🧮 Polynomial Theory — Remainder, Factor & Vieta
Advanced polynomial techniques essential for AMC12 and useful for AMC10.
🎯 Key Ideas
Remainder Theorem — If polynomial $f(x)$ is divided by $(x-a)$, the remainder is $f(a)$. This provides a quick way to evaluate remainders without long division.
Factor Theorem — If $f(a) = 0$, then $(x-a)$ is a factor of $f(x)$. Conversely, if $(x-a)$ is a factor, then $f(a) = 0$.
Vieta’s Formulas — Relationships between coefficients and roots: for $ax^2 + bx + c = 0$, sum of roots is $-\frac{b}{a}$ and product is $\frac{c}{a}$.
📊 Essential Formulas
Remainder & Factor Theorems
| Theorem | Statement | Example |
|---|---|---|
| Remainder | Remainder when $f(x) \div (x-a)$ is $f(a)$ | $f(x) = x^3 + 2x^2 - 5x + 1$, remainder when divided by $(x-2)$ is $f(2) = 8 + 8 - 10 + 1 = 7$ |
| Factor | $(x-a)$ is a factor of $f(x)$ if and only if $f(a) = 0$ | If $f(3) = 0$, then $(x-3)$ divides $f(x)$ |
Vieta’s Formulas
| Degree | Sum of Roots | Product of Roots | Other Relations |
|---|---|---|---|
| Quadratic $ax^2 + bx + c$ | $r_1 + r_2 = -\frac{b}{a}$ | $r_1 \cdot r_2 = \frac{c}{a}$ | — |
| Cubic $ax^3 + bx^2 + cx + d$ | $r_1 + r_2 + r_3 = -\frac{b}{a}$ | $r_1 \cdot r_2 \cdot r_3 = -\frac{d}{a}$ | $r_1r_2 + r_1r_3 + r_2r_3 = \frac{c}{a}$ |
🎯 Micro-Examples
Example 1: Find remainder when $x^3 - 2x^2 + 3x - 1$ is divided by $(x-2)$
- Method: Use remainder theorem: $f(2) = 8 - 8 + 6 - 1 = 5$
- Answer: Remainder is $5$
Example 2: Check if $(x+1)$ is a factor of $x^3 + 3x^2 + 3x + 1$
- Method: Use factor theorem: $f(-1) = -1 + 3 - 3 + 1 = 0$
- Answer: Yes, $(x+1)$ is a factor
Example 3: For $x^2 - 5x + 6 = 0$, find sum and product of roots
- Vieta’s: Sum = $-\frac{-5}{1} = 5$, Product = $\frac{6}{1} = 6$
- Check: Roots are $x = 2, 3$, so sum = $5$, product = $6$ ✓
⚠️ Common Traps & Fixes
Trap: Confusing remainder and factor theorems
- Fix: Remainder theorem finds remainder; factor theorem checks if remainder is zero
- Example: $f(2) = 5$ means remainder is $5$, not that $(x-2)$ is a factor
Trap: Sign errors in Vieta’s formulas
- Fix: Remember the negative sign: sum = $-\frac{b}{a}$
- Example: For $x^2 + 3x - 4 = 0$, sum = $-\frac{3}{1} = -3$, not $3$
Trap: Forgetting to divide by leading coefficient
- Fix: Vieta’s formulas use $\frac{b}{a}$, not just $b$
- Example: For $2x^2 - 6x + 4 = 0$, sum = $-\frac{-6}{2} = 3$, not $6$
🎯 AMC-Style Worked Example
Problem: If $r$ and $s$ are the roots of $x^2 - 3x + 1 = 0$, find $r^2 + s^2$.
Solution:
- Use Vieta’s: $r + s = 3$ and $rs = 1$
- Key identity: $r^2 + s^2 = (r+s)^2 - 2rs$
- Substitute: $r^2 + s^2 = 3^2 - 2(1) = 9 - 2 = 7$
- Answer: $r^2 + s^2 = 7$
Key insight: Vieta’s formulas often require algebraic manipulation to find desired expressions.
🔗 Related Topics
- Quadratics — Vieta’s formulas are most commonly used with quadratics
- Systems of Equations — Vieta’s can create systems to solve
- Sequences — Polynomial roots can define sequences
- Complex Numbers — Vieta’s works with complex roots too
📝 Practice Checklist
- Master remainder and factor theorems
- Practice Vieta’s formulas for quadratics
- Learn Vieta’s for cubics (AMC12)
- Practice algebraic manipulations with Vieta’s
- Understand when to use each technique
- Work on speed and accuracy
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