Show that $p(x)=2x^6+12x^5+30x^4+60x^3+8x^2+30x+45$ has no real roots

Question

The following question appears in a book for high school students who do not know any calculus but know the basic theory of polynomials:

Show that the equation $$p(x)=2x^6+12x^5+30x^4+60x^3+8x^2+30x+45=0$$ has no real roots.

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My thoughts:

Clearly it has no positive real roots. I guess the idea is to group the terms so that each grouping is a positive function for negative $x$. But I could not find such a grouping.

Another observation is that $p(x) - 2(x+1)^6 = 20x^3 + 50x^2 + 18x + 43$. If we could show $(1-x)^6 < -10x^3 + 25x^2 -9x + 21$ for all positive $x$ we will be done. But I could not show that as well.

Answer 1

The problem is unfortunately false; $p(-1) = -17$, but $p(0) = 45$, so there must be a root in $(-1,0)$.