Sometimes it is useful to express a single fraction such as the sum of 2 (or more in other cases) separate fractions. This is called decomposing a function, f(x), in partial fractions. Consider this example: This equation can be split into the sum of two single fractions. ThereforeTo find the value of the constants A and B depends on the factors in the denominator. Follow the examples below. When the denominator is linearUsing the example above: Adding the two fractions on the right hand side gives: As the denominators are now the same the numerators must match as well. Therefore: If we choose to substitute x = -1, this will eliminate the constant B. This gives: -2 − 1 = A (-3 + 2) -3 = -A A = 3 If we choose to substitute x = -2/3 to this will eliminate the constant A. This gives: -2/3 × 2 − 1 = B (-2/3 + 1) -4/3 − 1 = B (1/3) -7/3 = 1/3 B -7 = B B = -7 Therefore: When the denominator contains a quadratic equationLets take the example: As this has a quadratic factor in the denominator we need to write the partial equations in the following format: Once again we need to add the two fractions on the right hand side. This gives: Once again the denominators are now the same, so the numerators must be the same as well. Therefore: By substituting x = 0, we eliminate B and C. This gives: 02 + 3 = A (02 + 2) 3 = 2A A = 3/2 We can substitute two other values for x and solve B and C by simultaneous equations. When the denominator contains a ? repeated factor? Consider this example: The denominator contains (x − 2)2. Since, (x − 2)2 = (x − 2 )(x − 2) it is called a repeated factor. When there is a repeated factor in the denominator we need to write the partial equations in the following format: Once again we add the fractions on the right hand side together and thus the denominators are the same, so the numerators are the same, giving: We again proceed by substitution to obtain the values of the constants.