Question

What is the polynomial function `f` of least degree that has rational coefficients, a leading coefficient of 2, and the zeros `1, 2, 4+sqrt2`?

Answer 1

`f(x) = 2x^4-22x^3+80x^2-116x+56`

Explanation:

If `f(x)` has rational coefficients, then any irrational zeros will occur in conjugate pairs (assuming a conjugate exists).

So in our example, we can deduce that `4-sqrt(2)` must be a zero too.

Note that:

`(4-sqrt(2))(4+sqrt(2)) = 4^2-(sqrt(2))^2 = 16-2 = 14`

Each zero `a` corresponds to a linear factor `(x-a)`, so we can put:

`f(x) = 2(x-1)(x-2)(x-(4+sqrt(2)))(x-(4-sqrt(2)))`

`color(white)(f(x)) = 2(x-1)(x-2)(x^2-8x+14)`

`color(white)(f(x)) = 2(x-1)(x^3-10x^2+30x-28)`

`color(white)(f(x)) = 2(x^4-11x^3+40x^2-58x+28)`

`color(white)(f(x)) = 2x^4-22x^3+80x^2-116x+56`