Number of real roots of $x^3+9x^2-49x+49=0$ How many real roots does $x^3+9x^2-49x+49=0$ have in the open interval $1\lt x\lt 2$? I've applied the intermediate value theorem and found that it has at least 1 root. But what about other roots?  How can you find the exact numbers of real roots in a given interval. I would really appreciate if I could get some examples of higher degree as well.  
 A: Hint:
If $\;f(x)=x^3+9x^2-49x+49\;$ , then
$$f(1)>0\;,\;\;f(2)<0\,,\,f(3)>0\;,\;\;\lim_{x\to-\infty}f(x)=-\infty$$
A: You may use Sturm's theorem (I leave it to  wiki to explain  what the method does). In the present context, the calculation goes as follows:
$$ p_0(x)= x^3 + 9x^2-49 x+49, \ \ \ \ p_1(x) = p_0'(x) = 3x^2+18x-49$$
and then 
$$ p_2=-p_0 {\;\rm mod\;} p_1 = 50.666.. \;x - 98, \ \ p_3=-p_1 {\rm\; mod \;} p_2=2.96..$$
Then construct the vector $\sigma(x)={\rm sign } \; (p_0(x),p_1(x),p_2(x),p_3(x))$. You calculate the number of sign changes in 
$\sigma(1) = (+,-,-,+)$ (2 sign changes) and $\sigma(2)=(-,-,+,+)$ (1 sign change). Then $2-1=1$ is the exact number of real roots in the half-open interval $(1,2]$ (and $2$ is not a root).
A: The extrema appear at the roots of
$$3x^2+18x-49=0.$$
As none of them lies in the range $(1,2)$, the cubic has no other root.
A: The Sturm Chain for $x^3+9x^2-49x+49$ is
$$
\begin{array}{c}
p(x)&p(-14)&p(-13)&p(1)&p(2)&p(3)\\
x^3+9x^2-49x+49&-245&10&10&-5&10\\
3x^2+18x-49&287&224&-28&-1&32\\
\frac{152}3x-98&-\frac{2422}3&-\frac{2270}3&-\frac{142}3&\frac{10}3&54\\
\frac{17101}{5776}&\frac{17101}{5776}&\frac{17101}{5776}&\frac{17101}{5776}&\frac{17101}{5776}&\frac{17101}{5776}\\
\text{sign changes}&3&2&\color{#C00}{2}&\color{#C00}{1}&0
\end{array}
$$
The number of roots in $(1,2)$ is $\color{#C00}{2}-\color{#C00}{1}=1$.
The other two roots are in $(-14,-13)$ and $(2,3)$.
A: We can use the Sturm's theorem for finding the real roots in $(a,b)$ of a polynomial $p\in\mathbb{R}[x]$. Besides, if $f(z)=a_nz^n+\ldots +a_1z+a_0\in\mathbb{C}[z]$ and $c$ is a root of $f(z)$ then, $|c|\le M$ with $$M=\max\left \{\left(n\left| \frac{a_{i-1}}{a_n}\right|\right)^{1/i}:i=1,\ldots,n\right\}.$$
This bound allows to discard roots on $(-\infty,M)\cup (M,+\infty).$
A: Call the LHS $f(x)$. Then $f'$ has exactly two zeroes $a$ and $b$ (compute them!), one negative and the other positive (say $a<b$). So $f$ increases to the left of $a$, decreases between $a$ and $b$, and increases to the right of $b$. $f$ has a local maximum at  $a$ and a local minimum at $b$. Now since $f(a)>0$ and $f(b)<0$, $f$ must have three zeroes--one to the left of $a$, one between $a$ and $b$, and one to the right of $b$. 
In fact, since $1<2<b\approx 2.033$, and $f(1)=10$ and $f(2)=-5$, the middle zero of $f$ lies between $1$ and $2$. As mentioned above, the other zeroes lie to the left of $a\approx -8.033$ and to the right of $b\approx 2.033$.
