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## 'Weekly Challenge 10: Solve Me!' printed from http://nrich.maths.org/

Any sensible numerical method will lead to a solution
$-16.3(2)$.

In particular, an interval-halving method is efficient and
simple to implement. You can make it a little quicker by choosing a
sensible starting point: note that any solution would have to be
negative, since all of the coefficients are positive;
another moment of inspection will also show that the solution
must lie between $-10$ and $-100$, giving you a sensible starting
point for a computation.

To determine whether there are any other solutions, note that the
expression is a cubic and will therefore have either $1$ or $3$
real solutions.

To explore the properties of the cubic $y=2x^3+34x^2+567x+8901$,
look at the turning points. Differentiating, we find that

$$

\frac{dy}{dx} = 6x^2+68x+567

$$

The discriminiant of this quadratic is $68^2-4\times 6 \times
567 = -8984$. Since this is negative, there are no turning points
and the cubic consequently only has $1$ real solution.