSimpliFire Aug 3 '19 at 14:25
• Have you tried to factor the polynomial's derivative? – Robert Shore Aug 3 '19 at 14:28
• @RobertShore: is this simpler ? – Yves Daoust Aug 3 '19 at 14:32
• @YvesDaoust I don't know whether it's simpler. It just seemed like a relatively quick idea that was worth trying. My question mark was genuine, not a disguised hint. – Robert Shore Aug 3 '19 at 14:36

First note$$2x^6+12x^5+30x^4+60x^3+80x^2+30x+45=2(x^3+3x^2)^2+12\left(x^2+\tfrac52 x\right)^2+5(x+3)^2.$$Not only is this non-negative, but it could only be zero if$$x^3+3x^2=x^2+\tfrac52 x=x+3=0.$$The last condition simplifies to $$x=-3$$, which contradicts the second condition.
• @YvesDaoust I used up the $x^6,\,x^5$ terms with a perfect square, and noted the remaining $x^4$ coefficient was positive. I then mopped up the remaining $x^4,\,x^3$ terms in the same way, and got a perfect square left over. (I double-checked my arithmetic at the end on Wolfram Alpha.) Nothing magical about it. – J.G. Aug 3 '19 at 14:35
• As I was curious, I tried this method for some other polynomials to see if it was universal, either by subtracting $(x^n+ax^{n-1})^2$ or more terms $(x^n+ax^{n-1}+bx^{n-2}+...)^2$, but we are not systematically left with another positive polynomial. As if the method and the polynomial were destined to work together. – zwim Aug 3 '19 at 15:25
Factor $$x^4$$ and separate a non-negative part of the expression covering completely the terms $$x^6$$ and $$x^5,$$ then factor $$x^2,$$ ... \begin{aligned}P(x)=&2x^6+12x^5+30x^4+60x^3+80x^2+30x+45\\=&2x^4(x^2+6x+9)+12x^4+60x^3+80x^2+30x+45\\=&2x^4(x^2+6x+9)+3x^2(4x^2+20x+25)+30x^2+30x+45\\=&2x^4(x^2+6x+9)+3x^2(4x^2+20x+25)+15(2x^2+2x+3)\end{aligned} which is strictly positive for any real $$x.$$