Partial Fractions are the fractions that are formed when a complex rational expression is split into two or more simpler fractions. Generally, fractions with algebraic expressions are difficult to solve and hence we use the concepts of partial fractions to split the fractions into numerous subfractions. While decomposition, generally, the denominator is an algebraic expression, and this expression is factorized to facilitate the process of generating partial fractions. A partial fraction is a reverse of the process of the addition of rational expressions.
In the normal process, we perform arithmetic operations across algebraic fractions to obtain a single rational expression. This rational expression, on splitting in the reverse direction involved the process of decomposition of partial fractions and results in the two partial fractions. Let us learn more about partial fractions in the following sections.
| 1. | What are Partial Fractions? |
| 2. | Partial Fractions Formulas |
| 3. | Partial Fraction Decomposition |
| 4. | Partial Fractions of Improper Fraction |
| 5. | FAQs on Partial Fractions |
When a rational expression is split into the sum of two or more rational expressions, the rational expressions that are a part of the sum are called partial fractions. This is referred to as splitting the given algebraic fraction into partial fractions. The denominator of the given algebraic expression has to be factorized to obtain the set of partial fractions.
Every factor of the denominator of a rational expression corresponds to a partial fraction. For example, in the above figure, (4x + 1)/[(x + 1)(x - 2)] has two factors in the denominator, and hence there are two partial fractions, one with the denominator (x + 1) and the other with the denominator (x - 2).
In the above example, the numerators of partial fractions are 1 and 3. The numerator of a partial fraction is not always a constant. If the denominator is a linear function, the numerator is constant. And, if the denominator is a quadratic equation, then the numerator is linear. It means, the numerator's degree of a partial fraction is always one less than the denominator's degree. Further, the rational expression needs to be a proper fraction to be decomposed into a partial fraction. Listed below in the table are partial fraction formulas (here, all variables apart from x are constants).
| Type | Form of Rational Fraction | Partial Fraction Decomposition |
|---|---|---|
| Non-repeated Linear Factor | (px + q)/(ax + b) | A/(ax + b) |
| Repeated Linear Factor | (px + q)/(ax + b) n | A1/(ax + b) + A2/(ax + b) 2 + . An/(ax + b) n |
| Non-repeated Quadratic Factor | (px 2 + qx + r)/(ax 2 + bx + c) | (Ax + B)/(ax 2 + bx + c) |
| Repeated Quadratic Factor | (px 2 + qx + r)/(ax 2 + bx + c) n | (A1x + B1)/(ax 2 + bx + c) + (A2x + B2)/(ax 2 + bx + c) 2 + . (Anx + Bn)/(ax 2 + bx + c) n |
Let us look at a few examples of partial fractions.
In all these examples, A, B, and C are constants to be determined. Let's learn how to find these constants.
The partial fraction decomposition is writing a rational expression as the sum of two or more partial fractions. The following steps are helpful to understand the process to decompose a fraction into partial fractions.
Let us learn this process of decomposing a given fraction into partial fractions by an example.
Example: Find the partial fraction expansion of the expression(4x + 12)/(x 2 + 4x)
Solution:
Always remember to factor the denominator as much as possible before doing the partial fraction decomposition. (4x + 12)/(x 2 + 4x) = (4x + 12)/[x(x + 4)] ; The denominator has non-repeated linear factors. So, every factor corresponds to a constant in the numerator while writing the partial fractions.
Let us assume that: (4x + 12)/[(x)(x + 4)] = [A/x] + [B/(x + 4)] → (1)
The LCD (Least Common Denominator) of the sum (on the right side) is x(x + 4). Multiplying both sides by x(x + 4), 4x + 12 = A(x + 4) + Bx → (2)
Now we have to solve it for A and B. For that, we set each linear factor to zero.
Substitute x + 4 = 0 , or x = -4 in (2): 4(-4) + 12 = A(0) + B(-4); -4 = -4B; B = 1.
Substitute x = 0 in (2): 4(0) + 12 = A(0 + 4) + B(0); 12 = 4A; A = 3.
Substitute the values of A and B in (1), we get the partial fractions decomposition of the given expression: (4x + 12)/[x(x + 4)] = [3/x] + [1/(x + 4)]
Tips & Tricks on Partial Fractions Decomposition
The following tips are helpful to decompose a fraction into its partial fractions.
When we have to decompose an improper fraction into partial fractions, we first should do the long division. The long division is helpful to give a whole number and a proper fraction. The whole number is the quotient in the long division, and the remainder forms the numerator of the proper fraction, and the denominator is the divisor. The format of the result of the long division would be Quotient + Remainder/Divisor. Let us understand more of this with the help of the below example.
Example: Find the partial fraction decomposition of the expression (x 3 +4x 2 - 2x - 5)/(x 2 - 4x + 4)
Solution: Here, the degree of the numerator (3) is greater than the degree of the denominator (2). So the given fraction is improper. So we have to do the long division first.
Then write the given fraction as Quotient + Remainder/Divisor. Then we get: (x 3 + 4x 2 - 2x - 5)/(x 2 - 4x + 4) = x + 8 + (26x - 37)/(x 2 - 4x + 4).
Here, the fraction on the right side is a proper fraction and hence it can be split into partial fractions. (26x - 37)/(x 2 - 4x + 4) = (26x - 37)/(x - 2) 2 = A/(x - 2)+ B/(x - 2) 2
Now let us try to solve for A and B. Hint: Set each of (x - 2) and x one by one to zero to get A and B. You should get A = 26 and B = 15.
Substituting these values in we have: (26x-37)/(x 2 - 4x + 4) = [26/(x-2)] + [15/(x-2) 2 ] Further we have: (x 3 + 4x 2 - 2x - 5)/(x 2 - 4x + 4) = x + 8 + [26/(x - 2)] + [15/(x - 2) 2 ]
Important Notes on Partial Fractions
The following points would help in gaining a more clear understanding of partial fractions.
☛ Related Topics:
Example 1: Find the partial fraction decomposition of (x + 2) / [ (x + 1) (x - 2) ]. Solution: Since the denominator has non-repeating linear factors, by the partial fraction formulas, assume that: (x + 2) / [ (x + 1) (x - 2) ] = A / (x + 1) + B / (x - 2) . (1) Multiply both sides by (x + 1) (x - 2), (x + 2) = A (x - 2) + B (x + 1) Substitute x = -1: 1 = A (-3) ⇒ A = -1/3 Substitute x = 2: 4 = B (3) ⇒ B = 4/3 Substituting A and B values in (1): (x + 2) / [ (x + 1) (x - 2) ] = -1 / [3(x + 1)] + 4 / [3(x - 2)] Answer: -1 / [3(x + 1)] + 4 / [3(x - 2)]
Example 2: Decompose the following expression into partial fractions. (x 4 + x 3 + x 2 + 1)/(x 2 + x - 2) Solution: When we factorize the denominator, we get: x 2 + x - 2 = (x + 2)(x - 1). The degree of the numerator (4) is greater than that of the denominator (2). So it is an improper fraction. We need to first do the long division. So the given fraction can be written as: (x 4 + x 3 + x 2 + 1)/(x 2 + x - 2) = x 2 + 3 + (-3x + 7)/[(x + 2)(x - 1)]; Now we will decompose (-3x + 7)/[(x + 2)(x - 1)] into partial fractions using: (-3x + 7)/[(x + 2)(x - 1)] = A/(x + 2) + B/(x - 1) . (1) Multiplying both sides by the LCD (x + 2)(x - 1); -3x + 7 = A(x - 1) + B(x + 2). Substitute x - 1 = 0, or x = 1 we have -3 + 7 = 3B ;B = 4/3. Substitute x + 2 = 0, or x = -2 we have 6 + 7 = -3A ; A= -13/3. Substitute the values of A and B in the equation (1) we have:(-3x + 7)/[(x + 2)(x - 1)] = -13/[3(x + 2)]+ 4/[3(x - 1)]. Answer: Therefore the partial fractions decomposition of the given expression is: x 2 + 3 - 13/[3(x + 2)]+ 4/[3(x - 1)]
Example 3: Decompose the following rational expression into partial fractions. (4x 3 + x + 2)/[x 2 (x 2 + 1)] Solution: Look at the denominator. We have x 2 . It means the linear factor x is repeating. (x 2 + 1) is an irreducible (can't be factorized) quadratic factor. So the given fraction can be decomposed as follows: (4x 3 + x + 2)/[x 2 (x 2 + 1)] = [A/x] + [B/x 2 ] + [(Cx + D)/(x 2 + 1)] . (1) Multiplying both sides by the LCD x 2 (x 2 + 1); 4x 3 + x + 2 = Ax 3 + Ax + Bx 2 + B + Cx 3 + Dx 2 Setting the linear factor x to 0, i.e., x = 0, we get: 2 = B. Now we do not have any other linear factors to set to zero. So we will expand the right-hand side expression. Then we will compare the coefficients of x 3 , x 2 , x, and constant. By comparing the coefficients of x 3 , we get 4 = A + C. By comparing the coefficients of x 2 , we get 0 = B + D. By comparing the coefficients of x, we get 1 = A. By comparing the constants, we get 2 = B. By solving these equations, we get: A = 1, B = 2, C = 3, D = -2 . Further, substitute all these values in (1), the given expression becomes: [1/x] + [2/x 2 ] + [(3x - 2)/(x 2 + 1)] Answer: Therefore we have the resultant partial fractions as (4x 3 + x + 2)/[x 2 (x 2 + 1)] = [1/x] + [2/x 2 ] + [(3x - 2)/(x 2 + 1)].
