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第 11 週 · 多項式與餘式定理 · 第 6 節

綜合：用餘式／因式定理求系數 + 本週關卡

本週五件工具合起來用 — 給一個「餘數」或「因式」條件，倒推未知系數 a、b，再分解、判根。這是 DSE 多項式題的標準套路，最後一關打通全週。

▶ 第一關： 給「餘數／因式」條件 → 列方程求系數
given conditionremainder / factor① set equationP(a)=r or P(a)=0② solvefind a, b③ factorise / find roots / claim(x-1)(x+2)(x-3), roots 1,-2,3
condition: P(x) divided by (x-1) leaves remainder 5remainder = 5→P(1) = 5wordsan equation in the unknownremainder theorem: remainder = P(a)
condition: (x+2) is a factor of P(x)(x+2) is a factor→P(-2) = 0sign flips: +2 in bracket → use x = -2factor theoremfactor ⇔ P(a) = 0 (always equals zero)
▶ 第二關： 兩個條件 → 聯立解 a、b
two factors → two equations → eliminate(x-1) factor → P(1)=0(x+2) factor → P(-2)=0a + b = -72a - b = 1add → 3a = -6a = -2, b = -5so P(x) = x^3 - 2x^2 - 5x + 6
▶ 第三關： 解出系數後 → 分解、判根、claim 題
P(x) = (x-1)(x+2)(x-3)(x - 1)(x + 2)(x - 3)x = 1x = -2x = 3
claim: are the roots of x^2 - 3x + 2 = 0 integers?① factorise(x-1)(x-2)→② roots1 and 2→③ concludeTRUE + reasonboth 1 and 2 are integers → claim is TRUEalways write the reason, not just yes / no
🗺 本週知識地圖回顧
polynomial P(x)divide by (x - a)?want quotient + remainder?LONG DIVISIONonly want the remainder?REMAINDER THM: P(a)P(a) = 0 ?FACTOR THM → (x-a) factorfactor each → take commonHCF / LCMfactor top & bottom → cancelALGEBRAIC FRACTION
⚔ BOSS： 本週總關卡（長除／餘式／因式／HCF·LCM／分式）
this BOSS round covers all 5 tools of week 11long division+ remainder thmremainder thmP(1) = 4factor thmP(a) = 0HCF / LCMHCF = (x+1)algebraic fraction(x-2)/(x+3)5 tools → 1 final round → week 11 cleared