# Root to Poly

Find the polynomial p(x) with integer coefficients such that one solution of the equation p(x)=0 is $1+\sqrt 2+\sqrt 3$.

Find the polynomial $p(x)$ with integer coefficients such that one
solution of the equation $p(x)=0$ is $1+\sqrt{2}+\sqrt{3}$.

Start with the expression $x = 1 + \sqrt2 + \sqrt3$ square and simplify. Take it from there.

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.

Why do this problem?

It gives learners experience of algebraic manipulation of polynomials and working with surds. It is based on the fact that if you know one root of a polynomial then you know one of its factors.

Possible approach

This can be used as a lesson starter.

Key questions

We are looking for a polynomial in $x$, do you know any values of $x$ that satisfy the polynomial?

If you have an expression involving surds what can you try in order to get rid of the square roots?