Write a polynomial function of f

Question

Write a polynomial function of f pf the least degree that has rational coefficients, and a leading coefficient of 1, and the zeroes 1 and 5-√2

Answer 1

Viv, Mark’s procedure works for irrational roots as well as complex roots.

The way to construct the polynomial is simple: multiply several monomials together, one for each root, plus one for each “conjugate” (I’ll explain jn a moment).

We start with the two known roots: 1 and 5-sqrt(2).

The polynomial starts with

(x-1)(x-(5-sqrt(2))

however, because one root is irrational (it contains an irrational (sqrt(2)) ), we need to append another root. The other, “conjugate”, root has the opposite sign for the irrational: (5 + sqrt(2)).

So the complete polynomial is:

(x-1) * (x-(5-sqrt(2))) * (x-(5 + sqrt(2)))

Multiply out the “conjugates” first:

x²-x(5+sqrt(2)) -x(5-sqrt(2)) + (25-2)

= x² -x(10) + 23

= x² -10x + 23

now multiply by the simple root (x-1)

(x-1)(x² -10x + 23)

= x³ -11x² + 33x -23

So this is the answer for the problem you posed, while Mark’s is the answer for a similar problem with complex roots.

Answer 2

I wonder if you meant a zero (root) at 5 - i√2. As written, I don't think there could be rational coefficients in the polynomial.

Let's assume that there is a pair of complex roots : 5 ± i√2.

This means the factors are: (f - 5 + i√2)(f - 5 - i√2). The resulting polynomial is f² - 10f + 27

If there is also a root at f = 1, then the factor is f - 1

The polynomial is then (f - 1)*(f² - 10f + 27) = f³ - 11f² +37f - 27