A polynomial f(x) has a factor (x-a) if and only
if f(a)=0.
Thus, a polynomial cutting the x-axis at 10, 100, 1000 has
factors (x-10)(x-100)(x-1000). This defines a cubic polynomial up
to a multiplicative factor.
Thus f(x) = A(x-10)(x-100)(x-1000) = A\left(x^3 -(10+100+1000)x^2 + ..
\right)\,,
for some constant A.
Now, a point of inflection necessarily has f''(x) = 0. Only the
x^3 and x^2 terms of a cubic polynomial contributes to its
second derivative, so there is no need to expand the polynomial in
full to see that
f''(x) = 6Ax-2220A
This is zero at the single point x = \frac{2220}{6} = 370.
Therefore the point of inflection for the cubic is at x=370,
regardless of the choice of A.
For the second part, the polynomial must take the form
f(x) = A(x-10)(x-100)(x-a)\quad \mbox{for a constant } a \mbox{
where} \quad f''(a)=0
So, we need to take the second derivative to work out the
constraints on a. I will keep the form of the factors and use the
chain rule to make life simple, although you could expand the
brackets first if you wish
f''(x) = 2A\left((x-10)+(x-100)+(x-a)\right)
So,
f''(a) = 2A(2a-110)=0
Since A cannot be zero for a cubic polynomial, we must have
a=55.
The polynomial must therefore be of the form
f(x) = A(x-10)(x-55)(x-100)
Alternative, quick, method for
second part:
From the first part of the question I noticed a
generalisation that the point of inflection of a cubic is found at
one third of the sum of the roots r_1+r_2+r_3. If one of the
roots is the point of inflection then
r_1+a+r_2 = 3a
Thus, the point of inflection which is a root is found at one-half
of the sum of the other two roots.
Thus, in our special case,
a = \frac{1}{2}(10+100) = 55, \mbox{ as before}.
Isn't maths great!