1. ## Polynomials are irreducible

Use Eisenstein's criterion to show taht each of these polynomials is irreducible over the field of rational numbers. (You may need to make a substitution).

a) x^4 - 12x^2 + 18x - 24
b) 4x^3 - 15x^2 + 60x + 180
c) 2x^10 - 25x^3 + 10x^2 - 30
d) x^2 + 2x - 5 (substitute x-1 or x+1)

Here is what I have so far:
a) p = 2 or 3
when p = 2, then -24 ≡ 0 mod 2^2
and when p = 3, then -24 ≡ 3 mod 3^2
so, p = 3
also, 1 ≡ 0 mod 3 is not true

b) p = 3 or 5
when p = 3, then 180 ≡ 0 mod 3^2
and when p = 5, then 180 ≡ 5 mod 5^2
so, p = 5
also, 4 ≡ 0 mod 5 is not true

c) p = 5
so, -30 ≡ 6 mod 5^2
also, 2 ≡ 0 mod 5 is not true

d) I am not sure

Does what I have done look correct? Any help for part d would be greatly appreciated.

2. Originally Posted by page929
Use Eisenstein's criterion to show taht each of these polynomials is irreducible over the field of rational numbers. (You may need to make a substitution).

a) x^4 - 12x^2 + 18x - 24
b) 4x^3 - 15x^2 + 60x + 180
c) 2x^10 - 25x^3 + 10x^2 - 30
d) x^2 + 2x - 5 (substitute x-1 or x+1)

Here is what I have so far:
a) p = 2 or 3
when p = 2, then -24 ≡ 0 mod 2^2
and when p = 3, then -24 ≡ 3 mod 3^2
so, p = 3
also, 1 ≡ 0 mod 3 is not true

b) p = 3 or 5
when p = 3, then 180 ≡ 0 mod 3^2
and when p = 5, then 180 ≡ 5 mod 5^2
so, p = 5
also, 4 ≡ 0 mod 5 is not true

c) p = 5
so, -30 ≡ 6 mod 5^2
also, 2 ≡ 0 mod 5 is not true

d) I am not sure

Does what I have done look correct? Any help for part d would be greatly appreciated.
For a) $p=2$ is not allowed becuase $2^2=4 \text{ and } 4|24$

However $p=3$ does do the job because it divides all of the coefficients and $3^3=9 \implies 9 \not{|}24$

Therefore it is irreducible.

once again for b) You have used a prime that is not allowed $3^2=9$ and $9|180$ $p=5$ works.

For the last one Since both 2 and 5 are relatively prime the theorem does not apply directly. As suggested by the hint:

$p(x)=x^2+2x-5 \implies p(x+1)=(x+1)^2+2(x+1)-5=x^2+4x-2$

Now if we let $p=2$ the polynomial $p(x+1)$ is irreducible. So this implies that $p(x)$ is irreducible.

Lets see why. Suppose that $p(x+1)$ is irreduicble but $p(x)$ is then $p(x)=(ax+b)(cx+d)$ for $a,b,c,d \in \mathbb{Q}$, but then

$p(x+1)=[a(x+1)+d][c(x+1)+d]=(ax+a+d)(cx+c+d)$ but now $p(x+1)$ is reducible and this is a contradiction.