Partial fractions with distinct linear factors
| English | Español |
|---|---|
| partial fraction/ˈpɑːʃl ˈfrækʃn/ | partial fraction |
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.
Choose the mathematical structure
- 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.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Which description correctly defines partial fraction?
A simpler fraction in a sum that equals a rational expression on its original domain.
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 (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).
At x=−1 in the cleared identity, 2=A.
Test a tempting shortcut
- 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.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
Once denominators are cleared, their excluded inputs become valid inputs of the original fraction. This claim is false. Explain which definition or assumption it violates.
Find B in the same decomposition.
At x=−2 in the cleared identity, −1=−B, hence B=1.
Once denominators are cleared, their excluded inputs become valid inputs of the original fraction.
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.
Interpret a new situation
- 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.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
Find the coefficient of 1/x for (3x+1)/[x(x+1)(x+2)].
At x=0 in the cleared three-factor identity, 1=2A, hence A=1/2.
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 simpler fraction in a sum that equals a rational expression on its original domain. Choose the relationship, show the method, check its assumptions and interpret the result.