Difference between revisions of "Math 360, Fall 2013, Assignment 15"
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==Solutions:== |
==Solutions:== |
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+ | 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 $r=4x+3=4x-4$ |
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+ | (Note that in order to check my work, I had to redefine $3\in \mathbb{Z}_7$ to be -4, but it should work) |
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+ | 9.) We first observe that 1 is a root of $x^4+4\in \mathbb{Z}_5[x]$, since $1^4+4=1+4+5=0 (mod 5)$.<br> |
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+ | 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)<br> |
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+ | 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)<br> |
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+ | Now, we can factor by inspection. $(x+1)(x+2)=x^2+x+2x+2=x^2+3x+2$.<br> |
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+ | <br> |
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+ | So we conclude that $x^4+4=(x-1)(x-2)(x+1)(x+2)$<br> |
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+ | 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), $(x^2-1)(x^2-4)=(x^2-1)(x^2+1)=x^4-1=x^4+4$<br> |
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+ | Which confirms the answer. |
Revision as of 21:41, 14 December 2013
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):
- Theorem concerning polynomial long division.
- Factor theorem.
- Theorem bounding the number of roots of a given polynomial.
- Theorem on reducibility of quadratic and cubic polynomials.
- Theorem relating factorization in $\mathbb{Q}[x]$ to factorization in $\mathbb{Z}[x]$.
- Rational root theorem.
- Eisenstein's criterion.
Solve the following problems:
- Section 23, problems 1, 9, 15, 19, and 21.
Questions:
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.