Prove that a degree-$6$ polynomial has exactly $2$ real roots 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?
 A: We have:
$f(x)=7x^6+8x+2$. On differentiating, 
$f'(x)=42x^5+8$ which has one solution and the solution is $x=-0.718$. Now, you can be sure that your polynomial equation has at the most two roots. To prove exactly two roots, divide the real line in two intervals $(-\infty,-0.718]$ and $[-0.718,\infty)$. Now, you can check (using the sign of derivative) in the first Interval, the function is decreasing and in the second Interval function is increasing.
Also, note that the function takes positive values at the end of Interval and negative value in the neighborhood of your critical point which is $-0.718$. Can you guess what this means. This means, the function cuts the x-axis two times. In the first and second  Interval. Hence, function has exactly two roots. 
A: We just have to count the intersections between the graph of $g(x)=7x^6$ and the graph of $h(x)=-8x-2$. By convexity, they can be either $0,1$ or $2$. $f(x)$ has a sign change over $\left[-1,-\frac{1}{2}\right]$ and another sign change over $\left[-\frac{1}{2},0\right]$.
It follows that $f(x)$ has exactly $2$ real roots. One of them is pretty close to $-\frac{1}{4}$ since $7x^6$ is negligible in a small neighbourhood of the origin.
A: 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 $f$.
A: Hint: by Descartes' rule of signs the equation has no positive real roots, and at most $2$ negative ones. But you showed that it has at least one real root (and it's enough that $\,f(-1/2) \lt 0 \lt f(0)\,$ for that), then it must have a second real one, since non-real complex roots come in conjugate pairs.
A: $$f''(x)=7\cdot6\cdot5x^4\geq0,$$ which says that $f$ is a convex function.
Thus, $f$ has two roots maximum and by your work we get two roots exactly. 
A: Hint 
You proved so far that there are at least 2 solutions.
If the function would have 3 or more solutions, then by Rolle's Theorem, $f'$ would have at least 2 solutions.
A: $f'(x)$=42$x^5$+8
$f''(x)$ =210$x^4$
now you can observe that the double derivative of your function is always positive for all values of x in $R$ and also that 
$f'(x)$ is negative for x=-1 and positive for x=1, hence $f'(x)$ is zero atleast once (USING IMVT)
but since $f''(x)=0$ has no real root hence, $f'(x)=0$ has only a single real root.
And now using ROLLE'S THEOREM,we can say that $f(x)=0$ has at most two real roots or say atleast 4 complex roots.Now since $f(x)$ is negative at x=-1/2 and positive at x=-1. Thus, $f(x) =0$ has a real root. 
Since, there are atleast 4 complex roots but they occur in pairs so they can be either 6 or 4 in number. But we already have one solution for $f(x) =0$,there are 4 complex solution or exactly 2 real roots. 
