Repeated linear factors in partial fractions
| English | Français |
|---|---|
| repeated factor/rɪˈpiːtɪd ˈfæktə/ | facteur répété |
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.
Choose the mathematical structure
- 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.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Which description correctly defines repeated factor?
A denominator factor occurring more than once, requiring a term for each power.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
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.
At x=−1 in the identity, 1=B.
Test a tempting shortcut
- 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.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
For every denominator (x+1)²(x+2), one term over (x+1)² and one over x+2 are always enough. This claim is false. Explain which definition or assumption it violates.
Find C, the coefficient of 1/(x+2).
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 always enough.
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.
Interpret a new situation
- 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.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
Find A, the coefficient of 1/(x+1).
The x² coefficient gives 0=A+C; since C=−1, A=1.
Match each part of a complete solution to its purpose.
An assumption justifies the model; a check tests the result; interpretation connects it to the question.
Use this in your course
- 7357 · A-level · B. Match the target tier and specification before assigning extensions.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
A denominator factor occurring more than once, requiring a term for each power. Choose the relationship, show the method, check its assumptions and interpret the result.