Show Trinomial Is Prime

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mathdad

mathdad

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

Note: I think factoring is involved here. What about the quadratic formula?
 
According to google:

A polynomial with integer coefficients that cannot be factored into polynomials of lower degree , also with integer coefficients, is called an irreducible or prime polynomial.

Does the given polynomial have rational roots? Why am I asking about rational roots?
 
MarkFL said:
According to google:

A polynomial with integer coefficients that cannot be factored into polynomials of lower degree , also with integer coefficients, is called an irreducible or prime polynomial.

Does the given polynomial have rational roots? Why am I asking about rational roots?
Click to expand...

I do not know why you are asking about rational roots. A friend told me to use the discriminant. If the discriminant is negative, no real roots and thus prime.

Discriminant = b^2 - 4ac.

Let D = discriminant

D = (1)^2 - 4(1)(1)

D = 1 - 4

D = - 3

D is negative. So, no real roots. This means the trinomial is prime.

Yes?
 
If the given quadratic was reducible, then it could be expressed as the product to two linear functions having integer coefficients, both of which would have rational roots. The rational roots theorem tells us that the only possible rational roots here are:

[MATH]x=\pm1[/MATH]
But, neither of those values of \(x\) makes the polynomial zero. Since there are then no rational roots, the polynomial must be prime.

If you use the discriminant of a quadratic function, then you must find it is a non-negative perfect square.
 
MarkFL said:
If the given quadratic was reducible, then it could be expressed as the product to two linear functions having integer coefficients, both of which would have rational roots. The rational roots theorem tells us that the only possible rational roots here are:

[MATH]x=\pm1[/MATH]
But, neither of those values of \(x\) makes the polynomial zero. Since there are then no rational roots, the polynomial must be prime.

If you use the discriminant of a quadratic function, then you must find it is a non-negative perfect square.
Click to expand...

So, is my answer wrong?
 
mathdad said:
So, is my answer wrong?
Click to expand...

The polynomial is in fact prime (irreducible), and if the discriminant of a quadratic is negative, it is prime. But that doesn't mean if the discriminant is non-negative, the quadratic is composite. The discriminant must be a non-negative perfect square (so that the roots are rational) in order for it to be composite. I wanted to clear that point up, and provide another way to look at the problem.
 
MarkFL said:
The polynomial is in fact prime (irreducible), and if the discriminant of a quadratic is negative, it is prime. But that doesn't mean if the discriminant is non-negative, the quadratic is composite. The discriminant must be a non-negative perfect square (so that the roots are rational) in order for it to be composite. I wanted to clear that point up, and provide another way to look at the problem.
Click to expand...

Can you provide an example when the quadratic is composite?
 
Mark:

Can you check out my Solve for h and Solve Formula for h questions? I got stuck in one of the questions. I cannot isolate h.
 
mathdad said:
sqrt{1} = 1 because 1 x 1 = 1. It is a perfect square that is not negative.

What about 0?

sqrt{0} = 0 because 0 x 0 = 0.

Is 0 a non-negative perfect square?
Click to expand...

[MATH]1=1^2[/MATH]
[MATH]0=0^2[/MATH]
Both are non-negative (recall a negative number is less than zero), and both are perfect squares. When the disriminant is a non-negative perfect square, then we know the roots are rational (why?), and thus, we know the quadratic can be expressed as the product of linear factors, and must therefore be composite.
 
MarkFL said:
[MATH]1=1^2[/MATH]
[MATH]0=0^2[/MATH]
Both are non-negative (recall a negative number is less than zero), and both are perfect squares. When the disriminant is a non-negative perfect square, then we know the roots are rational (why?), and thus, we know the quadratic can be expressed as the product of linear factors, and must therefore be composite.
Click to expand...

Like you said:

"The discriminant must be a non-negative perfect square (so that the roots are rational) in order for it to be composite."
 
51Dxkytk6BL._SY445_.jpg
 
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