+0  
 
-2
20
2
avatar+819 

I cant solve this

 

Let $P(x) = 0$ be the polynomial equation of least possible degree, with rational coefficients, having $\sqrt[3]{2} + \sqrt[3]{3}$ as a root.  Compute the product of all of the roots of $P(x) = 0.$

 Aug 18, 2023
 #1
avatar+177 
+1

I think this problem is easier than it may first appear unless I am misreading the question.

 

First, I would consider the given information and determine if I can fit all the criteria in a first-degree polynomial. I think I can. Just do this: \(P(x) = x - \left(\sqrt[3]{2} + \sqrt[3]{3}\right)\). This is a first-degree polynomial, so this must be of least degree, and P(x) has rational coefficients. The coefficients of the x-term is 1, which is rational. Because of the way I have written this polynomial, it is guaranteed to have the root of \(\sqrt[3]{2} + \sqrt[3]{3}\).

 

There is only one root, so the product of the roots is just \(\sqrt[3]{2} + \sqrt[3]{3}\).

 Aug 19, 2023
 #2
avatar+177 
0

I did some research after posting this answer, and I am relatively certain that the answer I posted previously is not correct, but I will leave this here in case it helps someone else solve this problem. In a polynomial in the form \(P(x) = a_nx^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \cdots + a_2x^2 + a_1x + a_0\), it seems like there is some ambiguity about whether or not \(a_0\) fits the definition of a coefficient or not. I always considered \(a_0\) as classified as a constant, but some argue that \(a_0\) is a coefficient of the \(x^0\) term. If that interpretation fits this problem, then the answer I posted previously is not correct. I made a few attempts, but I was unable to solve this problem with the new interpretation.

The3Mathketeers  Aug 19, 2023

8 Online Users

avatar
avatar
avatar
avatar