Partial fractions with distinct linear factors
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| partial fraction/ˈpɑːʃl ˈfrækʃn/ | phân thức phân phần |
Can one fraction become a sum of simpler fractions?
- A product of two linear denominators can hide two simpler contributions. Separating them makes later integration and equation work easier.
- This lesson studies partial fraction 部分分式: A simpler fraction in a sum that equals a rational expression on its original domain.
The method
- For proper rational expressions, use A/(x+a)+B/(x+b) for distinct linear factors.
- Clear the denominator to obtain a polynomial identity, then substitute convenient roots or compare coefficients.
- Divide first if the numerator degree is not smaller.
After clearing denominators in 3x+5=A(x+2)+B(x+1), why may x=-1 be substituted despite the original fraction excluding it?
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Clearing denominators converts a rational equation into a polynomial identity holding for every input, so the convenient roots x=-1 and x=-2 give A=2 and B=1 instantly. The original fraction still excludes those inputs.
A worked case
For (3x+5)/[(x+1)(x+2)], write 3x+5=A(x+2)+B(x+1). At x=−1, A=2; at x=−2, B=1. Thus the decomposition is 2/(x+1)+1/(x+2), with x≠−1,−2. With three factors, (3x+1)/[x(x+1)(x+2)]=1/(2x)+2/(x+1)−5/[2(x+2)]. Clearing denominators and using x=0,−1,−2 gives A=1/2, B=2, C=−5/2.
Partial fractions with distinct linear factors
For proper rational expressions, use A/(x+a)+B/(x+b) for distinct linear factors
Classify the algebraic steps and identify the identity or domain condition behind each decision.
For (3x+5)/[(x+1)(x+2)], find A in A/(x+1)+B/(x+2).
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
At x=−1 in the cleared identity, 2=A.
Where it goes wrong
- The roots used after clearing denominators are allowed in the resulting polynomial identity, not in the original fraction.
- Retain all excluded inputs.
- A coefficient can be negative or zero.
Not true: Once denominators are cleared, their excluded inputs become valid inputs of the original fraction. Clearing denominators does not change the domain; inputs that make a denominator zero stay excluded.
Find B in the same decomposition.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
At x=−2 in the cleared identity, −1=−B, hence B=1.
Clearing denominators does not change the domain.
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
True. Clearing denominators does not change the domain; inputs that make a denominator zero stay excluded.
Applying it
- Check by recombining all terms and comparing coefficients, not just one numerical input.
- In 7357, numerators here are constant or linear and the decomposition has no more than three terms.
Find the coefficient of 1/x for (3x+1)/[x(x+1)(x+2)].
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
At x=0 in the cleared three-factor identity, 1=2A, hence A=1/2.
Match each partial-fraction step to its purpose.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Divide first if the fraction is improper; the decomposition equals the original on its whole original domain, exclusions included.
In your course
- AQA A-level Mathematics (7357), specification section B.
- Every paper allows a calculator, but method marks still need the expression or equation written before its value.
- Partial fraction: A simpler fraction in a sum that equals a rational expression on its original domain.
- For proper rational expressions, use A/(x+a)+B/(x+b) for distinct linear factors.
- Watch for: the roots used after clearing denominators are allowed in the resulting polynomial identity, not in the original fraction.