## 305 -Homework set 2

This set is due February 20 at the beginning of lecture. Consult the syllabus for details on the homework policy. I do not think this set is particularly difficult, but it is on the longish side of things, so make sure you leave yourself enough time to work on it.

1. Gauß’ fundamental theorem of algebra states that any equation $p(x)=0,$ where $p$ is a polynomial with complex coefficients, has at least one complex root $z.$ This means that $z$ is a complex number and $p(z)=0.$ Show that $p$ has at most $n$ roots, where $n$ is its degree, and that if we count roots up to multiplicity, then it has exactly $n$ roots. Since the multiplicity of a root $z$ is by definition the largest $m$ such that $(x-z)^m$ is a factor of $p(x),$ you may want to verify that $p(z)=0$ iff $(x-z)$ is a factor of $p.$

2. Let $p(x)$ be a polynomial with real coefficients, and let $z$ be a complex root of $p.$ Show that $p(\bar z)=0$ as well. Conclude that if the degree $n$ of $p$ is odd and the coefficients of $p$ are real, then $p$ has at least one real root. (You may use the fundamental theorem of algebra, if needed.) Conclude also that if $p$ is of degree four and has real coefficients, then $p$ can be factored as the product of two quadratic polynomials with real coefficients. (Does this follow “directly” from the argument described in lecture?)

3. Solve exercises 54-56 from Chapter 3 of the book.

4. Show directly that if $a,b,c$ are real numbers, then at least one of the solutions of $x^3+ax^2+bx+c=0$ is a real number. What I mean is that, rather than appealing to problem 2, you want to look at the solutions obtained by Cardano’s method as described in lecture, and argue directly from the formulas so obtained that at least one of the solutions must be real. Be careful, since your argument should not give you that all three roots are real, since this is not true in general.

5. Show directly that a quartic with complex coefficients admits only 4 roots. What I mean is that, rather than appealing to problem 1, you want to look at the solutions obtained by Ferrari’s method as described in lecture, and argue directly that they only produce 4 roots, even though, in principle, they produce 24 (since they involve solving a cubic and then taking a square root to obtain parameters from which four solutions are then found).