第 11 週 · 多項式與餘式定理 · 第 2 節 餘式定理 上節長除法能求餘數，但慢。這節給你一招「秒殺」捷徑 —— 除以 (x − a) 的餘數，就等於把 x = a 代進去算。代一個數，餘數立刻知道。 ▶ 第一關： 餘式定理長什麼樣（找 a + 代入） Remainder TheoremP(x) ÷ (x − a)remainder = ?put x = a= P(a)no division needed — just substitute one number how to find the number to plug inx − a = 0set divisor = 0→x = asolve for x→P(a)= remainder worked example: P(x) = x² + 3x + 5 ÷ (x + 1)x + 1 = 0→x = −1P(−1) = (−1)² + 3(−1) + 5= 1 − 3 + 5=3← remaindernote: (−1)² = +1 (negative squared is positive) ⚠ 請啟用 JavaScript 以參與互動題目。 ▶ 第二關： 為什麼成立（三行推導）+ 對比長除法 why it works: put x = aP(x) = (x − a)·Q(x) + Rdivision identity (R is a constant)↓ put x = aP(a) = (a − a)·Q(a) + R(a − a) = 0 → this whole term dies→P(a) = R same problem: (x³ + 2x² − 5x + 1) ÷ (x − 1)long division (slow)sort → divide→ multiply → subtract→ bring down → repeat…quotient x² + 3x − 2remainder = −1remainder theorem (fast)put x = 11 + 2 − 5 + 1= −1same answer — one step beats a whole page ⚠ 請啟用 JavaScript 以參與互動題目。 ▶ 第三關： 兩個坑 —— (ax − b) 型與負號 (ax − b) type: solve the equation firstdivide by (2x − 1)2x − 1 = 0 → 2x = 1x =1/2plug in 1/2NOT 1x has a coefficient the sign trap: (x + 2) → plug in −2(x + 2)→x + 2 = 0x = −2sign is + herebut you plug in − (flip the sign!) ⚠ 請啟用 JavaScript 以參與互動題目。 ⚔ BOSS： 求係數 + 因式預告 find the coefficient: turn the remainder into an equationP(x) = x² + kx + 1, divide by (x − 1), remainder 5P(1) = 1 + k + 1 = 5→k + 2 = 5k =3check: P(1) = 1 + 3 + 1 = 5 ✓ ⚠ 請啟用 JavaScript 以參與互動題目。 CLEARED · 全部過關 🎉 W11D2 通關！ 下一節：W11D3 · 因式定理（餘數 = 0 就是因式）+ 分解三次式。