8.3 Rational functions, partial fractions, and Vieta formulas

Recovering simpler pieces from a polynomial expression

Adding fractions combines their denominators; partial fractions reverse that operation. A factored denominator suggests the pieces, but repeated factors and irreducible quadratics impose different numerator shapes. Polynomial division removes any polynomial part, and a Bezout identity makes the split between coprime factors possible.

A factored polynomial also carries information about its roots without giving their individual values. In $x^3-33x^2+27x-3$, for example, the coefficients determine the sum and product of the three roots. Vieta's formulas express this relation by comparing coefficients with a factorization over $C$.

Rational functions and the polynomial part

Definition

Rational function

A rational function is a quotient

$$\frac{p(x)}{q(x)}$$

where $p(x),q(x)\in R[x]$ and $q(x)\ne0$.

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.

Here “nonzero denominator” means that the denominator is not the zero polynomial. It can still vanish at particular real inputs; those inputs are excluded from the domain of the displayed quotient. Clearing denominators produces a polynomial identity, but evaluating the original fractions still requires nonzero denominators. If a common factor is cancelled, remember any excluded input when interpreting the original expression as a function.

A fraction is proper when its numerator has smaller degree than its denominator; otherwise it is improper. The zero remainder is allowed, using the convention that its degree is negative infinity. A constant nonzero denominator leaves only a polynomial part. When the original fraction is already proper, the quotient in division is zero, so no preliminary arithmetic is needed. These cases belong to the same division theorem, rather than requiring different decomposition rules.

The polynomial part is unique because the quotient and remainder are unique. In particular, a nonzero polynomial cannot be hidden inside a sum of proper fractions: after putting that sum over a common denominator, its numerator still has smaller degree. This degree comparison will also close the existence proof below.

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.

Proof: Where existence and uniqueness use different hypotheses

The preliminary Bezout numerators need not have the required degrees. Write

$$rb=c_1q_1+r_1,\qquad ra=c_2q_2+r_2, \qquad \deg r_i\lt\deg q_i.$$

Each numerator is reduced modulo its corresponding denominator. Substituting and multiplying by the common denominator gives

$$r=(c_1+c_2)q_1q_2+r_1q_2+r_2q_1.$$

Both the original remainder and the last two terms have degree strictly below the degree of the product denominator. Thus, if the polynomial sum of the two quotients were nonzero, its product with that denominator would have degree at least that of the denominator, an impossibility. The quotient contributions therefore cancel exactly. This proves existence with the required degree bounds.

For uniqueness, suppose another pair of reduced numerators gives the same fraction. Subtraction gives the polynomial identity

$$(r_1-\widetilde r_1)q_2 =-(r_2-\widetilde r_2)q_1.$$

Consequently the first denominator divides the product on the left. Since the two denominator factors are relatively prime, Euclid's divisibility lemma shows that the first denominator divides the difference of the first numerators. That difference has smaller degree, so it must be zero. The identity then forces the other difference to be zero as well. Coprimality is used to remove the unwanted factor; the degree bound is used to turn divisibility into equality. Neither step can be omitted.

For several pairwise relatively prime factors, repeat this splitting argument. Group all occurrences of the same irreducible factor into a single power before splitting: distinct irreducible factors are coprime, and their powers remain coprime. Two copies of the same linear factor are not coprime, so the same argument cannot separate them directly.

Why repeated factors require more terms

Counterexample mode

A missing term can make the coefficient system impossible

A proposed shortcut for a repeated denominator is to keep only the highest power with a constant numerator. Test it on the rational function $(x^3+1)/(x-2)^4$. If this were equal to $D/(x-2)^4$, clearing denominators would say that the nonconstant polynomial $x^3+1$ equals a constant. This is impossible. The failure is in the chosen form, so solving more carefully for the same single unknown cannot repair it.

To find the required terms, set $y=x-2$, so $x=y+2$. The numerator becomes

$$x^3+1=(y+2)^3+1=y^3+6y^2+12y+9.$$

Division by the fourth power of the shifted variable gives four terms:

$$\frac{x^3+1}{(x-2)^4} =\frac1{x-2}+\frac6{(x-2)^2} +\frac{12}{(x-2)^3}+\frac9{(x-2)^4}.$$

This also illustrates why the coefficients above different powers have separate roles. Multiplying back by the common denominator recovers distinct powers of the shifted variable; suppressing one layer suppresses an available coefficient. All equalities of the fractions retain the exclusion $x=2$.

A different incomplete template is a constant numerator above every quadratic. The proper fraction $x/(x^2+1)$ cannot equal $B/(x^2+1)$, since clearing the nonvanishing denominator would require $x=B$ identically. The general form must allow a numerator of degree at most one. A coefficient may turn out to be zero after solving, but it must not be forced to be zero before the identity has been checked.

Theorem

Partial fraction shape over R

If the denominator factors as

$$K\prod_i(x-a_i)^{m_i}\prod_j(x^2+b_jx+c_j)^{n_j},$$

where $K\ne0$, the multiplicities $m_i,n_j$ are positive integers, the linear factors are distinct, and the monic quadratic factors are distinct and irreducible over $R$, then every rational function can be written uniquely as a polynomial plus terms of the forms

$$\frac{k}{(x-a)^m}, \qquad \frac{rx+s}{(x^2+bx+c)^n}.$$

After coprime splitting, the grouped numerators satisfy $\deg A\lt m$ and $\deg B\lt2n$ in the expansions below. 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:

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

For a real monic quadratic, irreducibility means that its discriminant is negative. A quadratic with two real roots must first be split into linear factors; a quadratic with a repeated real root belongs to the repeated-linear case. The constant factor in the denominator can be kept as an overall factor of its reciprocal or absorbed into the unknown coefficients, but the two conventions must not be mixed.

The numerator bound refers to the irreducible base factor, not to the full power in an individual final term. Over a linear factor, “degree less than one” means a constant. Over an irreducible quadratic, “degree less than two” means at most linear, so a constant or zero numerator is also allowed. Saying “linear numerator” describes the general template, not a requirement that its leading coefficient be nonzero.

To see why all powers occur, put $q=x^2+bx+c$ and divide $B=q\widetilde B+(rx+s)$. Then

$$\frac{B}{q^n}=\frac{rx+s}{q^n}+\frac{\widetilde B}{q^{n-1}}.$$

Repeat on the remaining numerator. Every division has a unique quotient and remainder; the degree drops sufficiently to stop at the first power. Division by a linear base works identically, with constant remainders. Thus the list of powers records repeated division, and its uniqueness comes from that same algorithm.

Worked example

Set up a partial fraction decomposition

Resolve

$$\frac{x^4+2x+4}{2x^3-4x^2+3x-6}$$

into partial fractions.

First divide:

$$\frac{x^4+2x+4}{2x^3-4x^2+3x-6} =\frac12x+1+ \frac{\frac52x^2+2x+10}{(2x^2+3)(x-2)}.$$

Then write

$$\frac{\frac52x^2+2x+10}{(2x^2+3)(x-2)} =\frac{Ax+B}{2x^2+3}+\frac{C}{x-2}.$$

Clear denominators before doing coefficient comparison:

$$\tfrac52x^2+2x+10=(Ax+B)(x-2)+C(2x^2+3).$$

Equating the quadratic, linear, and constant coefficients gives

$$A+2C=\tfrac52,\qquad -2A+B=2,\qquad -2B+3C=10.$$

Although substitution into the original fraction at the pole is forbidden, substitution into this polynomial identity is valid. Setting $x=2$ gives $24=11C$, and the first two coefficient equations then determine the remaining unknowns. Comparing coefficients gives

$$A=-\frac{41}{22},\qquad B=-\frac{19}{11},\qquad C=\frac{24}{11}.$$

Thus

$$\frac{x^4+2x+4}{2x^3-4x^2+3x-6} =\frac12x+1-\frac{41x+38}{22(2x^2+3)} +\frac{24}{11(x-2)}.$$

Worked example

Read a partial-fraction form before solving coefficients

Write the correct partial-fraction form for

$$\frac{2x^2+1}{x^2(x^2+1)^2}$$

over $R$.

The factor $x^2$ is a repeated linear factor, so it contributes

$$\frac{A}{x}+\frac{B}{x^2}.$$

The factor $x^2+1$ is an irreducible quadratic over $R$, and it is repeated to power $2$, so it contributes

$$\frac{Cx+D}{x^2+1}+\frac{Ex+F}{(x^2+1)^2}.$$

Thus the full form is

$$\frac{A}{x}+\frac{B}{x^2} +\frac{Cx+D}{x^2+1} +\frac{Ex+F}{(x^2+1)^2}.$$

Every denominator power is now present, with its permitted numerator degree. The original expression is even. Replacing the variable by its negative and using uniqueness forces the coefficients of all the odd terms to vanish, so $A=C=E=0$. To determine the other three coefficients, clear the common denominator:

$$2x^2+1=B(x^2+1)^2+Dx^2(x^2+1)+Fx^2.$$

The constant coefficient gives $B=1$. The fourth-degree coefficient gives $B+D=0$, so $D=-1$. Finally, the quadratic coefficient gives $2B+D+F=2$, so $F=1$. Thus the identity simplifies to

$$\frac{2x^2+1}{x^2(x^2+1)^2} =\frac1{x^2}-\frac1{x^2+1}+\frac1{(x^2+1)^2}.$$

This verification distinguishes a zero coefficient, justified by the completed identity, from an omitted term in the initial general template. The only real excluded input is zero, since the quadratic factor is positive everywhere.

Telescoping from partial fractions

Partial fractions often convert a complicated finite sum into a telescoping one. Consider

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

The coefficients can be checked by clearing the three denominators and substituting their roots into the resulting polynomial identity. At the three roots, respectively, this gives $1/2=8A$, $-1/2=-4B$, and $-3/2=8C$. The resulting identity holds whenever the original denominator is nonzero. All positive integer inputs meet this condition.

For complete boundary accounting, set $u_k=1/(2k-1)$ and denote the sum by $T_n$. Substituting the decomposition gives

$$16T_n=\sum_{k=1}^n(u_k+2u_{k+1}-3u_{k+2}).$$

Rewrite each summand as $(u_k-u_{k+1})+3(u_{k+1}-u_{k+2})$. Each of these two sums telescopes separately, so for every positive integer $n$, including $n=1$,

$$16T_n=(u_1-u_{n+1})+3(u_2-u_{n+2}) =2-\frac1{2n+1}-\frac3{2n+3}.$$

This arrangement avoids assuming that an interior range is nonempty. In the expanded sum, an interior odd denominator receives weights one, two, and negative three, whose sum is zero; the two initial and two final boundary contributions survive. Returning to the original sum gives

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

The cancellation comes from aligning the shifted denominators; both initial and final boundary terms must be retained.

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. To obtain the coefficient of $x^{n-j}$, choose a root term from exactly $j$ factors and choose $x$ from the others. Each selection contributes one product with sign $(-1)^j$. Increasing indices count each selection once. The argument counts factor positions, so repeated root values do not invalidate it; they must still appear as often as their multiplicities require.

Theorem

Vieta's formulas

For $j=1,2,\ldots,n$,

$$\sum_{1\le i_1\lt\cdots\lt i_j\le n} \alpha_{i_1}\alpha_{i_2}\cdots\alpha_{i_j} =(-1)^j\frac{a_{n-j}}{a_n}.$$

For a cubic

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

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

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

Worked example

Power sums for cubic roots

Let the roots of

$$a_0+a_1x+a_2x^2+a_3x^3=0$$

be $\alpha_1,\alpha_2,\alpha_3$, and write $S_r=\alpha_1^r+\alpha_2^r+\alpha_3^r$.

Then

$$S_2=(\alpha_1+\alpha_2+\alpha_3)^2 -2(\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3) =\frac{a_2^2-2a_1a_3}{a_3^2}.$$

For the cubic sum, let $e_1,e_2,e_3$ denote the three elementary symmetric sums just displayed. Expanding the cube of the first sum counts each mixed term with two equal indices three times, and the product of all three roots six times. Meanwhile, the product of the first and second elementary sums contains each of those mixed terms once and the three-root product three times. Eliminating the mixed terms yields

$$S_3=e_1^3-3e_1e_2+3e_3.$$

The plus sign on the last term compensates for over-subtraction. Substituting Vieta, including the negative sign in the root product, gives

$$S_3= \frac{-a_2^3+3a_1a_2a_3-3a_0a_3^2}{a_3^3}.$$

The useful habit is to rewrite nonsymmetric-looking expressions in terms of symmetric sums that Vieta controls.

There is also a useful independent check using the defining cubic. Every root satisfies the polynomial equation, so summing the three equations gives

$$a_3S_3+a_2S_2+a_1S_1+3a_0=0.$$

The constant term occurs three times because there are three roots counted with multiplicity. Insert the already computed first and second power sums and solve for the third; the same expression follows. This method never requires the roots to be distinct, real, or explicitly known. It checks both the denominator power and the signs in the expansion method.

For higher sums, multiply each root equation by the same nonnegative integer power of that root before summing. The resulting relation is

$$a_3S_{r+3}+a_2S_{r+2}+a_1S_{r+1}+a_0S_r=0, \qquad r\geq0,\quad S_0=3.$$

The initial value counts roots, including any zero root, so its definition is a count rather than an instruction to evaluate an ambiguous zero power. This relation explains why a small set of symmetric data can control later power sums. In the capstone below, only the first three sums are needed.

A trigonometric application of Vieta

Worked example

Use Vieta in the tangent capstone

To connect a trigonometric equation with its roots, first derive the polynomial. Let $t=\tan\theta$, with $\cos\theta\ne0$ and $\cos9\theta\ne0$. De Moivre's theorem gives

$$(1+it)^9=\frac{\cos9\theta+i\sin9\theta}{\cos^9\theta}.$$

Separate the even and odd powers in the binomial expansion. Their alternating signs come from successive powers of the imaginary unit:

$$N(t)=9t-84t^3+126t^5-36t^7+t^9,$$$$D(t)=1-36t^2+126t^4-84t^6+9t^8.$$

These are the imaginary and real parts, respectively. Factoring them, which can be checked by multiplication, gives

$$N(t)=t(t^2-3)(t^6-33t^4+27t^2-3),$$$$D(t)=(3t^2-1)(3t^6-27t^4+33t^2-1).$$

Taking the imaginary part divided by the real part therefore proves

$$\tan9\theta= \frac{t(t^2-3)(t^6-33t^4+27t^2-3)} {(3t^2-1)(3t^6-27t^4+33t^2-1)}.$$

The condition on the ninth-angle cosine is exactly what makes this denominator nonzero: the real-part identity says $D(t)=\cos9\theta/\cos^9\theta$. We have not extended the tangent formula through its poles by algebraic simplification.

Now take the angles $\pi/9,2\pi/9,4\pi/9$. They lie strictly between zero and $\pi/2$, and their ninth-angle cosines are respectively negative one, one, and one. The quotient is therefore defined, and its numerator vanishes because the corresponding ninth-angle sines vanish. Each tangent is positive, so the factor $t$ is nonzero. None of these angles equals $\pi/3$; strict monotonicity of tangent on this interval implies that none has tangent $\sqrt3$. Thus the factor $t^2-3$ is nonzero too. Only after checking these facts may we conclude that the degree-six factor vanishes.

Putting $x=t^2$ now reduces the three numbers

$$\tan^2(\pi/9),\quad \tan^2(2\pi/9),\quad \tan^2(4\pi/9)$$

to roots of $x^3-33x^2+27x-3=0$. The three positive tangents are distinct, so their squares are distinct too. A cubic has at most three roots; we have therefore identified its complete root list. Let these roots be $\alpha_1,\alpha_2,\alpha_3$. Vieta's formulas give

$$\alpha_1+\alpha_2+\alpha_3=33,\qquad \alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3=27,$$

and

$$\alpha_1\alpha_2\alpha_3=3.$$

Since the three tangent values are positive, their product is the positive square root of $3$:

$$\tan(\pi/9)\tan(2\pi/9)\tan(4\pi/9)=\sqrt3.$$

For the sixth-power sum, rewrite $\alpha_1^3+\alpha_2^3+\alpha_3^3$ as

$$(\alpha_1+\alpha_2+\alpha_3)^3 -3(\alpha_1+\alpha_2+\alpha_3)(\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3) +3\alpha_1\alpha_2\alpha_3.$$

Substitution gives $33^3-3(33)(27)+3(3)=33273$.

Optional interpolation viewpoint

Lagrange interpolation constructs a polynomial from prescribed values. 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.

To prescribe one value at a time, seek a polynomial that equals $1$ at $x_i$ and $0$ at every other node. The factors $x-x_j$ with $j\ne i$ give the required zeros; dividing by their values at $x_i$ normalizes the value there. Thus, 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.

Concept lensStructural

Point values replace coefficient data

The distinct-node condition makes every denominator in the basis polynomials nonzero. At its own node a basis polynomial equals one, because every factor is one; at another node one numerator factor vanishes. Thus each basis polynomial changes one prescribed value while contributing zero at the other nodes. One basis polynomial is assigned to each of the $n+1$ nodes.

This proves existence constructively. The difference argument proves uniqueness independently of that formula. The degree restriction is essential: adding any multiple of the product of all the node factors preserves every prescribed value, but generally raises the degree. Repeated nodes would require a different problem statement; the displayed formula is not defined there.

Common mistakes

Common mistake

Skipping denominator powers

For a denominator containing $(x-2)^4$, the general partial-fraction form includes a constant numerator over each power from $1$ through $4$. Some coefficients may be zero, but that must follow from the decomposition.

Common mistake

Using constant numerators over irreducible quadratics

Over $R$, a term over $x^2+1$ needs a numerator of degree less than $2$, so the general numerator is $Ax+B$, not just a constant.

Common mistake

Applying Vieta to non-symmetric expressions directly

Vieta gives sums of products of roots, not arbitrary individual roots. Before using it on a power sum or trigonometric product, rewrite the expression in terms of symmetric sums.

Summary

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.

Quick checks

Checkpoint

Why do we first divide $p(x)$ by $q(x)$ before doing partial fractions?

Check the degree requirement on the remaining fraction.

Solution · Answer

Partial fraction shapes are for proper rational functions. Division separates the polynomial part and leaves a remainder $r(x)$ with $\deg r\lt\deg q$.

Checkpoint

What numerator shape belongs above an irreducible quadratic factor such as $x^2+1$?

Use the degree rule for the numerator.

Solution · Answer

The numerator has degree at most one, so it has the form $Ax+B$; $A=0$ is allowed.

Checkpoint

For $x^3-33x^2+27x-3$, what is the product of the three roots?

Use the cubic Vieta formula.

Solution · Answer

For a monic cubic, the product is $-a_0$. Here $a_0=-3$, so the product is $3$.

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. Use root-coefficient relations to evaluate a trigonometric product. 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 $\tan(\pi/9)\tan(2\pi/9)\tan(4\pi/9)$.

Solution · Model solution 1

Since $x^2+x-6=(x+3)(x-2)$, the form is $A/(x+3)+B/(x-2)$.

Solution · Model solution 2

The form is $A/x+B/x^2+(Cx+D)/(x^2+1)+(Ex+F)/(x^2+1)^2$.

Solution · Model solution 3

The decomposition is $1/[16(2x-1)]+1/[8(2x+1)]-3/[16(2x+3)]$.

Solution · Model solution 4

Substitute $x=k$, sum, and align shifted odd denominators. The result is $1/16(2-1/(2n+1)-3/(2n+3))$.

Solution · Model solution 5

Vieta gives $33$, $27$, and $3$, respectively.

Solution · Model solution 6

The product of the three squared tangent values is $3$. Since each tangent is positive for the angles shown, the product of the tangents is $\sqrt3$.