How would you find the polynomial of the degree with rational coefficients with a root of nine minus six times the square root of three.

Question

I was wondering if you had to reverse the processes but I am unsure how to go from there. This is a lesson with the Conjugate Root theorem which I am sure I am missing the point of how it fits into the question.

Answer 1

Let x = 9 - 6√3 be one of the roots. By the complex conjugate root theorem, there should be a pair of complex numbers as roots such as x = 9 + 6√3. Now we have two complex roots.

According to the definition, you will need to convert the roots into factors and then multiply them to get the polynomial with real coefficients (rational numbers are considered real numbers).

[x - (9 - 6√3)]*[x - (9 + 6√3)] = [x - (9 - √108)]*[x - (9 + √108)]

= x² - x*(9 + √108) - x*(9 - √108) + (81 + 9√108 - 9√108 - 108)

= x² - 9x - x√108 - 9x + x√108 - 27

= x² - 18x - 27

This polynomial has rational coefficients with the root of 9 - 6√3.