Find the root of a polynomial of degree 5 I have the following equation:
$${138000\over(1+x)^5}+{71000\over(1+x)^4}+{54000\over(1+x)^3}+{37000\over(1+x)^2}+{20000\over1+x}-200000=0$$
And I need to find the real solution(s) to said equation, but I don't know how. It's for homework and I don't know how to find a solution to the equation, and Newton-Raphson seems like an unlikely solution if I'm going to solve this equation in a mid-term.
I think there's a way to know if an equation has more than one real root, but I don't remember how to, if anyone wants to explain to me if this is true I will be grateful.
I want to learn an "easy" way of solving these kind of equations without the use of computational aids. I don't want the solution but a way to get to the solution and with a simple calculator punch the numbers at the end and find the answer. Any ideas?
 A: This looks like an "internal rate of return" question, where usually you want $x>-1$. 
When $x>-1$, the function is strictly decreasing as $x$ increases. Clearly, if $x$ is very large, then the value is negative, and when $x=0$ it is $20000$, so you have one positive real root. You can use numerical methods to find the solution - binary search for example.
I don't know what you'd use on a mid-term - that would depend on what calculation tools you were allowed during the midterm.
One quick way is to write $t=\frac{1}{1+x}$ and note that we are solving $g(t)=0$ for some polynomial with $g(1)=20000$ and $g'(1)=730000$.  So an estimate is $t\approx 1-\frac{2}{73}$ or $x\approx 0.028$. This is Newton's method, just on a slightly easier formula when done relative to $t$, and it converges quickly because the root  $x$ is close to $0$. This might not work in general.
Wolfram alpha gives the root as $x\approx 0.02932$.
A: Letting $1+x=t$, multiplying with $t^5$ and sorting, we find
$$t^5=\frac{20t^4+37t^3+54t^2+71t+38}{200}.$$
This suggests an iterative method by letting $t_{n+1}=\sqrt[5]{\frac{20t_n^4+37t_n^3+54t_n^2+71t_n+38}{200}}$.
A suitable starting value is $t=1$, leading to a solution $t\approx1.0293$ and  $x\approx-0.02849$.
A: First make the substitution $t=\frac{1}{1+x}$ so you find a polinomial of degree $5$. Then make the derivative to study the function and see that there is only one solution (for $t$) which is between $0$ and $1$. Then use the bisection method to approximate your solution.
A: Let $t = \dfrac{1}{1+x}$. then you get
$138000t^5 + 71000t^4 + 54000t^3 + 37000t^2 + 20000t - 200000$
Which simplifies to $1000f(t)$ where 
$f(t) = 138t^5 + 71t^4 + 54t^3 + 37t^2 + 20t - 200 = 0$
Clearly $f'(t) > 0$ for all $t$, so $f(t)$ is strictly increasing. Hence it has exactly one real root.
We see that $f(0) = -200$ and $f(1) = 120$. So that one real root is between $t=0$ and $t=1$. 
There are all sorts of iterative methods that will approximate the value of that root to whatever precision you desire.
