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## 'Weekly Challenge 27: Cubic Roots' printed from http://nrich.maths.org/

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!