TITLE: Prove that a degree-$6$ polynomial has exactly $2$ real roots
QUESTION [10 upvotes]: I have the function $f(x)={7x^6+8x+2}$ and I'm trying to prove that $f$ has exactly 2 real roots.
What I've done:
The only kind of solution I have come up with is essentially guessing pairs of values for $x$ that give $f$ a different sign and then make use of Bolzano's Theorem.
More specifically:

$f(-1)=1>0$ and $f(-{1\over 2})=-{121\over 64}<0$, so according to Bolzano's Theorem, there is some $x \in (-1, -{1 \over 2})$ such that $f(x)=0$.
$f(-{1\over 2})=-{121\over 64}<0$ and $f(0)=2$, so according to Bolzano's Theorem, there is some $x \in (-{1 \over 2}, 0)$ such that $f(x)=0$.

Question:
The above solution looks kind of meh to me and I don't think it proves there are exactly 2 real roots, but rather that only 2 were found. Is there a better, more convincing way to prove the existence of exactly 2 roots?

REPLY [2 votes]: Reading all of the given answers and using elements from the proposed solutions, I came up with the following explanation:

By calculating the first derivative of $f$, which is $f'(x)=42x^5+8$, we get that:


$f$ has an extremum at $f'(x)=0\Leftrightarrow x=-\left({4 \over 21}\right)^{{1 \over 5}}≈-0.718$ with $f(-0.718)≈-2.785$.
$f$ is strictly decreasing in $(-\infty, -0.718)$.
$f$ is strictly increasing in $(-0.718, \infty)$.
The extremum is a minimum.



By calculating the second derivative of $f$, which is $f''(x)=210x^4$, we get that:


$f$ is strictly convex in $ℝ$, because $f''(x)=210x^4 ≥0,\ \  ∀\ \ x ∈ ℝ$.
$f$ has an undulation point at $x=0$, because $f''(0)=0$.


Taking into account that $f$ has a minimum at $(-0.718, -2.785)$, which is located below the real axis, and that $f$ is strictly convex in $ℝ$, it is evident that $f$ intersects with the real axis at exactly two points, which are the roots of $f$.