What’s the simplest way to prove that the polynomial $f(z)=z^4 + z^3 + z^2 +2z +3$ has no real zero?

Question

I have managed to write the polynomial mentioned in the title as a sum of squares as follows:

$$ z^4 + z^3 + z^2 +2z +3 = \left( z^2 + \frac12 z - \frac18 \right)^2 + \left( z + \frac{17}{16} \right)^2 + \frac{475}{256} $$

Is this the simplest way to prove that it has no real zero, or is there another way that is simpler?

Update. The first comment below, written by Piquito, seems interesting, but I don't understand why the inequality would be true for all real z. Can someone explain it?

Answer 1

Here is a completely elementary proof (without calculus, complicated calculations or tricky factorizations):

• For $z\geq 0$, we have $$z^4 + z^3 + z^2 +2z +3\geq 3.$$ Therefore, there is no real zero in the interval $[0,\infty)$.

• For $-1\leq z\leq 0$, we have (multiplying by $z^2$) $-z^2\leq z^3$. Then, $$z^4 + z^3 + z^2 +2z +3\geq z^4 + (-z^2) + z^2 +2(-1) +3=z^4+1>1.$$ Therefore, there is no real zero in the interval $[-1,0]$.

• For $z\leq -1$, we have (multiplying by $z^3$) $z^4\geq-z^3$. Then, $$z^4 + z^3 + z^2 +2z +3\geq(-z^3) + z^3 + z^2 +2z +3= z^2 +2z +3>0.$$ Therefore, there is no real zero in the interval $(-\infty,-1]$.

Answer 2

A solution(maybe more difficult but easier to think)

Let $f(z)=z^4+z^3+z^2+2z+3$, then $f'(z)=4z^3+3z^2+2z+2$.

It is easily to see that $f'(z)$ have only one real zero.

Let it be $z_0$, so $4z_0^3+3z_0^2+2z_0+2=0$.

And $f'(0)=2>0$, so $z_0<0$

(This is because f is a monotonically increasing function)

And $f(z)\geq f(z_0)=z_0^4+z_0^3+z_0^2+2z_0+3=z_0^4+z_0^3+z_0^2+1-(4z_0^3+3z_0^2)=z_0^4-3z_0^2-2z_0^2+1=z_0^2(z_0^2-3z_0+2)+1=z_0^2(z_0-1)(z_0-2)+1>1$

Done!

Answer 3

Rewrite the polynomial as follows :

$$ \begin{align}P(z):=\left(z^4+z^3+z+1\right)+\left(z^2+z+2\right)\end{align} $$

Then, you have :

$$ \begin{align}z^3\left(z+1\right)+(z+1)+\left(z^2+z+2\right)\end{align} $$

or

$$ \begin{align}(z+1)\left(z^3+1\right)+(z^2+z+2)\end{align} $$

Use the sum of cubes formula $z^3+1=(z+1)(z^2-z+1)$, which leads to :

$$ \begin{align}(z+1)^2\underbrace{\left(z^2-z+1\right)}_{\color{#c00}{\Delta_z<0}}+\underbrace {\left(z^2+z+2\right)}_{\color{#c00}{\Delta_z<0}}>0\end{align} $$

for all $z\in\mathbb R$ . Therefore, $P(z)>0,\thinspace \thinspace \forall z \in\mathbb R\thinspace .$

This completes the answer .

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Equivalently if you wish, convert both $\color{#c00}{\Delta_z}$ to completing the square, then you reach the following sum of squares :

$$ \begin{align}\left(z+1\right)^2\left(z-\frac 12\right)^2+\frac 34\left(z+1\right)^2+\left(z+\frac 12\right)^2+\frac 74\end{align} $$

which is equivalent to :

$$ \begin{align}\bbox[5px,border:2px solid #C0A000]{\left(z^2+\frac 12z-\frac 12\right)^2+\frac {1}{28}\left(7z+5\right)^2+\frac {13}{7}\thinspace .}\end{align} $$

Answer 4

First note $x=0$ is not a root. The idea is to split it into two parts and control the $\Delta=b^2-4ac<0$ for each part. For example,

$$f(z)=z^4 + \color{red}1\cdot z^3 + \alpha z^2+\beta z^2 +\color{blue}2\cdot z +3$$

where $\alpha+\beta=1$ and the coefficients of odd terms, which are $\color{red}1$ and $\color{blue}2$ will serve as the $b$ in our $\Delta$ for each part, namely $$\color{red}1^2-4\alpha<0~\land~\color{blue}2^2-12\beta<0~\land~\alpha+\beta=1$$ hence

$$\boxed{\alpha>\frac14~\land~\beta>\frac13~\land~\alpha+\beta=1}$$

So we have plenty of space to choose $\alpha$ and $\beta$, for example, $\alpha=\beta=\frac12$, then we have

$$\begin{align} f(z)&=z^4 + \color{red}1\cdot z^3 + \frac12z^2+\frac12z^2 +\color{blue}2\cdot z +3\\ \\ &=z^2\underbrace{\left(z^2+\color{red}1\cdot z+\frac12\right)}_{\Delta=\color{red}1^2-4\cdot\frac12<0}+\underbrace{\left(\frac12z^2 +\color{blue}2\cdot z +3\right)}_{\Delta=\color{blue}2^2-4\cdot\frac12\cdot3<0} \end{align}$$

Therefore, $f(z)>0, \forall z\in \mathbb R$

Answer 5

Remove the first two terms with $\left(z^2+\frac12z\right)^2$: $$ z^4+z^3+z^2+2z+3-\left(z^2+\frac12z\right)^2=\frac34z^2+2z+3\tag1 $$ Remove the first two terms of $\frac34z^2+2z+3$ with $\frac34\left(z+\frac43\right)^2$: $$ \frac34z^2+2z+3-\frac34\left(z+\frac43\right)^2=\frac53\tag2 $$ Thus, we get $$ z^4+z^3+z^2+2z+3=\left(z^2+\frac12z\right)^2+\frac34\left(z+\frac43\right)^2+\frac53\tag3 $$

Answer 6

I'll leave this answer here, if it works for you .

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Observe that :

$$ \begin{align}&z^4+z^3+z^2+2z+3\\ =\thinspace\thinspace\thinspace &\color{red}{z^2}(z^2+z+1)+2\color{red}{z}+3\end{align} $$

Then you have :

$$ \begin{align}\Delta_{\color{red}{z}}&=1-3(z^2+z+1)\\ &=-\underbrace {(3\color{blue}{z^2}+3\color{blue}{z}+2)}_{\Delta_{\color{blue}{z}}\thinspace<\thinspace0}\\ &<0\thinspace\thinspace,\forall z\in\mathbb R\thinspace .\end{align} $$

which completes the answer .

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$\rm {Explanation:}$

We rewrote the polynomial as a quadratic polynomial respect to the $\color{red}{z}$ . Next, we looked at the general discriminant of the equation with respect to $\color{red}{z}$ . We have shown that, this discriminant is always negative . Thus, you're done .

Answer 7

Alternative approach:

Let $P(x)$ a polynomial of even degree, (simple roots,) positive leading coefficient, such that its derivative $P'(x)$ has exactly one real root $\xi$ ( for instance, if $P''(x) \ge 0$ for all $x \in \mathbb{R}$). The minimum value of $P$ is $P(\xi)$, and, since the other roots are conjugate in pairs, it will have the sign of

$$\prod_{x_k, P'(x_k) = 0} P(x_k)$$

that is, the resultant of $P$, and $P'$, or, using the discriminant

$$(-1)^{\binom{n}{2}} \operatorname{Discr}(P(x))$$

Therefore, if the above is $>0$ then $P(x)>0$ on $\mathbb{R}$ ( this is a generalization of the case $P(x)$ of degree $2$).

In our case $n=4$, $P''(x)= 12 x^2 + 6 x + 2 >0$ on $\mathbb{R}$, $\binom{4}{2} = 6$ even, and the discriminant $$\operatorname{Discr}(P(x)) = 4345>0$$ so $P(x)> 0$ on $\mathbb{R}$.

Answer 8

It is not difficult to see that $z^3(z+1)=z^2z(z+1)\geq -1$ for all real numbers $z$. So, $$f(z)=z^4+z^3+z^2+2z+3=z^3(z+1)+(z+1)^2+2\geq -1+0+2=1.$$