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# Root to Poly

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Age 14 to 16

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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.

A group of 20 people pay a total of £20 to see an exhibition. The admission price is £3 for men, £2 for women and 50p for children. How many men, women and children are there in the group?