Prime and Factorable Polynomials

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November 26th 2011, 01:11 PM
xivivx

Prime and Factorable Polynomials

I'm new to factoring. Don't really understand the definition of "prime". I did 47 questions in my homework, but I want to make sure if some are right or wrong.

I will answer the questions. Tell me if they are right/ wrong. If they are wrong, then tell me how. At least show me some type of work , so I understand.

1) P^4-49= (P^2+7)(P^2-7)

2) X^4-1 = (X^2+1)(X^2-1)
(X^2-1) can be factored into (X+1) (X-1).
Total answer: (X^2+1)(X+1)(X-1)

3) 100x^2+49= PRIME

4) K^2-18= PRIME

5) 81W^2+16= Prime

6) 6t^3-2= Prime

7) M^2-12= Prime

I think I have a problem with primes, because it seemed weird I had too many primes.

Help me , please
November 26th 2011, 01:18 PM
Also sprach Zarathustra

re: Prime and Factorable Polynomials

"Prime", hm! Ok, let it be...

7) $M^2-12 = (M+\sqrt{12})(M-\sqrt{12})$

$\sqrt{12}=\sqrt{4\cdot 3}=2\sqrt{3}$

$M^2-12 = (M+\sqrt{12})(M-\sqrt{12})= (M+2\sqrt{3})(M-2\sqrt{3})$
November 26th 2011, 01:26 PM
xivivx

re: Prime and Factorable Polynomials

So all of them are correct except number 7?
November 26th 2011, 01:33 PM
alexmahone

re: Prime and Factorable Polynomials

Quote:

Originally Posted by xivivx

So all of them are correct except number 7?

I assume that the problem requires you to factorize into polynomials with integer coefficients, in which case all your answers are correct (including 7).
November 26th 2011, 01:34 PM
Also sprach Zarathustra

re: Prime and Factorable Polynomials

Quote:

Originally Posted by xivivx

So all of them are correct except number 7?

No.

In 1) you can factoring a little bit more, 2) correct, 3) correct, 4) incorrect , 5) correct, 6) incorrect.
November 26th 2011, 01:41 PM
xivivx

re: Prime and Factorable Polynomials

I get what you are saying. I need to square root them. For example, number 1 and 4 I can solve like this:

1) P^4-49= (P^2+7)(P^2-7)

(P+square root of 7) (P-square root of 7) (P^2+7)

4) K^2-18=

(K-3 square root of 2) (K+3 square root of 2)

I,however seriously have no idea on number 6.
November 26th 2011, 01:45 PM
Also sprach Zarathustra

Re: Prime and Factorable Polynomials

Quote:

Originally Posted by xivivx

I get what you are saying. I need to square root them. For example, number 1 and 4 I can solve like this:

1) P^4-49= (P^2+7)(P^2-7)

(P+square root of 7) (P-square root of 7) (P^2+7)

4) K^2-18=

(K-3 square root of 2) (K+3 square root of 2)

I,however seriously have no idea on number 6.

$a^3-b^3= (a-b)(a^2+ab+b^2)$
November 26th 2011, 01:48 PM
alexmahone

Re: Prime and Factorable Polynomials

Quote:

Originally Posted by Also sprach Zarathustra

No.

In 1) you can factoring a little bit more, 2) correct, 3) correct, 4) incorrect , 5) correct, 6) incorrect.

Out of curiosity, why do you refrain from factoring $(X^2+1)=(X+i)(X-i)$ ?
November 26th 2011, 02:04 PM
e^(i*pi)

Re: Prime and Factorable Polynomials

Quote:

Originally Posted by alexmahone

I assume that the problem requires you to factorize into polynomials with integer coefficients, in which case all your answers are correct (including 7).

I'm inclined to agree
November 26th 2011, 02:09 PM
xivivx

Re: Prime and Factorable Polynomials

(X^2+1)=(X+1)(X-1)

If you foil them, then you'll know they won't work.

X (multiplied by) X= X^2
X multiplied by -1=-x
1 multiplied by x= x
1 multiplied by -1= -1

X^2-x+x-1 = X^2+(-1) instead of what the formula says which is X^2+1
November 26th 2011, 02:12 PM
Also sprach Zarathustra

Re: Prime and Factorable Polynomials

Quote:

Originally Posted by xivivx

(X^2+1)=(X+1)(X-1)

If you foil them, then you'll know they won't work.

X (multiplied by) X= X^2
X multiplied by -1=-x
1 multiplied by x= x
1 multiplied by -1= -1

X^2-x+x-1 = X^2+(-1) instead of what the formula says which is X^2+1

Look here:

Complex number - Wikipedia, the free encyclopedia
November 27th 2011, 05:03 AM
HallsofIvy

Re: Prime and Factorable Polynomials

xivivx, please answer the question that was implied in allexmahone's response 4: What FIELD is this over? Complex numbers (in which $x^2+ 1= (x+ i)(x- i)$ , the real numbers, in which $x^2+ 1$ is prime but $x^2- 18= (x- 3\sqrt{2})(x+ 3\sqrt{2})$ , or the rational numbers in which both of those are prime? I suspect the rational numbers but you really need to tell us that.