SketchMath 數學團隊 SketchMath Maths Team ｜ 更新 Updated 2026-10-20
多項式長除法與數字相除相似：用首項相除、乘回、相減，再把下一項拉下。餘式定理：f(x) 除以 x − a 的餘式是 f(a)。因式定理：當且僅當 f(a) = 0，x − a 便是 f(x) 的因式。Divide polynomials like numbers: divide the leading terms, multiply back, subtract, then bring down. The remainder theorem says the remainder when f(x) is divided by x − a is f(a). The factor theorem: x − a is a factor exactly when f(a) = 0.
f(x) = (x − a) Q(x) + RR = f(a)ax − bDivisor ax − bR = f(b/a)f(a) = 0 ⟺ (x − a) is a factor把 x³ − 4x² + x + 6 除以 x − 2。Divide x³ − 4x² + x + 6 by x − 2.
x³ ÷ x = x²；x²(x − 2) = x³ − 2x²；相減：−2x² + xx³ ÷ x = x²; x²(x − 2) = x³ − 2x²; subtract: −2x² + x−2x² ÷ x = −2x；−2x(x − 2) = −2x² + 4x；相減：−3x + 6−2x² ÷ x = −2x; −2x(x − 2) = −2x² + 4x; subtract: −3x + 6−3(x − 2) = −3x + 6；餘式 0−3(x − 2) = −3x + 6; remainder 0答案： Answer: 商 x² − 2x − 3，餘式 0Quotient x² − 2x − 3, remainder 0
求 2x³ − 3x² + x − 5 除以 x + 1 的餘式。Find the remainder when 2x³ − 3x² + x − 5 is divided by x + 1.
x + 1 = x − (−1)，所以計算 f(−1)x + 1 = x − (−1), so evaluate f(−1)2(−1) − 3(1) + (−1) − 5 = −2 − 3 − 1 − 5答案： Answer: −11
x − 2 是 x³ + kx² − 7x + 2 的因式，求 k 並分解該多項式。x − 2 is a factor of x³ + kx² − 7x + 2. Find k and factorize the polynomial.
f(2) = 8 + 4k − 14 + 2 = 4k − 4 = 0, so k = 1x³ + x² − 7x + 2 除以 x − 2，得 x² + 3x − 1Divide x³ + x² − 7x + 2 by x − 2 to get x² + 3x − 1答案： Answer: k = 1; (x − 2)(x² + 3x − 1)
x³ − 7 寫成 x³ + 0x² + 0x − 7，使每一列對齊。Write x³ − 7 as x³ + 0x² + 0x − 7 before dividing, so every column lines up.x + 3 即 a = −3，餘式是 f(−3)。Dividing by x + 3 means a = −3, so the remainder is f(−3).溫習方法：點擊以下任何題目，即時在坐標網格載入並核對計算步驟。How to study: click any question to load it on the coordinate grid and check your steps.
SketchMath 題庫內的 DSE 風格練習，選擇題及長題目均附解答。 DSE-style practice from the SketchMath question bank, with solutions for every MC and long question.