SOLUTION: The question states: factor completely or state the polynomial is prime... 9x^2+25 and the answer says it is prime... BUT I HAVE NO CLUE WHY :( Could you please help me :

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: The question states: factor completely or state the polynomial is prime... 9x^2+25 and the answer says it is prime... BUT I HAVE NO CLUE WHY :( Could you please help me :      Log On


   

Question 635770: The question states: factor completely or state the polynomial is prime...
9x^2+25
and the answer says it is prime... BUT I HAVE NO CLUE WHY :(
Could you please help me :) God bless your evening :)
Found 2 solutions by solver91311, jsmallt9:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


"Prime" in this context is a relative term. Yes, this quadratic binomial is prime over the real numbers, but it is not prime over the complex numbers.

A second degree polynomial has two linear factors of the form and where and are the zeros of the polynomial.

Set your polynomial equal to zero and solve:









However, there is no real number that is the square root of a negative number, hence the solution set of the above equation over the real numbers is the null set. Since there are no real number zeros of the equation, there are no real number factors of the polynomial.

As an aside, if you define as the imaginary number such that , then the equation has the solutions:



and the factors of the polynomial, over the complex numbers are and


John

My calculator said it, I believe it, that settles it
The Out Campaign: Scarlet Letter of Atheism

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
9x%5E2%2B25
You conclude that a polynomial is prime after you have tried all the various factoring techniques you have learned and found that the only factors are 1 and the polynomial itself.

The factorsing techniques you have probably learned are:
    Greatest Common Factor (GCF)
  • Factoring with patterns:
    • a%5E2-b%5E2=%28a%2Bb%29%28a-b%29 (Difference of squares)
    • a%5E3%2Bb%5E3=%28a%2Bb%29%28a%5E2-ab%2Bb%5E2%29 (Sum of cubes)
    • a%5E3-b%5E3=%28a-b%29%28a%5E2%2Bab%2Bb%5E2%29 (Difference of cubes)
    • a%5E2%2B2ab%2Bb%5E2+=+%28a%2Bb%29%28a%2Bb%29+=+%28a%2Bb%29%5E2
    • a%5E2-2ab%2Bb%5E2+=+%28a-b%29%28a-b%29+=+%28a-b%29%5E2
  • Trinomial factoring: ax^2+bx+c
  • Factoring by grouping
  • Factoring by trial and error of the possible rational roots
All five of these must be attempted and you must fail to find any factors other than 1 and the polynomial itself before you declare a polynomial prime. Let's go through these one-by-one.

GCF.
9x^2 = 1 * 3 * 3 * x * x
25 = 1 * 5 * 3
The GCF is just 1.

Patterns.
Our expression has only two terms. So the only patterns we might be able to use are the one with two terms, the first three. Our expression could be described as a sum of squares: %283x%29%5E2%2B%285%29%5E2. We have patterns for difference of squares, sum of cubes and difference of cubes. Notably missing is a sum of squares. Our expression will not factor using a pattern.

Trinomial factoring.
With only two terms we do not have enough terms to factor with this method.

Factoring by grouping.
This method requires an even number of terms with at least 4 terms. With only two terms we do not have enough terms to factor with this method.

Factoring by trial and error of possible rational roots.
This method is only useful for polynomials of degree 3 or higher. Our polynomial is of degree 2.

So after all this we found only one factor: 1. So this polynomial is prime.