Show that the polynomial $x^8 -x^7+x^2-x+15$ has no real root

I am not getting on how to approach this problem. Clearly, this polynomial can have atleast 2 real roots. And using Descartes's rule of signs, it can have a maximum of 4 positive real roots. But after that, how should I proceed ?

Any help would be highly appreciated...

Note that $x^8-x^7$ and $x^2 - x$ look very similar. We can use this to factor some of the terms: $$x^8 - x^7 + x^2 - x + 15\\ = x^7(x-1) + x(x-1) + 15\\ = (x^7+x)(x-1) + 15\\ = x(x^6+1)(x-1) + 15$$ If the entire expression is to be zero, then at the very least, we must have $x(x^6+1)(x-1)< 0$, which happens for $x\in (0, 1)$. But for $x\in (0, 1)$, we have $$|x|<1\\ |x^6+1|<2\\ |x-1|<1$$ which implies $|x(x^6-1)(x-1)|<2$, which in turn implies that $x(x^6+1)(x-1) + 15$ is always larger than $13$.

• Using the fact that $15-2=13 > 11$. – Jeppe Stig Nielsen Oct 24 '17 at 12:54
• @JeppeStigNielsen The $11$ was a remnant from some editing. But yes, $x(x^6-1)(x-1)>-2$ is what I have shown (since $x(x^6-1)(x-1)$ is either positive, or has absolute value less than $2$), so adding $15$ to either side we get the result in my final paragraph. – Arthur Oct 24 '17 at 12:58
• I got to witness a new kind of approach. That should help me in the future. Highly appreciate everyone's help – Sayan Mukherjee Oct 24 '17 at 17:59

You can rewrite it as $$x^7(x-1) + x(x-1) + 15$$

The minimum of $x(x-1)$ is $-\frac14$, and the two local minima of $x^8-x^7$ are acchieved at $x=0$ and $x=\frac78$. One of them, obviously, is the global minimum, and it is equal to $\left(\frac{7}{8}\right)^8-\left(\frac{7}{8}\right)^7>-1$

So, the expression is equal to $$\left(\text{something bigger than }-\frac14\right) +\left(\text{something bigger than }-1\right) + 15$$

We have $p(x)=x^8-x^7+x^2-x+15=x(x-1)(x^6+1)+15$

If $x \le 0$ then $x-1<0$, hence $x(x-1) \ge 0$, which gives $p(x)>0$.

If $x \ge 1$, then it is clear that $p(x)>0$.

Now let $0<x<1$. Then $x(x-1) \ge -1/4$, hence $x(x-1)(x^6+1) \ge -\frac{1}{4}(x^6+1)$.

Now it is your turn to show that $-\frac{1}{4}(x^6+1)+15>0$ for $0<x<1$.

If so we have $$x^8-x^7+x^2-x+15=\frac{x^4(x^6-x^3+1)+x^2+1}{x^2+x+1}+14>14>0.$$

We have that $x^8-x^7$ and $x^2-x$ are bigger then zero for $x<0$ because $x^8>0$ and $-x^7>0$ similar logic for $x^2$ and $-x$ so $p(x)>15$ for $x<0$.

For $0<x<1$ we have that $x^8-x^7+x^2-x+15>-x^7-x+15>-1-1+15>13$ so $p(x)>13$

For $x\geq1$ we have that $x^8\geq x^7$ and $x^2\geq x$ so $x^8-x^7+x^2-x+15\geq 15$ so $p(x)\geq 15$.

Since $x=0$ is not a root we can conclude there aren't any roots.

This is not a very rigorous approach but I believe that it does the trick:

Let's assume that such a root $r \in \mathbb{R}$ exists, so $$r^8-r^7+r^2-r+15=0\Rightarrow\\r^2(r^6+1)-r(r^6+1)=-15\Rightarrow\\r(r-1)=\frac{-15}{r^6+1}\lt0$$

So $r(r-1)\gt0\Rightarrow r\in (0,1)$

We notice that if $p(x)$ is our polynomial, that $p(0)=p(1)=15$ so in $(0,1)$ its graph must consist of a deep downward spike that reaches for $0$.

Now let's consider the area covered by the rectangle with length $15$ and height $1$, which is obviously $15$ and that of the area which lies between the graph of $p(x)$, the $xx'$ axis and $x=0, x=1$.

We get $$\int_{0}^1(x^8-x^7+x^2-x+15)dx=14,824$$ which is quite close to $15$ indicating that the two shapes-the rectangle and the area under $p(x)$ are quite similar.

But this could not happen if $p(x)$ had a root in $(0,1)$ thus no such $r$ exists.