Explain how to conclude that x^2 - 59x + 6 is a prime polynomial without performing
Explain how to conclude that x^2 - 59x + 6 is a prime polynomial without performing any trials.
Since this is not my area of expertise, I searched Google under the key words "prime polynomial" to get these possible sources:
http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut7_factor.htm
http://answers.yahoo.com/question/index?qid=20080312210701AA2t3at
Use <Find> command to search for "prime."
In the future, you can find the information you desire more quickly, if you use appropriate key words to do your own search. Also see http://hanlib.sou.edu/searchtools/.
I hope this helps. Thanks for asking.
To determine whether the polynomial x^2 - 59x + 6 is prime without performing polynomial division or factoring, we can use the Rational Root Theorem.
The Rational Root Theorem states that any rational root of a polynomial equation with integer coefficients must be of the form p/q, where p is a factor of the constant term (in this case, 6), and q is a factor of the leading coefficient (in this case, 1).
For the polynomial x^2 - 59x + 6, we need to find the factors of 6 and 1. The factors of 6 are 1, 2, 3, and 6, and the factors of 1 are 1 and -1. This gives us a list of possible rational roots: ±1, ±2, ±3, and ±6.
We can now test each of these potential rational roots by substituting them into the polynomial equation and checking if the result is zero. If one of the rational roots gives a remainder of zero, then the polynomial is not prime.
Since there are only a few possible rational roots to test in this case, you can simply substitute each one into the polynomial equation and check if it equals zero. If you find a root that results in zero, then the polynomial is not prime.
In this case, we find that x = 1 is a rational root that makes the polynomial equal to zero when substituted. Therefore, the polynomial x^2 - 59x + 6 is not prime.