Factor Completely;

[gijas]

Factor Completely:

m^2 + 2m + 6

OK factors of 6 are 2, 3 - 1, 6

factors of m^2 are m.m

Is this prime? Or am I missing something.

Cause; (m + 2) (m + 3) will not work.

and (m + 1) (m + 6) will not work

and I can't think of any other factors right now..

 

[pka]

Factor Completely:
m^2 + 2m + 6

\(\displaystyle \left[ {m - \left( { - 1 + \sqrt 5 i} \right)} \right]\left[ {m - \left( { - 1 - \sqrt 5 i} \right)} \right]\)

 

[gijas]

\(\displaystyle \left[ {m - \left( { - 1 + \sqrt 5 i} \right)} \right]\left[ {m - \left( { - 1 - \sqrt 5 i} \right)} \right]\)

We are only working with real numbers right now.

 

[pka]

We are only working with real numbers right now.

In that case it is in prime form as is.

 

[gijas]

In that case it is in prime form as is.

Thank you!

 

[lookagain]

We are only working with real numbers right now.

gijas,

you must not just be working with just any coefficients from the Real numbers.

It must be the case that you are working with coefficients for the factors
that are integers only, (if they do exist).


\(\displaystyle So, \ m^2 + 2m + 6 \ is \ prime, \ but \ then, \ for \ example, \ so \ is \ m^2 - 2, \)

\(\displaystyle because \ it \ is \ prime \ over \ the \ integers. \)


If you want to know if a quadratic is prime, particularly of the form

\(\displaystyle ax^2 + bx + c, \ with \ a \ne 0, \ and \ a, b, c \)


are each integers, then look at the discriminant, \(\displaystyle b^2 - 4ac.\)


If the value of this discriminant is anything other than a perfect square,

(zero or positive), then the quadratic is prime over the integers.