Day 17 — Rational expressions

Day 17 — Rational expressions

Stage II · concept day
Goal: Find domains of rational expressions; simplify by canceling factors; multiply, divide, and add; simplify complex fractions; set up partial fractions for distinct linear factors (prep for recurrences later).

Why this matters

Rational expressions are ratios of polynomials—the algebraic form of transfer functions, probabilities as ratios of polynomials in generating functions, and many closed forms. Canceling incorrectly creates extraneous domains or loses holes. Partial fractions decompose for integration and for inverse Z-transforms / recurrence solving later.

Theory

Definition and domain

A rational expression is \(\dfrac{p(x)}{q(x)}\) where \(p,q\) are polynomials and \(q\not\equiv 0\).
Domain: all \(x\) such that \(q(x)\neq 0\) (over \(\mathbb{R}\)).

Even after simplifying, report excluded values from the original denominator.

Simplifying

Factor numerator and denominator; cancel common factors (not terms).
\[\frac{x^2-1}{x^2-x}=\frac{(x-1)(x+1)}{x(x-1)}=\frac{x+1}{x}\quad(x\neq 0,1).\] At \(x=1\) original undefined (hole if simplified expression would be defined).

Multiplication and division

\[ \frac{a}{b}\cdot\frac{c}{d}=\frac{ac}{bd},\qquad \frac{a}{b}\div\frac{c}{d}=\frac{a}{b}\cdot\frac{d}{c}, \] with all denominators nonzero. Factor first, cancel across.

Addition and subtraction

Common denominator: \[\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}.\] Prefer LCD of polynomial denominators (LCM of factorizations).

Complex fractions

Fractions within fractions: multiply top and bottom by LCD of all inner denominators, or rewrite as division.

Equations with rationals

Clear denominators by multiplying by LCD, then check solutions against excluded values (extraneous roots).

Partial fractions setup (distinct linear factors)

If \[R(x)=\frac{p(x)}{(x-r_1)(x-r_2)\cdots(x-r_n)}\] with \(\deg p<\deg\) denominator and distinct \(r_i\), seek \[R(x)=\frac{A_1}{x-r_1}+\cdots+\frac{A_n}{x-r_n}.\]

Cover-up / equate coefficients: multiply both sides by the full denominator: \[p(x)=A_1\prod_{j\neq 1}(x-r_j)+\cdots.\] Set \(x=r_i\) to isolate \(A_i\), or expand and compare coefficients.

Today: setup and solve small systems for \(A_i\); not full Heaviside theory.

Example form: \(\dfrac{3x+1}{(x-1)(x+2)}=\dfrac{A}{x-1}+\dfrac{B}{x+2}\).

If improper (\(\deg p\ge \deg q\)), first polynomial-divide, then decompose the proper remainder.

Holes vs vertical asymptotes (function view)

After canceling \((x-c)\) from \(\frac{(x-c)p_1(x)}{(x-c)q_1(x)}\) with \(q_1(c)\neq 0\), the simplified function \(\frac{p_1}{q_1}\) is defined at \(c\), but the original rational expression is not. Graphically this is a hole (removable discontinuity). If \(q_1(c)=0\) still, \(c\) remains a true singularity (often a vertical asymptote for the simplified form).

LCD construction algorithm

1. Factor every denominator completely over \(\mathbb{R}\) (linears/irreducible quadratics—today linears).
   
2. LCD = product of highest power of each distinct factor appearing.
   
3. Rewrite each fraction with the LCD; add numerators; factor/simplify; restore excluded zeros of original denominators.

Why partial fractions work (dimension count lite)

For distinct linear factors, the map \((A_1,\ldots,A_n)\mapsto \sum A_i\prod_{j\neq i}(x-r_j)\) produces all polynomials of degree \(<n\) as numerators when the product of \((x-r_i)\) is the denominator of degree \(n\). So a unique decomposition exists for every proper rational with those poles.

Worked examples

Example 1 — Domain

\(\dfrac{x+1}{x^2-4}\): domain \(x\neq \pm 2\).

Example 2 — Simplify

\(\dfrac{x^2-9}{x^2-x-6}=\dfrac{(x-3)(x+3)}{(x-3)(x+2)}=\dfrac{x+3}{x+2}\), \(x\neq 3,-2\).

Example 3 — Multiply

\(\dfrac{x^2-1}{x}\cdot\dfrac{2x}{x+1}=\dfrac{(x-1)(x+1)\cdot 2x}{x(x+1)}=2(x-1)\), \(x\neq 0,-1\).

Example 4 — Divide

\(\dfrac{x}{x-2}\div\dfrac{x^2}{x^2-4}=\dfrac{x}{x-2}\cdot\dfrac{(x-2)(x+2)}{x^2}=\dfrac{x+2}{x}\), \(x\neq 0,2,-2\).

Example 5 — Add

\(\dfrac{2}{x}+\dfrac{3}{x-1}=\dfrac{2(x-1)+3x}{x(x-1)}=\dfrac{5x-2}{x(x-1)}\).

Example 6 — Subtract

\(\dfrac{x}{x+1}-\dfrac{1}{x}=\dfrac{x^2-(x+1)}{x(x+1)}=\dfrac{x^2-x-1}{x(x+1)}\).

Example 7 — Complex

\(\dfrac{1+\dfrac{1}{x}}{1-\dfrac{1}{x}}=\dfrac{\frac{x+1}{x}}{\frac{x-1}{x}}=\dfrac{x+1}{x-1}\), \(x\neq 0,1\).

Example 8 — Equation

\(\dfrac{x}{x-1}=2\). Multiply by \(x-1\): \(x=2(x-1)\). \(x=2x-2\). \(x=2\). Check: \(x\neq 1\), OK. \(\frac{2}{1}=2\).

Example 9 — Extraneous

\(\dfrac{x}{x-1}=\dfrac{1}{x-1}\). Multiply: \(x=1\), but \(x\neq 1\) excluded. No solution. (Equivalently \(x-1=0\) impossible in domain.)

Example 10 — Partial fractions

\(\dfrac{3x+1}{(x-1)(x+2)}=\dfrac{A}{x-1}+\dfrac{B}{x+2}\).
\(3x+1=A(x+2)+B(x-1)\).
\(x=1\): \(4=A(3)\Rightarrow A=\frac{4}{3}\).
\(x=-2\): \(-5=B(-3)\Rightarrow B=\frac{5}{3}\).
Check: \(\frac{4/3}{x-1}+\frac{5/3}{x+2}=\frac{(4/3)(x+2)+(5/3)(x-1)}{(x-1)(x+2)}=\frac{\frac{4x+8+5x-5}{3}}{\cdots}=\frac{3x+1}{(x-1)(x+2)}\).

Example 11 — Three factors setup

\(\dfrac{1}{x(x-1)(x+1)}=\dfrac{A}{x}+\dfrac{B}{x-1}+\dfrac{C}{x+1}\).

Example 12 — Improper

\(\dfrac{x^2+1}{x-1}=x+1+\dfrac{2}{x-1}\) by division: \((x-1)(x+1)=x^2-1\), remainder \(2\).

Example 13 — LCD multi-factor

\(\dfrac{1}{x^2-1}+\dfrac{1}{x+1}=\dfrac{1}{(x-1)(x+1)}+\dfrac{1}{x+1}=\dfrac{1+(x-1)}{(x-1)(x+1)}=\dfrac{x}{(x-1)(x+1)}\).

Example 14 — Simplify nested

\(\dfrac{\dfrac{x}{y}-1}{\dfrac{x}{y}+1}=\dfrac{x-y}{x+y}\) (multiply by \(y/y\)).

Example 15 — Hole vs asymptote

\(\dfrac{x^2-1}{x^2-x}=\dfrac{(x-1)(x+1)}{x(x-1)}=\dfrac{x+1}{x}\) for \(x\neq 0,1\).
At \(x=1\): hole. At \(x=0\): still undefined after simplify (vertical asymptote of \(\frac{x+1}{x}\)).

Example 16 — Rational equation two candidates

\(\dfrac{x}{x-2}+\dfrac{2}{x+1}=1\). LCD \((x-2)(x+1)\).
\(x(x+1)+2(x-2)=(x-2)(x+1)\).
\(x^2+x+2x-4=x^2-x-2\).
\(3x-4=-x-2\). \(4x=2\). \(x=\frac12\).
Excluded: \(x\neq 2,-1\). Solution \(x=\frac12\) works.

Example 17 — Cover-up formula

For \(\dfrac{px+q}{(x-a)(x-b)}=\dfrac{A}{x-a}+\dfrac{B}{x-b}\),
\(A=\dfrac{pa+q}{a-b}\), \(B=\dfrac{pb+q}{b-a}\).
Apply: \(\dfrac{5x-1}{(x+1)(x-2)}\): \(A=\frac{5(-1)-1}{-1-2}=\frac{-6}{-3}=2\), \(B=\frac{10-1}{2-(-1)}=3\). So \(\frac{2}{x+1}+\frac{3}{x-2}\).

Example 18 — Difference for telescoping

\(\dfrac{1}{(k+1)(k+2)}=\dfrac{1}{k+1}-\dfrac{1}{k+2}\). Cover-up: \(A(k+2)+B(k+1)=1\); \(k=-1\): \(A=1\); \(k=-2\): \(B=-1\).

Example 19 — Domain of a sum after simplify

\(\dfrac{x}{x-1}-\dfrac{1}{x-1}=\dfrac{x-1}{x-1}=1\) for \(x\neq 1\) only—not the constant function \(1\) on all \(\mathbb{R}\).

Example 20 — LCD with powers

\(\dfrac{1}{x}+\dfrac{1}{x^2}=\dfrac{x+1}{x^2}\) (LCD \(x^2\); this is repeated linear factor—addition only today; full partial fractions for \(1/x^2\) later if needed).

Exercises

Easy

1. Domain of \(\dfrac{x-3}{x^2-9}\).
2. Simplify \(\dfrac{x^2-4}{x-2}\).
3. Multiply \(\dfrac{3}{x+1}\cdot\dfrac{x^2-1}{6}\).
4. Add \(\dfrac{1}{x}+\dfrac{1}{2}\).
5. \(\dfrac{x}{5}\div\dfrac{x}{10}\).

Medium

6. Simplify \(\dfrac{x^2+5x+6}{x^2-x-12}\) with domain notes.
7. Compute \(\dfrac{x}{x-1}-\dfrac{x}{x+1}\); simplify.
8. Simplify \(\dfrac{\frac{2}{x}-\frac{1}{x+1}}{\frac{3}{x}}\).
9. Solve \(\dfrac{2}{x}+\dfrac{1}{x-1}=1\); check domain.
10. Decompose \(\dfrac{5x-1}{(x+1)(x-2)}\) into partial fractions.
11. Decompose \(\dfrac{2x+3}{x^2-x}\).
12. Divide \(\dfrac{x^3+1}{x+1}\) as polynomials; write as polynomial + proper remainder if needed.
13. \(\dfrac{1}{x-1}-\dfrac{2}{x}+\dfrac{1}{x+1}\); combine carefully.

Hard

14. Prove that canceling a factor \((x-c)\) removes a potential root of the denominator and may introduce a removable discontinuity (hole) at \(x=c\).
15. Set up partial fractions for \(\dfrac{x^2+1}{(x-1)(x-2)(x-3)}\) (find \(A,B,C\)).
16. Explain why \(\dfrac{x-1}{x-1}=1\) is not an identity on all of \(\mathbb{R}\).
17. Solve \(\dfrac{x+1}{x-1}=\dfrac{x-1}{x+1}\); check.
18. If \(\dfrac{p(x)}{q(x)}\) is already simplified (\(\gcd(p,q)=1\)), show domain is exactly zeros of \(q\).
19. Complex fraction: \(\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{x}}}\) simplify for \(x\neq 0,-1,\ldots\).
20. Show addition formula using common denominator algebraically.

Challenge / CS-flavored

21. Probability rational: \(\dfrac{n(n-1)}{n^2}\) simplify; limit intuition as \(n\to\infty\) (no formal limits required).
22. Runtime ratio \(\dfrac{n^2+n}{2n^2}\); simplify and interpret for large \(n\).
23. Partial fractions prep: why \(\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}\) telescopes (Day 21).
24. Hash load \(\dfrac{n}{m}\); keep symbolic—when is expression undefined?
25. Extraneous roots in code: solving after multiplying by x-1 without checking x!=1.

Polynomial GCD (idea)

To reduce \(\frac{p}{q}\), cancel \(\gcd(p,q)\) in the polynomial ring (Euclidean algorithm with division). For hand work, factor both and cancel common polynomial factors.

Sign patterns and LCD

When denominators are \((x-1)(x+2)\) and \((x-1)\), LCD is \((x-1)(x+2)\). Keep factored form until the end when possible.

Partial fractions with three distinct lines

\(\frac{px^{2}+qx+r}{(x-a)(x-b)(x-c)}=\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}\) for proper numerator. Cover-up: \(A=\frac{p a^{2}+q a+r}{(a-b)(a-c)}\), etc.

Telescoping seed

\(\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}\) is partial fractions on integers—Day 21 uses this heavily.

Extra exercises

26. Simplify \(\frac{x^{3}-1}{x^{2}-1}\) with domain.
    
27. Add \(\frac{1}{x^{2}-1}+\frac{1}{x^{2}+2x+1}\).
    
28. Solve \(\frac{3}{x}+\frac{4}{x+1}=2\).
    
29. Decompose \(\frac{2x}{(x-1)(x-2)(x-3)}\).
    
30. Why is \(\frac{(x-1)^{2}}{x-1}\) not equal to \(x-1\) as functions on all \(\mathbb{R}\)?

CS connection

Simplifying rationals is like reducing fractions of big integers (Day 2) with polynomial gcd. Domain tracking is precondition checking. Partial fractions appear in summing series and solving linear recurrences. Removable discontinuities are “holes” analogous to optional values / undefined at a point.

Common pitfalls

Pitfall	What to do instead
Canceling terms \(+\)	Factor completely first
Losing excluded values after simplify	State original domain
Adding denominators	LCD / cross method
Extraneous solutions	Always check
Partial fractions without proper form	Divide first if improper
Forgetting factor multiplicity	Distinct linear only today; repeated needs higher powers

Fully worked extra examples

E15 — LCD with three denominators.
\(\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x-1}=\frac{(x+1)(x-1)+x(x-1)+x(x+1)}{x(x^{2}-1)}=\frac{x^{2}-1+x^{2}-x+x^{2}+x}{x(x^{2}-1)}=\frac{3x^{2}-1}{x(x^{2}-1)}\).

E16 — Extraneous root walkthrough.
\(\frac{2}{x-3}=\frac{x}{x-3}\). Multiply by \(x-3\): \(2=x\). But \(x=3\) is excluded and \(x=2\) works: \(\frac{2}{-1}=-2\) and \(\frac{2}{-1}=-2\). Wait check: for \(x=2\), both sides \(\frac{2}{-1}=-2\). Good. If equation were \(\frac{2}{x-3}=\frac{3}{x-3}\), then \(2=3\) never—empty.

E17 — Partial fractions three terms.
\(\frac{1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}\).
\(1=A(x^{2}-1)+Bx(x+1)+Cx(x-1)\).
\(x=0\): \(1=A(-1)\Rightarrow A=-1\).
\(x=1\): \(1=B(1)(2)\Rightarrow B=\frac12\).
\(x=-1\): \(1=C(-1)(-2)\Rightarrow C=\frac12\).
So \(-\frac{1}{x}+\frac{1/2}{x-1}+\frac{1/2}{x+1}\).

E18 — Improper then partial.
\(\frac{x^{2}+1}{x^{2}-1}=1+\frac{2}{x^{2}-1}=1+\frac{2}{(x-1)(x+1)}=1+\frac{1}{x-1}-\frac{1}{x+1}\) after partial fractions on \(\frac{2}{(x-1)(x+1)}\).

End-of-day synthesis problems

S1. Simplify \(\dfrac{x^{2}-5x+6}{x^{2}-9}\) with full domain notes from original.

S2. Compute \(\dfrac{x}{x+2}+\dfrac{3}{x-1}\) as a single fraction.

S3. Complex: \(\dfrac{1+\frac{2}{x}}{1-\frac{4}{x^{2}}}\) simplify.

S4. Solve \(\dfrac{x+1}{x-1}=\dfrac{2x}{x+1}\); check excluded values.

S5. Partial fractions \(\dfrac{4x}{(x-1)(x+3)}\).

S6. Polynomial divide \(\dfrac{x^{2}+0x+1}{x-1}\) to proper form.

S7. Why \(\dfrac{x-2}{x-2}=1\) fails at \(x=2\).

S8. Setup only: \(\dfrac{x^{2}+1}{(x-1)(x-2)(x-4)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-4}\).

Deep dive — Extraneous roots catalog

Move	Risk
Multiply by LCD that can be zero	Introduce candidates outside domain
Cancel factor \((x-c)\) in equation	May lose \(x=c\) if it was a solution of original (rare for rationals equal) or hide undefined
Square both sides (later abs)	Extraneous possible
Take reciprocal of both sides	Flip inequality direction; for equations need both nonzero and careful

Discipline: every time you multiply an equation by an expression, record “provided \(\ldots\neq 0\),” then check candidates against that list.

Synthesis

Rational expressions are polynomial ratios with domain hygiene. Arithmetic uses factor–cancel–LCD. Equations clear denominators then reject excluded values. Partial fractions reverse addition: they split a proper fraction into simple pole terms—seed for telescoping sums (Day 21) and recurrence closed forms later.

S9. Simplify \(\dfrac{x^3-8}{x^2-4}\) with full domain notes.

S10. Solve \(\dfrac{x+2}{x-3}=\dfrac{2x}{x+1}\); reject extraneous if any.

S11. Decompose \(\dfrac{x+5}{(x-1)(x+2)(x-3)}\) (find \(A,B,C\)).

S12. Explain hole vs asymptote for \(\dfrac{x^2-4}{x^2-5x+6}\).

Checkpoint

• State domain from denominator zeros
  
• Simplify, multiply, divide, add rational expressions
  
• Simplify complex fractions
  
• Solve a rational equation with extraneous check
  
• Set up and solve partial fractions for two distinct linear factors
  
• S1 and S5 exam-ready

Write two takeaways in your own words.

Quick reference card

Task	Rule
Domain	denominator \(\neq 0\) (original)
Simplify	factor; cancel factors only
Multiply	numerators \(\times\); dens \(\times\)
Divide	multiply by reciprocal
Add	LCD of dens
Partial fracs	proper; distinct linear dens → sum \(A_i/(x-r_i)\)
Equations	clear LCD; reject excluded roots

Tomorrow

Day 18 — Functions. Definition as rule/pairs; domain/range; vertical line test; piecewise; graph reading; even/odd; inverse existence idea.