8.3 Rational functions, partial fractions, and Vieta formulas

Why the final part of the polynomial chapter matters

Polynomial division and gcds are not isolated techniques. They support two large tools that appear throughout later mathematics:

• partial fraction decomposition, which rewrites rational functions into simpler pieces;
• root-coefficient relations, especially Vieta's formulas, which let us compute symmetric expressions in roots without solving for the roots one by one.

Both tools depend on earlier results. Partial fractions need polynomial division, factorization, and relative primality. Vieta's formulas need the factorization of a polynomial over $C$.

Rational functions and the polynomial part

The first step in any partial fraction problem is to separate the polynomial part. If $\deg p\ge \deg q$, the division algorithm gives

$$p(x)=q(x)b(x)+r(x),\qquad \deg r\lt\deg q.$$

Therefore

$$\frac{p(x)}{q(x)}=b(x)+\frac{r(x)}{q(x)}.$$

Only the proper rational function $r(x)/q(x)$ still needs decomposition.

Splitting relatively prime denominator factors

Suppose

$$q(x)=q_1(x)q_2(x)$$

and $\gcd(q_1,q_2)=1$. For a proper rational function $r(x)/q(x)$, Bezout's identity gives polynomials a(x),b(x) with

$$a(x)q_1(x)+b(x)q_2(x)=1.$$

Multiplying by r(x) and dividing by $q_1q_2$ gives

$$\frac{r(x)}{q_1(x)q_2(x)} =\frac{r(x)b(x)}{q_1(x)} +\frac{r(x)a(x)}{q_2(x)}.$$

Then divide the numerators by $q_1$ and $q_2$ to reduce their degrees. The result is a unique decomposition

$$\frac{r(x)}{q(x)} =\frac{r_1(x)}{q_1(x)} +\frac{r_2(x)}{q_2(x)}, \qquad \deg r_i\lt\deg q_i.$$

This is the structural reason partial fractions work.

Repeated linear factors require one constant numerator for each power:

$$\frac{A(x)}{(x-a)^m} =\frac{k_1}{x-a}+\frac{k_2}{(x-a)^2}+\cdots+\frac{k_m}{(x-a)^m}.$$

Repeated irreducible quadratic factors require linear numerators:

$$\frac{B(x)}{(x^2+bx+c)^n} =\sum_{j=1}^n\frac{r_jx+s_j}{(x^2+bx+c)^j}.$$

Telescoping from partial fractions

Partial fractions often convert a complicated finite sum into a telescoping one. The slides use

$$\frac{x}{(2x-1)(2x+1)(2x+3)} =\frac{1}{16(2x-1)}+\frac{1}{8(2x+1)} -\frac{3}{16(2x+3)}.$$

Substituting $x=k$ and summing from $k=1$ to n makes most middle terms cancel. The remaining boundary terms are

$$\sum_{k=1}^{n} \frac{k}{(2k-1)(2k+1)(2k+3)} =\frac1{16}\left(2-\frac1{2n+1}-\frac3{2n+3}\right).$$

The main lesson is not the particular numbers; it is the method. Decompose first, then check whether shifted denominator patterns cancel in a finite sum.

Vieta's formulas

Let

$$p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\in C[x], \qquad a_n\ne0.$$

By the fundamental theorem of algebra,

$$p(x)=a_n(x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_n),$$

where the roots are counted with multiplicity.

For a cubic

$$a_3x^3+a_2x^2+a_1x+a_0,$$

with roots $\alpha_1,\alpha_2,\alpha_3$, this says

$$\alpha_1+\alpha_2+\alpha_3=-\frac{a_2}{a_3},$$ $$\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3=\frac{a_1}{a_3},$$ $$\alpha_1\alpha_2\alpha_3=-\frac{a_0}{a_3}.$$

Optional interpolation viewpoint

The slide chapter ends with Lagrange interpolation. If

$$(x_1,y_1),\ldots,(x_{n+1},y_{n+1})$$

have distinct $x_i$, then there is a unique polynomial of degree at most n passing through those $n+1$ points.

For each i, define

$$L_i(x)= \prod_{j\ne i}\frac{x-x_j}{x_i-x_j}.$$

Then $L_i(x_j)=\delta_{ij}$, so

$$f(x)=\sum_{i=1}^{n+1}y_iL_i(x)$$

has the required values. Uniqueness follows from the root bound: the difference of two such polynomials has degree at most n and $n+1$ roots, so it must be zero.

Proof sketch or proof idea

Partial fraction decomposition is not a memorized list of templates. The templates are the final form of three earlier facts working together. First, the division algorithm removes the polynomial part. If the fraction is improper, there is a polynomial part b(x) and a smaller proper fraction $r(x)/q(x)$.

Second, relatively prime denominator factors can be separated by Bezout's identity. If $q=q_1q_2$ and $\gcd(q_1,q_2)=1$, then $a q_1+b q_2=1$. Multiplying by the numerator and dividing by q splits the fraction into a term over $q_1$ and a term over $q_2$. This explains why distinct relatively prime factors receive separate partial-fraction terms.

Third, repeated powers are peeled off by division. A numerator above $(x-a)^m$ can be divided by $x-a$, leaving a constant remainder over the top power and a smaller-power fraction. Repeating this produces one term for each power. The same idea works for irreducible quadratics, except the remainder has degree less than 2, so it has the form $rx+s$.

Vieta's formulas have a different but equally structural proof. Once the fundamental theorem of algebra writes $p(x)=a_n\prod_{i=1}^{n}(x-\alpha_i)$, the coefficient of each power of x is an elementary symmetric sum in the roots. Comparing coefficients gives the formulas. This is why Vieta controls symmetric expressions in roots, but not arbitrary expressions unless they can be rewritten symmetrically.

Common mistakes

Summary

This note uses the algebraic machinery from the first two polynomial notes. Division separates polynomial and proper-rational parts. Bezout identities and relative primality explain why denominator factors split. Repeated powers determine the number of partial-fraction terms. Vieta's formulas compare the root-factor form of a complex polynomial with its coefficient form, allowing root sums and products to be computed without solving the polynomial.

Study guide for the exercises

For partial fractions, always decide the form before solving for coefficients. This prevents two common mistakes: omitting a repeated power and using a constant numerator where a linear numerator is needed. A repeated linear factor $(x-a)^m$ contributes one term for each denominator power. A repeated irreducible quadratic contributes one linear numerator for each denominator power. Only after the form is correct should you clear denominators and compare coefficients.

When coefficient comparison becomes messy, organize it by powers of x. After clearing denominators, expand the right-hand side and collect the coefficient of each power. A cubic identity gives four scalar equations, one for each of $x^3$, $x^2$, x, and the constant term. The system may look long, but it is ordinary linear algebra in the unknown coefficients of the partial fractions.

For telescoping sums, do not expect cancellation until after the summand has been decomposed. The original rational expression may show no obvious cancellation. Once it is written as a combination of shifted simple fractions, write out the first few and last few terms. The middle terms should reveal which pieces cancel and which boundary terms remain.

For Vieta problems, identify the polynomial and name its roots before doing any computation. Then list the elementary symmetric sums that Vieta gives: the sum of roots, the sum of pairwise products, and the full product in the cubic case. If the target expression is a power sum, rewrite it using these symmetric sums. If the target expression involves square roots of the roots, as in the tangent exercise, decide the sign separately from the product of the squared values.

The optional interpolation section is included because it uses the same zero-count theorem from 8.1. It is not a new algebraic technique to memorize; it is an application of the idea that a low-degree polynomial is determined by enough point values. This same uniqueness logic appears in numerical methods, curve fitting, and approximation theory.

One final habit helps throughout the chapter: after every decomposition or Vieta computation, state what has actually been proved. A partial fraction form proves equality of rational functions only where the original denominator is nonzero. A Vieta computation proves a relation among roots counted with multiplicity. Keeping these conditions visible prevents algebraic formulas from being used outside their valid domain.

Quick checks

Exercises

1. Write the correct partial fraction form for $5/(x^2+x-6)$.
2. Write the correct partial fraction form for $(2x^2+1)/(x^2(x^2+1)^2)$.
3. Resolve $x/((2x-1)(2x+1)(2x+3))$ into partial fractions.
4. Use the result of exercise 3 to evaluate $\sum_{k=1}^n k/((2k-1)(2k+1)(2k+3))$.
5. Let $\alpha_1,\alpha_2,\alpha_3$ be the roots of $x^3-33x^2+27x-3$. Compute $\alpha_1+\alpha_2+\alpha_3$, $\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3$, and $\alpha_1\alpha_2\alpha_3$.
6. Following Exercise 8.3 from the slides, suppose $\alpha_1,\alpha_2,\alpha_3$ are $\tan^2(\pi/9)$, $\tan^2(2\pi/9)$, and $\tan^2(4\pi/9)$, and they are the roots of $x^3-33x^2+27x-3$. Deduce $\tan(\pi/9)\tan(2\pi/9)\tan(4\pi/9)$.