# Problem of the MonthHint for Problem 6: March 2024

1. The Rational Root Theorem could come in handy: If a polynomial $$p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$$ 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 $$f(x)=(x-r)^2p(x)$$, then $$f(x+y) = [(x-r)+y]^2p(x+y)$$. Expand $$p(x+y)$$ and $$f(x+y)$$ as described in part (i) and compare "coefficients" of $$y$$.

1. 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 $$g(x)$$.

2. 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).