部分分数分解の計算機
この計算機は (x + 5)/(x² + x − 2) のような有理関数を部分分数 2/(x − 1) − 1/(x + 2) に分解します。異なる 1 次因数、重複する 1 次因数、既約な 2 次因数に対応し、仮分数のときは先に多項式の割り算を行い、まとめ直して答えを確かめます。
結果
値を入力して「計算」を押すと結果が表示されます。
公式
Distinct linear factors: one constant over each factor.
A repeated factor gets one term for each power.
An irreducible quadratic factor gets a linear numerator Bx + C.
この計算ツールの使い方
- Type the numerator and the denominator as polynomials, e.g. x + 5 and x^2 + x - 2.
- Press Calculate. The calculator divides first if the numerator’s degree is not smaller, factors the denominator, and solves for the unknown numerators.
- Read the decomposition and the steps; the check multiplies everything back together.
解き方付き例題
Easy: two distinct linear factors
Factor the denominator: x² + x − 2 = (x + 2)(x − 1).
Write (x + 5)/((x + 2)(x − 1)) = A/(x + 2) + B/(x − 1), so x + 5 = A(x − 1) + B(x + 2).
x = 1 gives 6 = 3B, so B = 2; x = −2 gives 3 = −3A, so A = −1.
Answer: −1/(x + 2) + 2/(x − 1).
Medium: a repeated factor
The denominator is (x + 1)², so use A/(x + 1) + B/(x + 1)².
3x + 5 = A(x + 1) + B. Comparing x terms: A = 3; constants: 3 + B = 5, so B = 2.
Answer: 3/(x + 1) + 2/(x + 1)².
Hard: a quadratic factor
Factor: x³ + x = x(x² + 1), and x² + 1 has no real roots, so it gets a numerator Bx + C.
2x² + 3 = A(x² + 1) + (Bx + C)x. x = 0 gives A = 3.
Then 2x² + 3 = (3 + B)x² + Cx + 3, so B = −1 and C = 0.
Answer: 3/x − x/(x² + 1).
Improper fraction: divide first
The numerator’s degree is not smaller, so divide: x³ = x(x² − 1) + x, giving x + x/(x² − 1).
x/((x + 1)(x − 1)) = A/(x + 1) + B/(x − 1): x = 1 gives B = 1/2 and x = −1 gives A = 1/2.
Answer: x + 1/(2(x + 1)) + 1/(2(x − 1)).
よくある間違い
- Skipping long division: if the numerator’s degree is not smaller than the denominator’s, divide first.
- Using only one term for a repeated factor: (x + 1)² needs both A/(x + 1) and B/(x + 1)².
- Giving a quadratic factor a constant numerator: it needs Bx + C.
- Not factoring the denominator completely before setting up the terms.
よくある質問
What is partial fraction decomposition?
Rewriting one rational function as a sum of simpler fractions whose denominators are the factors of the original denominator. It is used to integrate rational functions and in Laplace transforms.
How do I find the constants A and B?
Multiply both sides by the denominator, then either substitute the roots of each factor (the cover-up method) or compare coefficients. The calculator solves the coefficient equations exactly.
Which denominators can this calculator handle?
Any polynomial up to degree 6 that factors into linear factors with rational roots (repeated or not) and at most one irreducible quadratic factor. It says so when a denominator falls outside that.
How is the answer checked?
All the fractions are multiplied back over the common denominator, and the new numerator must equal yours exactly.