3

I have the following polynomial : $x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1$

I must determine if this polynomial has at least 1 real solution and justify why. We have a theorem which says that all polynomials with real coefficients can be decomposed in a product of polynomials of real coefficients with degree 1 or 2. So this means we have four scenarios :

Factors : 2+2+2+1 , 2+2+1+1+1, 2+1+1+1+1+1, 1+1+1+1+1+1+1

In all these cases, we have atleast one factor of degree 1, so there is atleast one real solution in each case. What do you think ?

  • 3
    Yes, this is fine, although a simpler solution for this particular case is simply noting that $x = -1$ is a solution. Note that your proof extends to any polynomial of odd degree that has real coefficients (if you write it a tiny bit more generally). – stochasticboy321 Nov 01 '15 at 02:41
  • 2
    What if you put $x = -1$ ? – Shailesh Nov 01 '15 at 02:41
  • Yeah, but I don't think my teacher wants me to simply find x=-1. I think he really wants a reason why my polynomial would have one real solution. Anyway, thank you ! –  Nov 01 '15 at 02:42
  • The Fundamental Theorem of Algebra is quite a powerful thing to use here. Maybe, if you have seen it, you could use the Intermediate Value Theorem. – André Nicolas Nov 01 '15 at 02:47
  • 1
    If the teacher really didn't want the root just to be identified, it would have been easy to change the polynomial to one that doesn't have a nice rational root. – hmakholm left over Monica Nov 01 '15 at 02:53
  • If you want to justify finding $x=-1$, it's the first thing you'd check from the rational root theorem. – Théophile Nov 01 '15 at 02:55
  • Another way is to look at the factorization. You could notice that the polynomial is $x^4(x^3+x^2+x+1)+x^3+x^2+x+1$ and proceed from there, or start from $\frac{x^8-1}{x-1}$ then break down $x^8-1$ as a difference of squares, etc. – Théophile Nov 01 '15 at 02:58
  • See cyclotomic polynomial. – Lucian Nov 01 '15 at 07:38

4 Answers4

7

Your equation is guaranteed to have at least one real solution because it is a polynomial of odd degree. We can prove this by the Fundamental Theorem of Algebra or by the Intermediate Value Theorem.

FTOA- All polynomials of degree $n$ have $n$ roots, real or complex, and complex roots come in pairs, therefore a polynomial of odd $n$ must have at least one real root, because that one root will be the "odd one out" so to speak. IVT- for all polynomials of odd $n$, as $x \rightarrow \infty, \ \ x^n \rightarrow \infty$ and as $x \rightarrow -\infty, \ \ x^n \rightarrow -\infty$. And, since all polynomials are continuous, any polynomial of odd degree must cross the $x$-axis at least once, somewhere. This guarantees at least one root exists.
Rob
  • 6,727
3

Since $$x^7+x^6+x^5+x^4+x^3+x^2+x+1=\frac{x^8-1}{x-1}$$ we can use the fact that $x^8-1$ has roots at $-1$ and $1$ to get the root of $x=-1$.

robjohn
  • 345,667
0

Astroman.

Let $ f(x) = x^7+x^6+x^5+x^4+x^3+x^2+x+1 $. Being a polinomial function, $f$ is continuous and differentiable along it's domain, which is $ \mathbb{R} $.

Now, let's take $ f(-2) $ and $ f(0) $:

$ \begin{cases} f(-2) = 85 \\ f(0) = 1 \end{cases} $, or, in another words: $ f(-2) < 0 < f(0) $

Thus, according to the Intermediate Value Theorem, because $ f $ is continuous on $[-2,0] $ and $ f $ is differentiable on $ (-2,0 ) $, it exists a $ c \in (-2,0) $ such that $ f(c) = 0 $, which proves that $ f $ has at least one real root.

Kind regards, Pedro.

Pedro Cunha
  • 233
0

$\begin{align*} x^7+x^6+x^5+x^4+x^3+x^2+x+1 &= (x^4+1)(x^3+x^2+x+1) \\ &= (x^4+1)(x^2+1)(x+1) \end{align*}$

Eric Towers
  • 67,037