Partial Fraction¶
- To express a single rational fraction as a sum of two or more rational fractions is called Partial fraction.
- Expressing a rational function as a sum of partial fraction is called Partial Fraction Resolution.
- It plays an important part in the study of calculus
Terms Regarding Partial Fraction¶
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Conditional Equation: It is an equation is which two two algebraic expressions are equal for particular values of variables.
Example: \(2x=3\)
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Identity: An equation that is true for all the value of \(x\) is called identity.
Example: \((x+3)(x+4)=x^2+7x+12\)
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Rational Fraction: The quotient of two polynomials \(\frac{P(x)}{Q(x)}\), \(Q\rlap{/}{=}0\) with no common factors, is called a rational fraction. Rational fractions are of two types.
Example: $$ \frac{x^3 - x^2 + x+1}{x^2 + 5} $$
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Proper Fraction: The rational fraction \(\frac{P(x)}{Q(x)}\) is called proper if degree of numerator (\(P(x)\)) is less than degree of denominator (\(Q(x)\)).
Example: $$ \frac{2x-5}{x^2+4} $$
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Improper Fraction: The rational fraction \(\frac{P(x)}{q(x)}\) is called proper if degree of numerator (\(P(x)\)) is greater than or equal to degree of denominator (\(Q(x)\)).
Example: $$ \frac{x}{2x-3} $$
Partial Fraction Resolution.¶
Following are the main points of resolving a rational fraction \(\frac{P(x)}{Q(x)}\) into partial fraction:
- The degree of \(P(x)\) should be less than the degree of \(Q(x)\). If not, divide and work with reminder theorem.
- Clear the given equation of fractions.
- Equate the coefficients of like terms (power of \(x\))
- Solve the resulting equation for coefficients.
There are following cases of partial fraction resolution from a rational fraction $$ \frac{P(x)}{Q(x)} $$
Case 1¶
When \(Q(x)\) has only non-repeating linear factors
Example¶
Solution¶
This case can be solved easily by using cover up method:
- As we known the above equation will be written as: $$ \frac{x-7}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2} $$
- To find \(A\) cover \(x-1\) and put \(x=1\) on right side: $$ \frac{(1)-7}{(1)+2}=\frac{-6}{3}=-2 $$
- To find \(B\) cover \(x+2\) and put \(x=-2\) on right side: $$ \frac{(-2)-7}{(-2)-1}=\frac{-9}{-3}=3 $$
- So $$ \frac{x-7}{(x-1)(x+2)}=\frac{-2}{x-1}+\frac{3}{x+2} $$
Case 2¶
When \(Q(x)\) has repeating linear factors
Example¶
Solution¶
Checkout textbook solution.
Case 3¶
When \(Q(x)\) has non-repeating irreducible quadratic factors
Example¶
Solution¶
Checkout textbook solution.
Case 4¶
When \(Q(x)\) has repeating irreducible quadratic factors
Example¶
Solution¶
Checkout textbook solution.