Showing that $2z^4-3z^3+3z^2-z+1=0$ has no Real Roots

Question

I am trying to show that $$p(z)=2z^4-3z^3+3z^2-z+1=0$$ has no real roots.

From a previous related post, I incorrectly proposed the following:

Suppose $p(z)$ has a root on the real axes, then $p(x)=2x^4-3x^3+3x^2-x+1=0$ for some $x\in\mathbb{R}$. Now by the rational root theorem, the only possible roots that $p(x)$ can have is when $x=1$. But $p(1)\neq 0$ and hence $p(z)$ cannot have a root on the real axes by contradiction.

Alternatively, if we can show that $p(x)\geq 0$ or $p(x)\leq 0$, this would be sufficient to show that no real roots exist. But this is not obvious. I have also shown that $p(z)$ has no imaginary roots.

I am unsure of how to show that $p(z)$ does not have a real root.

Answer 1

For a real $z$ we have: $$ 2z^4-3z^3+3z^2-z+1 = \underbrace{z^2(2z^2-3z+2)}_{\ge 0}+ \underbrace{z^2-z+1}_{>0} >0\ . $$

Answer 2

I see the two threes in there, and I immediately think binomial theorem. To make those threes appear in the right degrees and with the right sign, you would need $$ z(z-1)^3 = z^4 - 3z^3 + 3z^2 - z $$ Thus we have $$ p(z) = z(z-1)^3 + z^4 + 1 $$ Clearly $z^4 + 1$ is always greater than or equal to $1$. It remains to show that $z(z-1)^3>-1$. Since the only place $z(z-1)^3$ is negative is for $x\in (0,1)$, and on that interval, it's the product of one term between $0$ and $1$, and one term between $-1$ and $0$, we have our result.

Answer 3

Note that$$P\left(x+\frac38\right)=2 x^4+\frac{21 x^2}{16}+\frac{13x}{32}+\frac{1901}{2048}.$$Since the discriminant of $\frac{21 x^2}{16}+\frac{13x}{32}+\frac{1901}{2048}$ is negative, $\frac{21 x^2}{16}+\frac{13x}{32}+\frac{1901}{2048}>0$ for each real number $x$. Therefore$$(\forall x\in\mathbb{R}):P\left(x+\frac38\right)>0$$and this is equivalent to $(\forall x\in\mathbb{R}):P(x)>0$.

Answer 4

Hint: Note that $$2z^4-3z^3+3z^2-z+1=2\,\left(z^2-\frac{3}4z\right)^2+\frac{15}{8}\,\left(z-\frac{4}{15}\right)^2+\frac{13}{15}\,.$$