3. Polynomial and rational functions

Learning goals

Use the division algorithm to locate polynomial roots, solve rational inequalities without changing signs incorrectly, and distinguish removable holes from genuine asymptotes. Coefficients are real unless stated otherwise.

Polynomial algebra

A nonzero polynomial $p(x)=a_nx^n+\cdots+a_0$ has degree $n$ when $a_n\ne0$. The zero polynomial has no degree under our convention. For a nonzero divisor polynomial $d$, polynomial long division gives unique $q,r$ with $p=dq+r$ and either $r=0$ or $\deg r<\deg d$.

Proof of the division algorithm. If the degree of the current remainder is at least $\deg d$, subtract the multiple of $d$ that cancels its leading term. Each subtraction strictly reduces the degree, so the process terminates. If two representations exist, $d(q_1-q_2)=r_2-r_1$. A nonzero left side would have degree at least $\deg d$, whereas the right side has smaller degree. Therefore both sides are zero, proving uniqueness.

Theorem: factor and root bounds

For every real $a$, $p(a)=0$ if and only if $x-a$ divides $p$. A nonzero polynomial of degree $n$ has at most $n$ distinct real roots.

Proof. Divide by $x-a$; the remainder is a constant $r$, and evaluating at $a$ gives $r=p(a)$. If $p(x)=(x-a)s(x)$ and $b\ne a$ is another root, then $0=p(b)=(b-a)s(b)$ and $b-a\ne0$, so $s(b)=0$. Thus removing one root factor preserves every other distinct root, and the removal can be repeated. After removing $k$ distinct factors the remaining degree is $n-k$, so $k$ cannot exceed $n$. This is a bound, not a claim that all roots are real.

A root has multiplicity $m$ if $p(x)=(x-a)^m s(x)$ with $s(a)\ne0$. The sign changes at $a$ when $m$ is odd and stays the same when $m$ is even, because $s$ has constant sign sufficiently close to $a$.

Rational functions and inequalities

For $R=p/q$, the original domain excludes every zero of $q$. Cancelling a common factor gives equal values only on that domain; it does not restore excluded inputs. At an uncancelled denominator root with nonzero numerator, the magnitude becomes unbounded nearby. Long division describes behavior at infinity: if $R=s+r/q$ and $\deg r<\deg q$, then $R-s\to0$ as $|x|\to\infty$. A constant $s$ is a horizontal asymptote; a linear $s$ is an oblique asymptote. Higher-degree $s$ describes a polynomial asymptote.

To solve a rational inequality, mark all numerator and denominator roots, test the sign on each interval, include numerator zeros only for non-strict inequalities, and never include denominator zeros. Multiplying by an expression of unknown sign without splitting cases is invalid.

Proof bridge: polynomial signs and rational limits

Let $f(x)=\sum_{j=0}^d c_jx^j$ have real coefficients. When $|x-a|\le1$, the finite identity $x^j-a^j=(x-a)\sum_{\ell=0}^{j-1}x^{j-1-\ell}a^\ell$ gives

$$|f(x)-f(a)|\le C_a|x-a|,\qquad C_a=\sum_{j=1}^d j|c_j|(|a|+1)^{j-1}.$$

If $C_a=0$, $f$ is constant. Otherwise, when $f(a)\ne0$, taking $|x-a|<\min(1,|f(a)|/(2C_a))$ makes $f(x)$ have the same sign as $f(a)$ and gives $|f(x)|>|f(a)|/2$. This proves the local sign assertion used for multiplicities, without silently assuming a continuity theorem.

If $p(a)\ne0$ and $q(a)=0$ for a nonzero polynomial $q$, the preceding bound gives $|p(x)|>|p(a)|/2$ near $a$, while $|q(x)|\le C_q|x-a|$ for a finite positive $C_q$. Therefore $|p(x)/q(x)|\ge |p(a)|/(2C_q|x-a|)$ wherever the quotient is defined near $a$. For any $M>0$, sufficiently small nonzero $|x-a|$ makes this exceed $M$. The root bound permits an interval with no other zeros of $q$, so the quotient is defined throughout that punctured interval. This proves the vertical-asymptote claim in its stated nonzero-numerator case.

At infinity, let $q(x)=b_dx^d+\cdots+b_0$ with $b_d\ne0$, and let $r=0$ or $\deg r=k<d$. For $|x|\ge1$,

$$\left|\frac{q(x)}{x^d}-b_d\right|\le\frac{\sum_{j<d}|b_j|}{|x|}.$$

For sufficiently large $|x|$, $|q(x)/x^d|\ge|b_d|/2$. If $r\ne0$ and $C_r$ is the sum of the absolute values of its coefficients, $|r(x)|\le C_r|x|^k$ for $|x|\ge1$. Thus $|r(x)/q(x)|\le 2C_r/(|b_d||x|^{d-k})\to0$. The case $r=0$ is immediate. All estimates concern the original denominator-nonzero domain; cancelling factors never inserts an excluded point.

Worked examples

Cancellation with a retained hole. For $R(x)=(x^2-1)/(x^2-x)$ the original domain excludes $0$ and $1$. On this domain $R=(x+1)/x=1+1/x$. Thus $x=0$ is a vertical asymptote, $y=1$ a horizontal asymptote, and $(1,2)$ a missing point. Calling $x=1$ a vertical asymptote would be incorrect.

A sign chart. Solve $(x-1)/(x+2)\ge0$. The critical values are $-2$ and $1$. The quotient is positive on $(-\infty,-2)$ and $(1,\infty)$, negative on $(-2,1)$, and zero at $1$. The solution is $(-\infty,-2)\cup[1,\infty)$.

Graded practice

1. Foundation. Divide $x^3-2x^2+4$ by $x-2$, and decide whether $2$ is a root.
2. Core. Solve $(x-2)^2(x+1)/(x-3)<0$.
3. Core. Find the domain, holes and asymptotes of $(x^2+2x-3)/(x-1)$.
4. Proof. If two polynomials of degree at most $n$ agree at $n+1$ distinct real numbers, prove that they are identical. Explain why “degree exactly $n$” is not needed.

Full solutions

1. Long division gives quotient $x^2$ and remainder $4$: $x^3-2x^2+4=(x-2)x^2+4$. The value at $2$ is $4$, so $2$ is not a root.
2. Away from $2$, the squared factor is positive, so the sign is that of $(x+1)/(x-3)$. This is negative for $-1<x<3$. The strict inequality excludes the zeros $-1,2$ and the pole $3$. The answer is $(-1,2)\cup(2,3)$.
3. Factor the numerator as $(x+3)(x-1)$. The domain is $\mathbb R\setminus\{1\}$ and the graph is the line $y=x+3$ with the point $(1,4)$ removed. There is no vertical asymptote; $y=x+3$ is the oblique asymptote (indeed the formula agrees with it everywhere on the domain).
4. Their difference has degree at most $n$ if nonzero and has $n+1$ distinct roots, contradicting the root bound. Hence the difference is zero. Allowing lower-degree polynomials is essential because leading coefficients may cancel in the difference.

Summary and term definitions

A factor divides a polynomial with zero remainder; multiplicity counts repeated linear factors. A rational function is a quotient on the denominator-nonzero domain. A hole is a missing point with a finite limiting value, not an unbounded branch. An asymptote describes limiting behavior near a point or infinity. Exact factorization and domain bookkeeping should precede graph sketches or inequality calculations.