Identifying a Prime Polynomial

Oculus8596 · 2024-09-21 02:46:46

Show that x^2 + 9 is prime.

I will first list the pairs of integers whose product is 9.

1, 9

-1, -9

3, 3

-3, -3

2. Compute their sums.

1 + 9 = 10

-1 + (-9) = -10

3 + 3 = 6

-3 + (-3) = -6

I can also say x^2 + 9 = x^2 + 0x + 9 and none of the sums added above equal 0. This leads me to conclude that x^2 + 9 is prime.

1. Do you agree?

2. How is this done for a trinomial?

Sample:

Show that x^1 + x + 1 is prime.

Bob · 2024-09-21 09:12:30

If you are able to graph the expression then a prime expression will not cross the x axis.

For f(x) = x^2 + x + 1 df/dx = 2x+1 so the turning point is at x = -1/2

d2f/dx^2 = 2 implies the turning point is a minimum. f(-1/2) = 3/4

This means that the lowest point on the graph is at (-1/2, 3/4) so it never crosses the x axis.

Your method for the first part is ok but, as you have discovered, another approach is needed for more complicated quadratics.

Bob

Oculus8596 · 2024-09-21 09:21:10

Bob wrote:

If you are able to graph the expression then a prime expression will not cross the x axis.
For f(x) = x^2 + x + 1 df/dx = 2x+1 so the turning point is at x = -1/2
d2f/dx^2 = 2 implies the turning point is a minimum. f(-1/2) = 3/4
This means that the lowest point on the graph is at (-1/2, 3/4) so it never crosses the x axis.
Your method for the first part is ok but, as you have discovered, another approach is needed for more complicated quadratics.
Bob

I had no idea that this kind of problem could be solved by graphing. By the way, I have not never taken calculus. So, please no derivative and/or integration methods for now.

Bob · 2024-09-21 09:38:09

Oh, tricky. There's a graph plotter here: https://www.mathsisfun.com/data/function-grapher.php

But is that acceptable as a proof of primeness? Not sure.

All quadratics have a single minimum or maximum point (ie. the bottom of the U shape if the x^2 term is positive)

So I think an acceptable non calculus method would be (1) use a graph plotter to find the minimum point then (2) show it is a minimum by evaluating the quadratic just left and just right of that point, because bigger values there show you really have identified the minimum.

I don't know how else you could do this, sorry.

edit: I'm being stupid. Of course there's a way. Evaluate b^2 - 4ac. If it's negative there's no real square root so no factors.

Bob

Oculus8596 · 2024-09-21 17:55:29

Bob wrote:

Oh, tricky. There's a graph plotter here: https://www.mathsisfun.com/data/function-grapher.php
But is that acceptable as a proof of primeness? Not sure.
All quadratics have a single minimum or maximum point (ie. the bottom of the U shape if the x^2 term is positive)
So I think an acceptable non calculus method would be (1) use a graph plotter to find the minimum point then (2) show it is a minimum by evaluating the quadratic just left and just right of that point, because bigger values there show you really have identified the minimum.
I don't know how else you could do this, sorry.
edit: I'm being stupid. Of course there's a way. Evaluate b^2 - 4ac. If it's negative there's no real square root so no factors.
Bob

Ok. The graphing method as you suggested works. You also said to evaluate the discriminant b^2 -"4ac. I will try both methods to see if the trinomial is prime or not.

Math Is Fun Forum

#1 2024-09-21 02:46:46

Identifying a Prime Polynomial

#2 2024-09-21 09:12:30

Re: Identifying a Prime Polynomial

#3 2024-09-21 09:21:10

Re: Identifying a Prime Polynomial

#4 2024-09-21 09:38:09

Re: Identifying a Prime Polynomial

#5 2024-09-21 17:55:29

Re: Identifying a Prime Polynomial

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