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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.

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    $\begingroup$ @Hendrix: mh, as the leading coefficient is positive and the polynomial is said to have no real roots, this is not a great discovery. $\endgroup$ – Yves Daoust Aug 3 '19 at 14:25
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    $\begingroup$ @Hendrix That's a rather pointless statement, since that is precisely what needs to be shown. $\endgroup$ – TheSimpliFire Aug 3 '19 at 14:25
  • $\begingroup$ Have you tried to factor the polynomial's derivative? $\endgroup$ – Robert Shore Aug 3 '19 at 14:28
  • $\begingroup$ @RobertShore: is this simpler ? $\endgroup$ – Yves Daoust Aug 3 '19 at 14:32
  • $\begingroup$ @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. $\endgroup$ – Robert Shore Aug 3 '19 at 14:36
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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.

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    $\begingroup$ How did you get this magical decomposition ? $\endgroup$ – Yves Daoust Aug 3 '19 at 14:34
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    $\begingroup$ @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. $\endgroup$ – J.G. Aug 3 '19 at 14:35
  • $\begingroup$ Thanks. Actually I was trying to make perfect cube and completely forgot about perfect square $\endgroup$ – user456230 Aug 3 '19 at 14:46
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    $\begingroup$ @user456230 Perfect cubes aren't non-negative unless you're cubing a non-negative, so it sounds like you were trying to cube a quadratic of non-positive discriminant. With problems like these, you should always start by trying to write a sum of squares. $\endgroup$ – J.G. Aug 3 '19 at 14:48
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    $\begingroup$ 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. $\endgroup$ – zwim Aug 3 '19 at 15:25
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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 $x.$

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