Prove that $x^4 + x^3 + x^2 + x + 1$ has no roots

Question

Prove that $x^4 + x^3 + x^2 + x + 1$ has no roots.

I can see two ways of proving it.

The first one is to notice, that $x^5 - 1 = (x - 1)(x^4 + x^3 + x^2 + x + 1)$. It has the only root 1. And it is not the root of $x^4 + x^3 + x^2 + x + 1$. So, $x^4 + x^3 + x^2 + x + 1$ does not have roots.

Another way is to solve it as a palindromic polynomial. It does not have roots.

But is there any way to directly manipulate the expression to show that it is always greater than zero?

Can you show that it is always positive ? . That will prove it very easily. The case when x > 0 is obvious. So your task is to show for x < 0. — Shailesh, Aug 09 '15 at 09:46
I imagine you mean that there are no real roots. Polynomials always have complex roots. — Mark Bennet, Aug 09 '15 at 09:55
@MarkBennet Yes of course. And that is probably what OP meant too. Didn't even see that he missed that word 'real'. — Shailesh, Aug 09 '15 at 09:58

score 5 · Answer 1 · edited Feb 28 '20 at 21:18

5

$$f(x)=x^4+x^3+x^2+x+1$$ if $x=1$ then $f(1)=5$ hense $f(1)>0$

if $x\neq 1$ then $f(x)=\frac{x^5-1}{x-1}$

$(x^5-1)$ and $(x-1)$ have the same sign any $x$. Hence $f(x)>0$

edited Feb 28 '20 at 21:18

Bernard

175,478

answered Aug 09 '15 at 10:02

JJacquelin

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Awesome solution! – Harshal Gajjar Aug 09 '15 at 10:07
Is this not slightly circular? Having the same sign shows that $f(x) \ge 0$ not that $f(x)>0$. To get a strict inequality, you need to know that $f(x) \ne 0$. – Mathmo123 Aug 09 '15 at 11:07
The case $x=1$ being considered before, in the case $x\neq 1$ it is obvious that $x^5-1\neq0$ hense $f(x)\neq0$. – JJacquelin Aug 09 '15 at 11:16
@JJacquelin that's precisely my point. Your proof that $f(x) >0$ requires knowing that $f(x) \ne 0$, and therefore cannot be used to prove that $f(x) \ne 0$. – Mathmo123 Aug 09 '15 at 11:24
Read again my prove. Never $f(x)\neq 0$ is required. – JJacquelin Aug 09 '15 at 13:12

score 4 · Answer 2 · answered Aug 09 '15 at 09:54

There is a theorem about positive polynomials being expressible as the sum of squares. Here $$f(x)=\left(x^2+\frac x2\right)^2 +\frac 34\left(x+\frac 23\right)^2+\frac 23$$

obtained by successively completing the square. Personally I prefer some of the simpler methods. For example one can observe that the roots of $f(x)$ are the complex fifth roots of $1$, so there are no real zeros.

score 3 · Answer 3 · answered Aug 09 '15 at 11:11

Another proof by contradiction.

Suppose $x^4+x^3+x^2+x+1$ has a real root. Necessarily, $x<0$. Set $x=-t,\enspace t>0$. Rewrite the equation as $$t^4-t^3+t^2-t=t(t-1)(t^2+1)=-1$$ It shows $t-1<0$, hence $\;0<t<1$.

On another hand, $\;t=\dfrac1{(1-t)(1+t^2)}=\dfrac1{1-t+t^2-t^3}$, and since $0<t<1$, we have $0<1-t+t^2-t^3<1-t+t^2=1-t(1-t)<1$, hence $$t=\dfrac1{1-t+t^2-t^3}>1,\quad\text{contradiction}.$$

Marco Cantarini · Answer 4 · 2015-08-09T10:05:49.717

Assuming that exists $a $ such that $$a^{4}+a^{3}+a^{2}+a+1=0. $$ Now it's obvious that $a\neq0 $. So we have $$0=\frac{\left(a^{4}+a^{3}+a^{2}+a+1\right)}{a}=a^{3}+a^{2}+a+1+\frac{1}{a} $$ so $$a^{3}+a^{2}+a+1+\frac{1}{a}=a^{4}+a^{3}+a^{2}+a+1\Longrightarrow\frac{1}{a}=a^{4}\Rightarrow1=a^{5} $$ so we have $$a=1 $$ and this is absurd. Now since $f\left(x\right)=x^{4}+x^{3}+x^{2}+x+1 $ is a continuous function, to see if it is positive it is sufficient see what happens at one point.

Prove that $x^4 + x^3 + x^2 + x + 1$ has no roots

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