Problem of the Month
Hint for Problem 6: March 2024

1. The Rational Root Theorem could come in handy: If a polynomial $p(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x+a_0$ has integer coefficients and $r$ is a rational root of $p(x)$, then it must be of the form $r=\frac{u}{v}$ where $u$ and $v$ are integers, $u$ is a divisor of $a_0$, and $v$ is a divisor of $a_n$.

   For the polynomial in (a), this means the only possible rational roots are $\pm 1$, $\pm 2$, $\pm 4$, and $\pm 8$, the divisors of $8$ (the leading coefficient is $1$).

2. This polynomial is a perfect square.

3. This polynomial has no rational roots, but it does have a real root that is easy to find.

4. A polynomial has a rational root if and only if it has a rational linear factor. If $p(x)$ and $q(x)$ are polynomials of degree $m$ and $n$, then what is the degree of $p(x)q(x)$?

5. To show that a number is algebraic, you need to find a rational polynomial with that number as a root. For $1+\sqrt[3]{2}$, let $r=1+\sqrt[3]{2}$ so that $r-1=\sqrt[3]{2}$. Now cube both sides.

6. If $p(x)$ has degree $d$ and factors as the product of two polynomials of degree at least $1$, then both of these polynomials have degree less than $d$. Read the definition of "degree" carefully.

   For uniqueness, suppose that two polynomials, $p(x)$ and $q(x)$, have the described properties. What can be said about the degree of their difference?

7. No hint given.

8. (i) Warm up by trying this with a polynomial of lower degree. It turns out that the polynomial $f_1(x)$ is the derivative of $f(x)$. This is not important for the problem, but it is interesting.

   (ii) If you know some calculus, then there is a nice proof of this involving the product rule. Otherwise, if $f(x)=(x-r{)}^{2}p(x)$, then $f(x+y)=[(x-r)+y{]}^{2}p(x+y)$. Expand $p(x+y)$ and $f(x+y)$ as described in part (i) and compare "coefficients" of $y$.

10. By definition, the shared root is algebraic and so has a minimal polynomial, $m(x)$. Show that each of $p(x)$ and $q(x)$ is a scalar multiple of $m(x)$.

    Division Algorithm for Polynomials: For polynomials $f(x)$ and $g(x)$ with $g(x)$ not the zero polynomial, there exist unique polynomials $h(x)$ and $r(x)$ such that $$f(x)=h(x)g(x)+r(x)$$ where the degree of $r(x)$ is less than the degree of $g(x)$. This is essentially the result of doing polynomial or synthetic division of $f(x)$ by $g(x)$.

11. Convince yourself that every polynomial can be expressed as a product of irreducible polynomials. As well, as long as $f(x)$ has degree at least $1$, the polynomial $f_1(x)$ is not the zero polynomial and has degree less than that of $f(x)$ (in fact, the degree is one less).