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

Question

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

As I have learnt from my previous post, using Descarte's Sign rule I am getting $4$ positive and $0$ negative roots.

So no of nonreal roots are $N-(p+q) = 8-(4+0)= 4$.So rest $4$ roots must be real! It is violating the question's condition.Can anybody help me out! I am not understanding the derivative concept regarding this!

Descarte's rule of signs does not tell you there are $4$ positive roots. It only says there are an even number of positive roots, and there are at most $4$. As the answers show, your objective here is rather to show that this function cannot be zero for real $x$. — Izaak van Dongen, Jun 20 '20 at 13:18
https://math.stackexchange.com/questions/2487480/show-that-the-polynomial-x8-x7x2-x15-has-no-real-root?rq=1 I guess this helps! — , Jun 20 '20 at 14:21
Does this answer your question? Show that the polynomial $x^8 -x^7+x^2-x+15$ has no real root — tryst with freedom, Jun 14 '21 at 18:16

score 10 · Accepted Answer · edited Jun 20 '20 at 13:18

10

If $x<0$ every term is positive, hence the polynomial is $>0$.

If $x>1$ , then $x^2-x>0$ , $x^8 - x^7 >0$ , $15>0$ hence the function is $>0$.

If $0<x<1$, then $1-x>0$ , $x^2 - x^6>0$, $14+x^8>0$ hence the function is $>0$

Hence for every real $x$ the function is $>0$ , hence no real root.

edited Jun 20 '20 at 13:18

Henno Brandsma

242,131

answered Jun 20 '20 at 13:09

aryan bansal

1,923

1

One suggestion for this post (and possibly others): try to implement different phrases. In this post alone, you’ve used the word “hence” 5 times in 4 sentences. Just a tip to keep in mind while writing :) – Clayton Jun 20 '20 at 13:24
1

@Clayton, on the contrary. In mathematical writing similar content is better expressed in similar form. See for instance https://books.google.com/books?id=EYHFObaUIdIC&pg=PA52&lpg=PA52&dq=mathematical+writing+parallel+phrases&source=bl&ots=i0OQfW3lg2&sig=ACfU3U3VfRrkDUJAgjU8OwzEJJQBmUIc_w&hl=en&sa=X&redir_esc=y#v=onepage&q=mathematical%20writing%20parallel%20phrases&f=false – lhf Jun 20 '20 at 13:29

score 1 · Answer 2 · answered Jun 20 '20 at 13:18

Descarte's Sign rule is saying that you can have $4$,$2$ or $0$ positive real roots. The change of sign is giving the upper limit to the number of real roots not the exact value.

So it is not really helpful in deciding if there are no real roots at all. You would still have to argue that the polynomial has no negative evaluation at all.

So, it is very easy to see that this polynomial is always positive. For $|x|>1$ it is always $x^{2n}-x^{n} > 0$, because it is the same as $x^{2n} > x^{n}$, and this covers two parts of it. For $|x| \leq 1$, the sum $|x^{8}-x^{7}+x^{2}-x^{1}|$ cannot exceed $15$ as it is $4$ at worst.

The polynomial is then positive all the time.

score 0 · Answer 3 · edited Jul 11 '20 at 16:03

0

$f(x)=x^8-x^7+x^2-x+15$ has atmost $4$ positive real roots due to $4$ sign changes in its coefficients.

$f(-x)=x^8+x^7+x^2+x+15$ has no sign changes so no negative real root. (using Descartes rule of signs)

Further, $f(x)=(x-1)(x^7+x)+15$ and observe that $f(x)>0$ for all $x>0$ so there will be no positive root in $(0,\infty)$.

edited Jul 11 '20 at 16:03

answered Jun 20 '20 at 16:38

Nitin Uniyal

7,946

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

3 Answers3