Math Is Fun Forum

mathxyz

Member

From: Brooklyn, NY

Registered: 2024-02-24

Posts: 1,053

Show Trinomial Is Prime

Show that x^2 + x + 1 is prime.

Problem 134; page 58.

I need help here.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,627

Re: Show Trinomial Is Prime

Say  x^2 + x + 1 =  (x-p))x-q) = x^2 -x(p+q) +pq

pq = 1  => p and q must both be negative or both positive

But p + q = 1   so they must be positive.

p = 1/q so if p >1 then q < 1 and if p < 1 then q > 1

So one of p and q must be > 1 so they cannot add to make 1.  => p and q don't exist.

Bob

Further thoughts: This doesn't help with a general method to tackle these.  I think the following does:

Pretend it's a quadratic equation ie.  x^2 + x + 1 = 0.  If this has factors then we've got a factorisation of the expression.

So try the formula or rather just the b^2 - 4ac bit.  b^2 - 4ac = 1 - 4 = -3. As this has no square root in real numbers the quadratic cannot have any real solutions so the expression cannot be factored.

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

mathxyz

Member

From: Brooklyn, NY

Registered: 2024-02-24

Posts: 1,053

Re: Show Trinomial Is Prime

Say  x^2 + x + 1 =  (x-p))x-q) = x^2 -x(p+q) +pq

pq = 1  => p and q must both be negative or both positive

But p + q = 1   so they must be positive.

p = 1/q so if p >1 then q < 1 and if p < 1 then q > 1

So one of p and q must be > 1 so they cannot add to make 1.  => p and q don't exist.

Bob

Further thoughts: This doesn't help with a general method to tackle these.  I think the following does:

Pretend it's a quadratic equation ie.  x^2 + x + 1 = 0.  If this has factors then we've got a factorisation of the expression.

So try the formula or rather just the b^2 - 4ac bit.  b^2 - 4ac = 1 - 4 = -3. As this has no square root in real numbers the quadratic cannot have any real solutions so the expression cannot be factored.

I will definitely get back to this later today or tomorrow morning. Thank you for your notes. I will certainly do as you said. I also will search YouTube for video clips showing this stuff.

mathxyz

Member

From: Brooklyn, NY

Registered: 2024-02-24

Posts: 1,053

Re: Show Trinomial Is Prime

Say  x^2 + x + 1 =  (x-p))x-q) = x^2 -x(p+q) +pq

pq = 1  => p and q must both be negative or both positive

But p + q = 1   so they must be positive.

p = 1/q so if p >1 then q < 1 and if p < 1 then q > 1

So one of p and q must be > 1 so they cannot add to make 1.  => p and q don't exist.

Bob

Further thoughts: This doesn't help with a general method to tackle these.  I think the following does:

Pretend it's a quadratic equation ie.  x^2 + x + 1 = 0.  If this has factors then we've got a factorisation of the expression.

So try the formula or rather just the b^2 - 4ac bit.  b^2 - 4ac = 1 - 4 = -3. As this has no square root in real numbers the quadratic cannot have any real solutions so the expression cannot be factored.

Sullivan does not have an example for prime trinomials.

So, this is what I did.

A prime polynomial has two factors: 1 and itself just like a prime number has 1 and itself as prime.

The trinomial x^2 + x + 1 cannot be factored over the real numbers.

The trinomial x^2 + x + 1 cannot be factored into more than 1 and itself.

I know that this does not show mathematically that the given trinomial is prime or not.

My conclusion is that x^2 + x + 1 is prime because it cannot be factored over the real numbers.

Bob, I searched online for a sample involving prime trinomials but didn't find one single example.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,627

Re: Show Trinomial Is Prime

I think my 'quadratic formula' method above is what you're seeking.

Here's why it works:

You're trying to factorise ax2 + bx + c.  Let's say it can be factored as (ax + p)(x + q)

Now consider the quadratic equation ax^2 + bx + c = 0.  If it can be factored you will have (ax + p)(x + q) = 0 so you will have found p and q for the trinomial.

Conversely if you can show the quadratic has no solution this means that p and q don't exist so the trinomial is prime.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

mathxyz

Member

From: Brooklyn, NY

Registered: 2024-02-24

Posts: 1,053

Re: Show Trinomial Is Prime

I think my 'quadratic formula' method above is what you're seeking.

Here's why it works:

You're trying to factorise ax2 + bx + c.  Let's say it can be factored as (ax + p)(x + q)

Now consider the quadratic equation ax^2 + bx + c = 0.  If it can be factored you will have (ax + p)(x + q) = 0 so you will have found p and q for the trinomial.

Conversely if you can show the quadratic has no solution this means that p and q don't exist so the trinomial is prime.

Bob

Ok. I will work this out on paper.