Contents / Precalculus / Polynomial & Rational Functions

Chapter 2

Polynomial & Rational Functions

Polynomial behavior, zeros and roots, rational functions, asymptotes, and partial fractions.

Introduction

Polynomial behavior, zeros and roots, rational functions, asymptotes, and partial fractions.

2.1Polynomial

Polynomials are the building blocks of algebra. Their simple structure (finite sum of power terms) gives them nice properties: defined everywhere, smooth, and completely determined by finitely many coefficients.

Polynomials

Definition 2.1 (Polynomial). A polynomial is an expression of the form

p (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0},

where the $a_{i}$ are constants (coefficients) and $n$ is a nonnegative integer called the degree (provided $a_{n} \neq = 0$ ).

The degree tells the highest power, and it governs the end behavior.

Proposition 2.2 (End behavior is decided by the leading term). For large $∣ x ∣$ , the leading term $a_{n} x^{n}$ dominates every other term of $p$ , so the far-left and far-right shape of the graph is decided entirely by the sign of $a_{n}$ and whether $n$ is even or odd.

Proof. Divide through by $x^{n}$ . Every remaining term then carries a positive power of $x$ underneath, so all the lower-order terms shrink to $0$ and only $a_{n}$ survives.∎

Polynomials are defined for all real $x$ , continuous, and differentiable everywhere — there are no denominators to vanish and no roots or logs to restrict the domain. They form a vector space and are dense in the continuous functions on closed intervals (Weierstrass approximation theorem).

Intuition. A polynomial is a "power sum recipe": you take powers of your input, multiply by fixed numbers (the coefficients), and add them up. Higher degree means the graph can wiggle more.

Example 2.3 (Identifying degree, leading coefficient, and constant term). For $p (x) = - 3 x^{4} + 7 x^{2} - 5 x + 9$ :

Solution. Highest power with a nonzero coefficient is $4$ , so the degree is $4$ . Leading coefficient $a_{4} = - 3$ . Constant term $a_{0} = 9$ . Note that the $x^{3}$ term is missing, meaning its coefficient is implicitly $0$ ; this does not change the degree. End-behavior check: degree $4$ is even and $a_{4} = - 3 < 0$ , so both ends point down — $p (x) \to - \infty$ as $x \to \pm \infty$ . (Sanity check: $p (10) = - 30000 + 700 - 50 + 9 < 0$ and $p (- 10) = - 30000 + 700 + 50 + 9 < 0$ , both large and negative. ✓)□

Common Mistakes

Leading term mistakes: Forgetting that the degree is determined only by the highest power with nonzero coefficient (lower terms do not affect degree or end behavior).
Missing implicit zero coefficients: x^3 term absent means coeff 0; does not change degree.
Confusing degree with number of terms: A degree 5 poly can have only 2 terms (e.g. x^5 + 1).
End behavior errors for negative leading coeff: For even degree negative leading, both ends go to -∞.
Thinking constant is degree 0 or 1: Nonzero constant poly has degree 0; the zero polynomial degree is often defined as -∞ or undefined.
Forgetting polynomials have no vertical asymptotes or holes: They are defined and smooth on all reals.

Exercises for §2.1 →

2.2Root of a function

Roots (zeros) tell where the function crosses or touches the x-axis. For polynomials they reveal factors and are guaranteed by the Fundamental Theorem of Algebra.

Roots of a Function

Definition 2.4 (Root and multiplicity). A root (or zero) of a function $f$ is a value $r$ such that

f (r) = 0.

If $(x - r)^{k}$ is the highest power of $(x - r)$ dividing a polynomial $p (x)$ , we say $r$ is a root of multiplicity $k$ and count it $k$ times.

For polynomials, roots correspond to x-intercepts and linear factors, and their number is fixed in advance.

Theorem 2.5 (Fundamental Theorem of Algebra). A degree- $n$ polynomial with $n \geq 1$ has exactly $n$ roots in the complex numbers, counting multiplicity.

Over the reals, non-real roots of a polynomial with real coefficients always come in conjugate pairs $a \pm bi$ , so a real polynomial of odd degree always has at least one real root.

Multiplicity also decides the graph's shape at a real root: an odd multiplicity means the graph crosses the x-axis there, while an even multiplicity means it touches and bounces back without crossing.

Theorem 2.6 (Factor Theorem). For a polynomial $p$ , the number $r$ is a root of $p$ if and only if $(x - r)$ is a factor of $p$ .

That equivalence is why factoring helps find roots, and why finding one root lets you divide it out and continue with a polynomial of smaller degree. For a first guess at what the roots might be, there is a finite list to try.

Theorem 2.7 (Rational Root Theorem). If a polynomial with integer coefficients has a rational root $\frac{p}{q}$ in lowest terms, then $p$ divides the constant term $a_{0}$ and $q$ divides the leading coefficient $a_{n}$ .

Intuition. Roots are the "sweet spots" where the output is exactly zero — the places the graph crosses or touches the x-axis. For a quadratic like a parabola, there are at most two such places.

Example 2.8 (Finding all roots of $x^{3} - 6 x^{2} + 11 x - 6 = 0$ ). Possible rational roots by the Rational Root Theorem are $\pm \frac{p}{q}$ where $p ∣ 6$ (constant term) and $q ∣ 1$ (leading coefficient), giving candidates $\pm 1, \pm 2, \pm 3, \pm 6$ .

Solution.

Test $x = 1$ : $1 - 6 + 11 - 6 = 0$ . So $x = 1$ is a root and $(x - 1)$ is a factor.
Divide out $(x - 1)$ (synthetic division on coefficients $1, - 6, 11, - 6$ with root $1$ ): bring down $1$ ; $1 \cdot 1 = 1$ , $- 6 + 1 = - 5$ ; $- 5 \cdot 1 = - 5$ , $11 - 5 = 6$ ; $6 \cdot 1 = 6$ , $- 6 + 6 = 0$ . The quotient is $x^{2} - 5 x + 6$ with remainder $0$ , so $(x - 1) (x^{2} - 5 x + 6) = 0$ .
Factor the quadratic: $x^{2} - 5 x + 6 = (x - 2) (x - 3)$ .
Roots: $1, 2, 3$ (all real, each with multiplicity $1$ ).
Sanity check (Vieta): the sum of the roots should equal $- \frac{a _{n - 1}}{a _{n}} = - \frac{- 6}{1} = 6$ , and indeed $1 + 2 + 3 = 6$ . ✓ The product should equal $- \frac{a _{0}}{a _{n}} = 6$ (odd degree flips the sign), and $1 \cdot 2 \cdot 3 = 6$ . ✓

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Common Mistakes

Missing multiplicity: When counting roots or factoring, repeated roots (x-r)^2 count as two roots for the Fundamental Theorem.
Rational Root Theorem misapplication: Possible rational roots are p/q where p divides constant term, q divides leading coeff; sign ± must be considered.
Assuming all roots are real: High degree polys often have complex roots; they come in conjugate pairs for real coeffs.
Forgetting Factor Theorem converse: If r is root then (x-r) divides; useful for factoring.
Graphing without finding roots: Roots help sketch; but multiplicity tells if it crosses or bounces.
Confusing root of poly with root of equation in general: For non-polys, may have no closed form or infinitely many (trig).

Exercises for §2.2 →

2.3Rational function

Rational functions are ratios of polynomials. They bring asymptotes, holes, and richer end behavior than pure polynomials, while still being algebraic.

Rational Functions

Definition 2.9 (Rational function). A rational function is a quotient

r (x) = \frac{p ( x )}{q ( x )}

of two polynomials $p$ and $q$ , where $q$ is not the zero polynomial. Its domain is all real $x$ except where $q (x) = 0$ .

The intuition for its behavior: a zero of $q$ makes the denominator tiny, so the fraction blows up — unless the numerator vanishes there too. So before drawing any conclusions, factor both and cancel common factors. A zero of $q$ that survives cancellation gives a vertical asymptote; one that cancels leaves a hole (a removable gap in the graph).

Proposition 2.10 (End behavior from the degrees). The end behavior of $r = p / q$ as $x \to \pm \infty$ is determined by comparing the degrees of $p$ and $q$ , because only the leading terms matter for large $∣ x ∣$ :

$de g p < de g q$ : horizontal asymptote $y = 0$ (denominator grows faster).
$de g p = de g q$ : horizontal asymptote $y = \frac{leading coeff of p}{leading coeff of q}$ .
$de g p = de g q + 1$ : oblique (slant) asymptote, found by polynomial division.
$de g p \geq de g q + 2$ : no horizontal or slant asymptote; the graph grows like the polynomial quotient of $p$ by $q$ .

Intuition. Rational functions are like "division of two power-sum recipes." They usually have breaks (asymptotes) where the bottom recipe is zero, and their far-left and far-right behavior is predictable from the "sizes" (degrees) of top and bottom.

Example 2.11 (Analyzing $r (x) = \frac{x ^{2} - 1}{x ^{2} - 4}$ ). Factor: numerator $x^{2} - 1 = (x - 1) (x + 1)$ , denominator $x^{2} - 4 = (x - 2) (x + 2)$ .

Solution.

No common factors, so no holes; the zeros of the denominator, $x = \pm 2$ , are genuine vertical asymptotes.
Degrees are equal (both $2$ ), so the horizontal asymptote is the ratio of leading coefficients, $y = \frac{1}{1} = 1$ .
Holes: none (nothing cancelled in step 2).
x-intercepts (where the numerator is zero and the denominator is not): $x = \pm 1$ .
y-intercept: $r (0) = \frac{0 - 1}{0 - 4} = \frac{- 1}{- 4} = \frac{1}{4}$ .
Sanity check on the horizontal asymptote: at $x = 100$ , $r (100) = \frac{10000 - 1}{10000 - 4} = \frac{9999}{9996} \approx 1.0003$ , close to $1$ as predicted. ✓

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Common Mistakes

Vertical asymptote vs hole: If a factor cancels, the common root produces a hole, not an asymptote. Always simplify first.
Horizontal asymptote degree rule misremembered: When numerator degree is larger by exactly 1 there is a slant asymptote, not "no asymptote."
End behavior for rationals: Students sometimes think the horizontal asymptote is always y=0; it depends on the degrees.
Forgetting y-intercept or x-intercepts after asymptote analysis: Full picture needs intercepts too.
Not checking if num and den have common factors before stating asymptotes: Simplification changes the graph (holes vs asymptotes).
Oblique asymptote requires polynomial division: Just knowing deg tells you it exists; you must divide to find the actual line equation.

Exercises for §2.3 →

2.4Asymptote

Asymptotes are "ghost lines" that the graph approaches at the edges or at forbidden points. They are key to understanding the long-term behavior of rational and other functions.

Asymptotes

Definition 2.12 (Asymptote). Vertical asymptotes occur where the function heads to $\pm \infty$ as $x$ approaches a finite value $c$ (usually where the denominator tends to $0$ while the numerator stays nonzero).

Horizontal asymptotes describe the limiting value of $f (x)$ as $x \to \pm \infty$ .

Oblique (slant) asymptotes appear when the function behaves like a non-horizontal line at infinity (when $de g p = de g q + 1$ ).

The three kinds answer different questions — one is about a forbidden input, the other two about the far ends of the graph — but for a rational function all three are found by the same short routine.

Method 2.13 (Finding the asymptotes of a rational function).

Factor numerator and denominator and cancel any common factors; a cancelled zero is a hole, not a wall.
To find the vertical asymptotes: solve $q (x) = 0$ after simplification and check the numerator is nonzero there.
To find horizontal or oblique asymptotes: use degree comparison, or divide and take the limit of the remainder term to $0$ .

Intuition. Think of a car driving next to a straight wall: a vertical asymptote is a wall the car races alongside and shoots up (or down) forever as it nears a forbidden spot; a horizontal asymptote is a cruising speed the car settles into far down the road; an oblique asymptote is a steady straight ramp the car ends up hugging. The graph can hug or even briefly cross a horizontal/slant asymptote in the middle, but it can never cross a vertical one — that spot is simply off-limits.

Example 2.14 (Finding all asymptotes of $f (x) = \frac{2 x ^{2} + 3 x - 1}{x - 1}$ ). Vertical: the denominator is zero at $x = 1$ ; the numerator there is $2 (1)^{2} + 3 (1) - 1 = 4 \neq = 0$ , so there is a vertical asymptote at $x = 1$ .

Solution.

Degrees: numerator $2$ , denominator $1$ , so $de g p = de g q + 1$ — expect an oblique (slant) asymptote and no horizontal one.
Perform long division of $2 x^{2} + 3 x - 1$ by $x - 1$ :
$\frac{2 x ^{2}}{x} = 2 x$ ; then $2 x (x - 1) = 2 x^{2} - 2 x$ ; subtract: $(2 x^{2} + 3 x - 1) - (2 x^{2} - 2 x) = 5 x - 1$ .
$\frac{5 x}{x} = 5$ ; then $5 (x - 1) = 5 x - 5$ ; subtract: $(5 x - 1) - (5 x - 5) = 4$ .
Quotient $2 x + 5$ , remainder $4$ .
Thus $f (x) = 2 x + 5 + \frac{4}{x - 1}$ . As $x \to \pm \infty$ the term $\frac{4}{x - 1} \to 0$ , so the oblique asymptote is $y = 2 x + 5$ .
Sanity check on the division: $(2 x + 5) (x - 1) + 4 = 2 x^{2} + 3 x - 5 + 4 = 2 x^{2} + 3 x - 1$ , which matches the original numerator. ✓

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Common Mistakes

Vertical asymptote vs hole: If a factor cancels, the common root produces a hole, not an asymptote. Always simplify first.
Horizontal asymptote degree rule misremembered: When numerator degree is larger by exactly 1 there is a slant asymptote, not "no asymptote."
End behavior for rationals: Students sometimes think the horizontal asymptote is always y=0; it depends on the degrees.
Forgetting to check numerator when finding vertical: If num also zero, it may cancel to hole.
Slant asymptote calculation errors: Must perform the division correctly to get the exact linear expression.
Assuming oblique only for rationals: Other functions (e.g. with logs or exp) can have slant or other asymptotic behavior too.

Exercises for §2.4 →

2.5Partial fraction decomposition

Partial fractions turn a complicated rational into a sum of simpler pieces. This is the key technique for integrating rationals by hand and for many inverse transforms.

Partial Fraction Decomposition

Definition 2.15 (Partial fraction decomposition). Partial fraction decomposition rewrites a rational function with a factored denominator as a sum of simpler fractions.

This is invaluable for integration and inverse Laplace transforms. It requires the fraction to be proper ( $de g p < de g q$ ); if it is not, first divide to split off a polynomial part, then decompose the proper remainder.

Method 2.16 (Setting up a partial fraction decomposition). Factor the denominator completely into linear and irreducible-quadratic factors, then write one group of terms per factor.

Distinct linear factors: $\frac{p ( x )}{( x - r _{1} ) \dots ( x - r _{k} )} = \frac{A _{1}}{x - r _{1}} + \dots + \frac{A _{k}}{x - r _{k}}$ .
A repeated linear factor $(x - r)^{m}$ : include one term for every power up to $m$ , e.g. for $m = 2$ , $\frac{A}{x - r} + \frac{B}{( x - r ) ^{2}}$ .
An irreducible quadratic factor $a x^{2} + b x + c$ (one with no real roots, i.e. discriminant $< 0$ ): its numerator must be linear, giving $\frac{B x + C}{a x ^{2} + b x + c}$ .

Then clear denominators and solve for the coefficients, either by equating coefficients of like powers or by substituting convenient $x$ -values (the cover-up method, which works cleanly for distinct linear factors).

The number of unknown coefficients always equals $de g q$ ; omitting any required term leaves the system underdetermined.

Intuition. Partial fractions is "undoing the addition of fractions with different bottoms" so you can work with each piece separately — like separating a mixed salad back into individual ingredients for easier digestion (integration).

Example 2.17 (Decompose $\frac{5 x + 7}{( x - 1 ) ( x + 3 )}$ ). The fraction is proper ( $de g = 1 < 2$ ) and the denominator is already factored into distinct linear factors, so write $\frac{5 x + 7}{( x - 1 ) ( x + 3 )} = \frac{A}{x - 1} + \frac{B}{x + 3}$ .

Solution.

Multiply both sides by $(x - 1) (x + 3)$ : $5 x + 7 = A (x + 3) + B (x - 1)$ .
Set $x = 1$ (kills the $B$ term): $5 (1) + 7 = 12 = A (1 + 3) = 4 A \Rightarrow A = 3$ .
Set $x = - 3$ (kills the $A$ term): $5 (- 3) + 7 = - 8 = B (- 3 - 1) = - 4 B \Rightarrow B = 2$ .
Thus $\frac{5 x + 7}{( x - 1 ) ( x + 3 )} = \frac{3}{x - 1} + \frac{2}{x + 3}$ .
Verify by recombining: $\frac{3 ( x + 3 ) + 2 ( x - 1 )}{( x - 1 ) ( x + 3 )} = \frac{3 x + 9 + 2 x - 2}{( x - 1 ) ( x + 3 )} = \frac{5 x + 7}{( x - 1 ) ( x + 3 )}$ . ✓

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Example 2.18 (Decompose $\frac{x + 3}{( x - 2 ) ^{2}}$ (repeated linear factor)). The fraction is proper, and the denominator is a repeated linear factor $(x - 2)^{2}$ , so include a term for each power: $\frac{x + 3}{( x - 2 ) ^{2}} = \frac{A}{x - 2} + \frac{B}{( x - 2 ) ^{2}}$ .

Solution.

Multiply both sides by $(x - 2)^{2}$ : $x + 3 = A (x - 2) + B$ .
Set $x = 2$ (kills the $A$ term): $2 + 3 = 5 = B$ , so $B = 5$ .
To find $A$ , compare coefficients of $x$ : the left side has coefficient $1$ , the right side $A (x - 2) + B$ has coefficient $A$ , so $A = 1$ .
Thus $\frac{x + 3}{( x - 2 ) ^{2}} = \frac{1}{x - 2} + \frac{5}{( x - 2 ) ^{2}}$ .
Verify by recombining: $\frac{1 \cdot ( x - 2 ) + 5}{( x - 2 ) ^{2}} = \frac{x + 3}{( x - 2 ) ^{2}}$ . ✓

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Common Mistakes

Partial fractions setup: Forgetting to include terms for every factor (including powers for repeated linear factors) leads to underdetermined systems.
Not clearing the denominator completely: When setting up the equation, multiply both sides by the full factored denom before plugging values or equating coeffs.
Cover-up method only for simple distinct linear: For repeated or quadratic factors you need full system or derivatives for repeated.
Forgetting to factor the denominator first: You must have linear and irreducible quad factors completely before starting.
Sign errors when plugging the roots: When x=r_i , only the A_i term survives; easy arithmetic slip.
Not verifying the decomposition by recombining: Always combine back over common denom to check coefficients match original.

Exercises for §2.5 →

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

Practice this chapter