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If we know $f(x) = (x-a)^k\cdot q(x)$, and $b\neq a$ is a zero of $q$, prove that $b$ is a zero of $f$ with the exact same multiplicity. Why do we need to prove this? What does this tell us about unique factorization.
Let's practice testing if polynomials are reducible or irreducible over $\mathbb{Q}$. These polynomials come straight out of Gallian.
- $21x^3-3x^2+2x+9$
- $x^5+2x+4$ (check mod 2 and mod 3 - we'll have 9 cases to check. Divide up the work.)
- $3x^5+15x^4-20x^3+10x+20$
- $x^5+9x^4+12x^2+6$
- $x^4+x+1$
- $x^4+3x^2+3$
- $(5/2)x^5 +(9/2)x^4+15 x^3+(3/7(x^2+6x + 3/14$
Show that $x^4+1$ is irreducible over $\mathbb{Q}$, but reducible over $\mathbb{Z}_p$ for every prime $p$.
Find all the zeros of $3x^2+x+4$ over $\mathbb{Z}_7$. Do so in two ways. First, find all the zeros by substitution. Then use the quadratic formula. Then repeat this with $2x^2+x+3$ over $\mathbb{Z}_5$. When does the quadratic formula work?
Show that the ideal $\left<x^2+1\right>$ is prime in $\mathbb{Z}[x]$ but not maximal in $\mathbb{Z}[x]$.
For more problems, see AllProblems