Repeated linear factors in partial fractions
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| repeated factor/rɪˈpiːtɪd ˈfæktə/ | nhân tử lặp lại |
Why does a squared factor need two terms?
- A squared factor needs two separate terms. Keeping only the squared term can leave too few coefficients to reproduce the numerator.
- This lesson studies repeated factor 重因式: A denominator factor occurring more than once, requiring a term for each power.
The method
- For a repeated factor (x+a)², include A/(x+a)+B/(x+a)².
- With a distinct factor x+b, add C/(x+b).
- Clear denominators, find convenient coefficients and use another input or coefficient comparison for the remaining unknown.
Why does (x+1)²(x+2) need BOTH A/(x+1) and B/(x+1)², not just one of them?
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
With only B/(x+1)²+C/(x+2), clearing denominators cannot match every coefficient. The full set A/(x+1)+B/(x+1)²+C/(x+2) makes 2x+3=A(x+1)(x+2)+B(x+2)+C(x+1)² solvable: B=1, C=-1, A=1.
A worked case
For (2x+3)/[(x+1)²(x+2)], the identity is 2x+3=A(x+1)(x+2)+B(x+2)+C(x+1)². At x=−1, B=1. At x=−2, C=−1. The x² coefficient gives A+C=0, hence A=1. The result is 1/(x+1)+1/(x+1)²−1/(x+2), with x≠−1,−2. At x=0 both sides are 3/2; recombination gives numerator 2x+3 for every allowed x.
Repeated linear factors in partial fractions
For a repeated factor (x+a)², include A/(x+a)+B/(x+a)²
Classify the algebraic steps and identify the identity or domain condition behind each decision.
Find B, the squared-factor coefficient in the worked decomposition.
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At x=−1 in the identity, 1=B.
Where it goes wrong
- Do not omit the first-power term or place Ax+B over each linear factor.
- A numerical agreement at one point is only a check, not an identity proof.
- Keep the square on its own denominator.
Not true: For every denominator (x+1)²(x+2), one term over (x+1)² and one over x+2 are enough. A repeated factor needs terms over both (x+1) and (x+1)², plus one over x+2.
Find C, the coefficient of 1/(x+2).
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
At x=−2, −1=C.
For every denominator (x+1)²(x+2), one term over (x+1)² and one over x+2 are enough.
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
A repeated factor needs terms over both (x+1) and (x+1)², plus one over x+2.
Applying it
- 7357 permits squared linear factors and at most three partial-fraction terms.
- Verify all numerator coefficients after recombination.
- Higher powers and irreducible quadratic factors are outside this lesson’s specification scope.
Find A, the coefficient of 1/(x+1).
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The x² coefficient gives 0=A+C; since C=−1, A=1.
Match each denominator factor to its partial-fraction terms.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Each power of a repeated factor gets its own constant numerator; a linear Ax+B belongs over an irreducible QUADRATIC, not here.
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.
- Repeated factor: A denominator factor occurring more than once, requiring a term for each power.
- For a repeated factor (x+a)², include A/(x+a)+B/(x+a)².
- Watch for: do not omit the first-power term or place Ax+B over each linear factor.