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

Question

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.

Answer 1

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.

Answer 2

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 $x.$

Answer 3

$2x^6+12x^5+30x^4+60x^3+80x^2+30x+4$

Can be factored into:

$ 2x^4(x^2 + 6x + 9) + 12x^2(x^2 + 5x + \frac{25}{4}) + 5(x^2+6x+9) = 0 $

$=2x^4(x+3)^2 +12x^2(x+\frac{5}{2})^2 +5(x+3)^2 = 0$

But this is not possible for real $x$