Zeroes and factorization of a general polynomial

Task

Suppose $p(x)$ is a polynomial of degree $d > 0$.

1. If $p(0)=0$, show that $p(x)$ is evenly divisible by $x$.
2. If $p(1)=0$, show that $p(x)$ is evenly divisible by $x-1$.
3. If $r$ is a real number such that $p(r)=0$, show that $p(x)$ is evenly divisible by $x-r$.
4. Using part (c) show that $p$ can have at most $d$ distinct roots, that is, there can be at most $d$ numbers $r_1,\ldots,r_d$ with $p(r_1) = \cdots = p(r_d) = 0$.

IM Commentary

This task builds on ''Zeroes and factorization of a quadratic function'' parts I and II. The teacher may wish to recall the result from the first of these tasks, generalized to the polynomials of degree $d$ considered here. This result, the division algorithm applied to polynomials, says that if $p(x)$ is a polynomial of degree $d$ and $l(x)$ is a linear polynomial then $$ p(x) = q(x)l(x) + r $$ where $r$ is the remainder of the division process, a real number in this case.

Although a polynomial of degree $d$ can have at most $d$ roots, it is not easy, once the polynomial is of degree larger than two, to find those roots. In fact, a closed form for the roots of a polynomial function is only available for degrees 2,3, and 4. The level of complexity of these expressions grows rapidly.

This task is related to a very deep theorem in mathematics, the Fundamental Theorem of Algebra, which says that a polynomial of degree $d$ always has exactly $d$ roots, provided complex numbers are allowed as roots and provided roots are counted with the proper ''multiplicity'' (for example the polynomial $f(x) = x^2$ has only one root, $0$, but this root occurs with ''multiplicity'' $2$). The teacher may wish to take this occasion to mention this important theorem. The teacher may also wish to discuss the inductive structure of the argument in part (d): the impact of each root on the factorization of the polynomial $p(x)$ builds upon the factorization found in the previous step.

Solution

1. If $0$ is a root of the function $p$ this means that $p(0) = 0$. Since $p$ is a polynomial of degree $d$ it is given by a formula $$ p(x) = a_dx^d + a_{d-1}x^{d-1} + \ldots + a_0, $$ where $a_d\neq 0$. If we divide $p(x)$ by $x$, using long division of polynomials, we will find $$ p(x) = q(x)x + s $$ where $q(x)$ is a polynomial of degree $d-1$ and the remainder $s$ is a number. Plugging in $x = 0$ we find

   \begin{align} 0 &= p(0) \\ & = q(0)(1-1) + s \\ & = s. \end{align}
   Thus $s = 0$ and so $p(x) = xq(x)$ and $p(x)$ is evenly divisible by $x$ as desired.

   Alternatively, we see from inspection that $p(0) = a_0$ and so $a_0$ is $0$ when $p(0) = 0$. Since all other terms of $p(x)$ have at least one power of $x$, we can conclude that $p(x)$ is evenly divisible by $x$. This argument is quicker than the preceding but does not generalize as readily to the other parts of the problem.

2. If $1$ is a root of the function $p$ this means that $p(1) = 0$. If we divide $p(x)$ by $x-1$, using long division of polynomials, we will find $$ p(x) = q(x)(x-1) + s $$ where $q(x)$ is a polynomial of degree $d-1$ and the remainder $s$ is a number. Plugging in $x = 1$ we find
   \begin{align} 0 &= p(1) \\ & = q(1)(1-1) + s \\ & = s. \end{align}
   Thus $s = 0$ and so $p(x) = (x-1)q(x)$ and $p(x)$ is evenly divisible by $(x-1)$ as desired.
3. If $r$ is a real number which is a root of $p$ we have $p(r) = 0$. Performing long division, as in part (b), this time dividing $p(x)$ by $x-r$ gives $$ p(x) = a(x)(x-r) + s $$ where $a(x)$ is polynomial of degree $d-1$ and $s$ is a real number. Plugging in $r$ we find $$ p(r) = a(r)(0) + s. $$ Since $p(r) = 0$ we conclude that $s = 0$ and so $x-r$ divides $f$ evenly.

4. If there were $d+1$ different real numbers $r_1,\ldots,r_{d+1}$ which are all roots of $p$, we need to apply the argument of part (c) $d+1$ times and will find that this is not possible because it would give too many factors, of the form $x-r_i$, of $p$. Concretely, applying part (c) to $r_1$ we find $$ p(x) = p_1(x)(x-r_1) $$ where $p_1(x)$ has degree $d-1$. Now, evaluating at $r_2$ gives $$ 0 = p(r_2) = p_1(r_2)(r_2-r_1). $$ Since $r_1$ and $r_2$ are distinct, this means that $r_2 -r_1 \neq 0$ so we must have $p_1(r_2) = 0$.

   Repeating the above argument with $p_1(x)$ in place of $p(x)$ we conclude that $$ p_1(x) = p_2(x)(x-r_2) $$ where $p_2(x)$ is a polynomial of degree $d-2$. So we have $$ p(x) = p_2(x)(x-r_1)(x-r_2). $$ Continuing in the same way we find $$ p(x) = a_d(x-r_1)(x-r_2) \ldots (x-r_d). $$ But now we see that when we plug in $r_{d-1}$ we get $$ p(r_{d+1}) = a_d(r_{d+1}-r_1)(r_{d+1}-r_2) \ldots (r_{d+1}-r_d) $$ and this is not zero because $a_d$ is not zero and $r_{d+1}$ is distinct from the other roots $r_1,\ldots, r_d$.

   Part (d) of this problem is closely related to a very important result in mathematics called the Fundamental Theorem of Algebra. This says that a polynomial of degree $d$ has $d$ roots provided that complex numbers are allowed for solutions and provided the roots are counted with the appropriate ''multiplicity.'' It is interesting to note that limiting the number of possible solutions can be done, as in this task, within the high school curriculum: on the other hand, showing that $d$ solutions to a polynomial of degree $d$ always exist (when counted appropriately) remains to this day a very difficult problem.