+304 Let \(f(x)\) be a polynomial of degree \(4\) with rational coefficients which has \(1+2\sqrt{3}\) and \(3-\sqrt{2}\) as roots, and such that \(f(0) = -154.\) Find \(f(1).\)
Thank you for all your help <3
+118608 You did not write it on one question. You wrote a general comment.
You did thank your answerers, which was nice, but even so, the comment sounded arrogant.
The comment should have been on the thread that you were talking about.
It was not appropriate to place it as a new 'question'.
There was a reply left saying you should have shown 'us' how to do it.
This reply was whited out because the person was using 'language' to display his/her annoyance.
But he was right.
If you ask a question here, then find out how to do it elsewhere, you should share your new knowledge here.
https://web2.0calc.com/questions/scratch-the-polynomial-help-i-figured-it-out
+118608 Thanks Alan,
I understand that roots come in pairs for parabolas but I did not realize that was still true for degrees higher than 2.
I am still confused by it. Why is it so? What happens if the degree is odd?
+33614 The question specified that the polynomial had rational coefficients. The only way that could occur here is if the irrational terms in sqrt(3) and sqrt(2) were eliminated by complementary terms. In general, an n'th order polynomial will have n roots (counting possible repeated ones). Therefore with an odd degree polynomial with rational coefficients you couldn't have an odd number of roots that all contained irrational terms; the irrational ones would have to appear in complementary pairs.
Alan i don't think thats right. There are 3rd degree polynomials with rational coefficients that have only irrational roots (for example x3-3x+1)
+118608 Hi guest,
I looked here....... but how do you know for sure that they are all irrational ?
+33614 I guess I shouldn't have put the emphasis on the word "irrational" in my earlier reply. However, I think the terms with the surds should come in complementary pairs. For example, the cubic can be rewritten as follows:
All solutions are (I suspect) irrational, but two are in the form a + b√3 and a - b√3.
