π― Polynomials & Rational Functions
Polynomials and rational functions are central to AMC problems. Master factoring, Vieta’s formulas, and asymptote analysis.
π― Key Ideas
Polynomials: Sums of power terms $p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$
Rational Functions: Ratios of polynomials $r(x) = \frac{p(x)}{q(x)}$ where $q(x) \neq 0$
Vieta’s Formulas: Relate polynomial coefficients to sums and products of roots.
π’ Polynomial Basics
Degree and Leading Coefficient
- Degree: Highest power of $x$ with non-zero coefficient
- Leading Coefficient: Coefficient of highest degree term
- Constant Term: Coefficient of $x^0$ term
Example: $p(x) = 3x^4 - 2x^2 + 5x - 1$
- Degree: 4
- Leading coefficient: 3
- Constant term: -1
βοΈ Factoring Techniques
Common Patterns
- Difference of squares: $a^2 - b^2 = (a-b)(a+b)$
- Perfect squares: $a^2 \pm 2ab + b^2 = (a \pm b)^2$
- Sum/difference of cubes: $a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$
- Grouping: Factor common terms from groups
Example: Factor $x^3 - 8$
- Recognize as difference of cubes: $x^3 - 2^3$
- Apply formula: $(x-2)(x^2 + 2x + 4)$
Pitfall: Incomplete Factoring
Always check if further factoring is possible. Look for common factors first.
π― Vieta’s Formulas
For polynomial $ax^2 + bx + c = 0$ with roots $r_1$ and $r_2$:
- Sum of roots: $r_1 + r_2 = -\frac{b}{a}$
- Product of roots: $r_1 r_2 = \frac{c}{a}$
For cubic $ax^3 + bx^2 + cx + d = 0$ with roots $r_1, r_2, r_3$:
- Sum: $r_1 + r_2 + r_3 = -\frac{b}{a}$
- Sum of products: $r_1 r_2 + r_1 r_3 + r_2 r_3 = \frac{c}{a}$
- Product: $r_1 r_2 r_3 = -\frac{d}{a}$
Example: If $x^2 - 5x + 6 = 0$ has roots $r$ and $s$, find $r^2 + s^2$
Solution:
- From Vieta’s: $r + s = 5$ and $rs = 6$
- Use identity: $r^2 + s^2 = (r+s)^2 - 2rs$
- Substitute: $r^2 + s^2 = 5^2 - 2(6) = 25 - 12 = 13$
π Remainder and Factor Theorems
Remainder Theorem: When polynomial $p(x)$ is divided by $(x-a)$, the remainder is $p(a)$.
Factor Theorem: $(x-a)$ is a factor of $p(x)$ if and only if $p(a) = 0$.
Example: Find remainder when $x^3 - 2x^2 + 3x - 1$ is divided by $(x-2)$
Solution:
- By Remainder Theorem: remainder = $p(2) = 2^3 - 2(2^2) + 3(2) - 1 = 8 - 8 + 6 - 1 = 5$
π Rational Functions
Asymptotes
- Vertical asymptotes: Where denominator is zero (but numerator isn’t)
- Horizontal asymptotes: Compare degrees of numerator and denominator
- Oblique asymptotes: When degree of numerator = degree of denominator + 1
Asymptote Rules
| Numerator Degree | Denominator Degree | Horizontal Asymptote |
|---|---|---|
| $< n$ | $n$ | $y = 0$ |
| $= n$ | $n$ | $y = \frac{\text{leading coeff of num}}{\text{leading coeff of denom}}$ |
| $> n$ | $n$ | None (oblique asymptote) |
Example: Find asymptotes of $f(x) = \frac{x^2 - 1}{x^2 - 4}$
Solution:
- Vertical asymptotes: Set denominator = 0
- $x^2 - 4 = 0$ β $x = \pm 2$
- Horizontal asymptote: Degrees are equal (both 2)
- $y = \frac{1}{1} = 1$
π― AMC-Style Worked Example
Problem: If $x^3 - 6x^2 + 11x - 6 = 0$ has roots $a, b, c$, find $a^2 + b^2 + c^2$.
Solution:
Apply Vieta’s formulas:
- $a + b + c = 6$ (sum of roots)
- $ab + ac + bc = 11$ (sum of products)
- $abc = 6$ (product)
Use the identity: $a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab+ac+bc)$
Substitute: $a^2 + b^2 + c^2 = 6^2 - 2(11) = 36 - 22 = 14$
Answer: 14
π Common Traps & Fixes
Trap: Wrong Vieta’s formula signs
Fix: Remember the negative sign: sum of roots = $-\frac{b}{a}$
Trap: Confusing holes and asymptotes
Fix: Holes occur when both numerator and denominator have the same zero; asymptotes when only denominator is zero.
Trap: Forgetting to check domain restrictions
Fix: Always identify where rational functions are undefined.
Trap: Incomplete factoring
Fix: Check for common factors and use multiple techniques.
π Quick Reference
Essential Factoring Formulas
- $a^2 - b^2 = (a-b)(a+b)$
- $a^2 \pm 2ab + b^2 = (a \pm b)^2$
- $a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$
- $a^3 \pm 3a^2b + 3ab^2 \pm b^3 = (a \pm b)^3$
Vieta’s for Quadratic
- Sum: $r_1 + r_2 = -\frac{b}{a}$
- Product: $r_1 r_2 = \frac{c}{a}$
Common Identities
- $r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1 r_2$
- $r_1^3 + r_2^3 = (r_1 + r_2)^3 - 3r_1 r_2(r_1 + r_2)$
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