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## 'Root to Poly' printed from http://nrich.maths.org/

Congratulations to Fok Chi Kwong from Yuen Long Merchants
Association Secondary School, Hong Kong on this solution.

We may find the required polynomial by starting from the
expression :

$$x = 1 + \sqrt 2 + \sqrt 3$$.

Squaring both sides and simplifying, we get

\[x - 1 = \sqrt 2+ \sqrt 3 \] \[x^2 - 2x + 1 = 5 + 2\sqrt 6 \]
\[ x^2 - 2x - 4 = 2\sqrt 6 \] \[(x^2 - 2x - 4)^2 = 24 \] \[x^4 -
4x^3 + 4x^2 - 8x^2 + 16x + 16 = 24 \] \[x^4 - 4x^3 - 4x^2 + 16x - 8
= 0 \]

Thus $p(x) = x^4 - 4x^3 - 4x^2 + 16x - 8$ is the required
polynomial.

Tony Cardell, State College Area High School, PA, USA, also sent
in a good solution.