# Thread: Prime and Factorable Polynomials

1. ## 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).

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.

2. ## re: Prime and Factorable Polynomials

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

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

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

$\displaystyle M^2-12 = (M+\sqrt{12})(M-\sqrt{12})= (M+2\sqrt{3})(M-2\sqrt{3})$

3. ## re: Prime and Factorable Polynomials

So all of them are correct except number 7?

4. ## re: Prime and Factorable Polynomials

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).

5. ## re: Prime and Factorable Polynomials

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.

6. ## 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.

7. ## Re: Prime and Factorable Polynomials

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.

$\displaystyle a^3-b^3= (a-b)(a^2+ab+b^2)$

8. ## Re: Prime and Factorable Polynomials

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 $\displaystyle (X^2+1)=(X+i)(X-i)$?

9. ## Re: Prime and Factorable Polynomials

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

10. ## 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

11. ## Re: Prime and Factorable Polynomials

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

12. ## 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 $\displaystyle x^2+ 1= (x+ i)(x- i)$, the real numbers, in which $\displaystyle x^2+ 1$ is prime but $\displaystyle x^2- 18= (x- 3\sqrt{2})(x+ 3\sqrt{2})$, or the rational numbers in which both of those are prime? I suspect the rational numbers but you really need to tell us that.