Math 360, Fall 2013, Assignment 15

I must study politics and war that my sons may have liberty to study mathematics and philosophy.

- John Adams, letter to Abigail Adams, May 12, 1780

Carefully state the following theorems (you need not prove them):

1. Theorem concerning polynomial long division.
2. Factor theorem.
3. Theorem bounding the number of roots of a given polynomial.
4. Theorem on reducibility of quadratic and cubic polynomials.
5. Theorem relating factorization in $\mathbb{Q}[x]$ to factorization in $\mathbb{Z}[x]$.
6. Rational root theorem.
7. Eisenstein's criterion.

Solve the following problems:

1. Section 23, problems 1, 9, 15, 19, and 21.

--------------------End of assignment--------------------

Solutions:

1.) Dividing $x^6+3x^5+4x^2-3x+2$ by $x^2+2x-3$, with all polynomials in $\mathbb{Z}_7[x]$ yields $q=x^4+x^3+x^2+x-2$ and $r=4x+3=4x-4$ (Note that in order to check my work, I had to redefine $3\in \mathbb{Z}_7$ to be -4, but it should work)
9.) We first observe that 1 is a root of $x^4+4\in \mathbb{Z}_5[x]$, since $1^4+4=1+4+5=0 (mod 5)$.
So we divide $x^4+4$ by $(x-1)$, yielding $x^3+x^2+x+1$ (confirm that the remainder upon division is 0. I checked and it was. Remember everything is mod 5)
Now, we observe that 2 is a root of $x^3+x^2+x+1$, since $2^3+2^2+2+1=8+4+2+1=15=0(mod 5)$. Dividing $x^3+x^2+x+1$ by $(x-2)$ yields $x^2+3x+2$ (once again, check that the remainder is 0)
Now, we can factor by inspection. $(x+1)(x+2)=x^2+x+2x+2=x^2+3x+2$.

So we conclude that $x^4+4=(x-1)(x-2)(x+1)(x+2)$
To confirm, we recall that multiplication is commutative, so long as it is modulo 5. Thus $(x-1)(x-2)(x+1)(x+2)=(x-1)(x+1)(x-2)(x+2)=(x^2-1)(x^2-4)$, and then noting that -4=1(mod 5), $(x^2-1)(x^2-4)=(x^2-1)(x^2+1)=x^4-1=x^4+4$
Which confirms the answer.
19.) Our polynomial is $8x^3+6x^2-9x+24$. Let $p=3$. Clearly $p\in \mathbb{Z}$, and $p$ is also prime. Furthermore $8 \neq 0 (mod 3)$, because $8=2(mod 3)$. Also, $6=0(mod 3)$ and $-9=0(mod 3)$, since $6=2*3$ and $-9=-3*3$. Finally $24 \neq 0(mod 3^3)$ because $24 = 6 (mod 3^3)$. Therefore, by Eistenstein's Criterion, the polynomial is irreducible in $\mathbb{Q}[x]$

21.) Our polynomial is $2x^{10}-25x^3+10x^2-30$. Let $p=5$. Clearly $p\in \mathbb{Z}$ and $p$ is also prime. Furthermore, $2 \neq 0(mod 5)$ because $2=2(mod 5)$. Also, $-25=0 (mod 5)$ and $10 = 0(mod 5)$, since $-25=-5*5$ and $10=2*5$. Finally $-30\neq 0 (mod 5^5)$ because $-30 = 20(mod 5^5)$. Therefore, by Eisenstein's Criterion, the polynomial is irreducible in $\mathbb{Q}[x]$