Solve $x$ for $0.0000004x^3 -0.001x^2 +1.6 x - 400 = 200$ The question reads as follows: 

A accountant tells a writer that he can calculate the gain (or loss)
  in function of the sold books with the following formula: 
$$f(x)= 0.0000004x^3-0.001x^2+1.6x-400$$

The question is "how many books does the writer need to sell to make a $200\,$€ profit"?
I already got this far (if I am correct).
\begin{align}
    &0.0000004x^3-0.001x^2+1.6x-400 = 200\\
\iff &0.0000004x^3-0.001x^2+1.6x = 600\\
\iff &0.001x ( 0.0004x^2 -x +1600) = 600\\
\iff &x ( 0.0004x^2+1600 -x ) = 600\,000 \\
\iff &x( (0.02x+40)(0.02x-40) -x ) = 600\,000
\end{align}
I probably already made a mistake somewhere but I can't see where. I also know that the answer should be $500$, but I don't find the right way.
 A: We start from this:
$$
0.0000004x^3−0.001x^2+1.6x−400=200
$$
Subtract $200$ to both sides to get
$$
0.0000004x^3−0.001x^2+1.6x−600=0
$$
Substitute $u = 0.001x$ to obtain*
$$
400u^3-1000u^2+1600u-600 = 0
$$
Divide by $200$:
$$
2u^3-5u^2+8u-3 = 0
$$
Using the rational root theorem, possible rational roots are $\left\{\pm 3, \pm \frac{3}{2}, \pm 1, \pm \frac{1}{2}\right\}$ (although physically, only positive roots make sense).  We find that $u = \frac{1}{2}$ is a solution:
$$
2\left(\frac{1}{2}\right)^3-5\left(\frac{1}{2}\right)^2+8\left(\frac{1}{2}\right)-3 = \frac{1}{4}-\frac{5}{4}+4-3 = 0
$$
Since $u = 0.001x$, $x = 1000u = 500$.

*It's worth pointing out a motivation for this particular substitution.  One is, as Will Jagy points out, to get rid of the annoying decimals.  But further, we observe that the coefficients go down geometrically as the exponent goes up; this suggests that a substitution for a constant ratio will resolve the disparity in scale.  We have a cubic ($x^3$) coefficient of $4 \times 10^{-7}$ and a constant ($x^0$) coefficient of $6 \times 10^2$; this suggests a constant (log) ratio of $\frac{(-7)-2}{3-0} = -3$—i.e., $u = 10^{-3}x$.  Indeed, in this case, this substitution "evens out" the coefficients and leads to a tractable cubic (with respect to rational roots, that is).
A: $$x^3 -2500 x^2 + 4 000 000 x -1 500 000 000 =0 $$
Let
$$  x = 1000 t    $$
$$ 1000000000 t^3 - 2500000000 t^2 + 4000000000 t - 1500000000 = 0   $$
$$ 10 t^3 - 25 t^2 + 40 t - 15 = 0   $$
Divide by $5,$
$$ 2 t^3 - 5 t^2 + 8 t - 3 = 0.   $$
NOW you need to find the "rational roots" for this $t$
A: To clear the decimals, you can define $x=100y$.  It is usually not useful to factor with a constant on the right, so you should start with 
$$0.0000004x^3-0.001x^2+1.6x-400 = 200\\
.4y^3-10y^2+160y-600=0\\y^3-25y^2+400y-1500=0$$
Now you can try factors of $1500$ looking for integer roots, and find $y=5, x=500$ is the only real solution.  Clearly if $y \ge 25$ the left side will be positive, so there aren't too many to try.
