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Steps to doing partial fractions
Step 1: Make sure given fraction is proper (degree of power of x for numerator is less than degree of power of x for denominator). If fraction is not proper, do long division to express fraction as proper fraction before breaking it into its partial fractions. <See first video below for an example whereby the fraction is not proper and LONG DIVISION MUST BE DONE before breaking the fraction into partial fractions.>
Step 2: Check if denominator of given fraction is completely factorised. If not, the denominator has to be factorised first before one can determine how the partial fractions are formed.
Step 3: Express fraction into its partial fractions. There are three different cases depending on what is given as denominator.
Case 1: Two linear factors are given. The example below shows how to break two linear factors into its partial fractions.
Step 2: Check if denominator of given fraction is completely factorised. If not, the denominator has to be factorised first before one can determine how the partial fractions are formed.
Step 3: Express fraction into its partial fractions. There are three different cases depending on what is given as denominator.
Case 1: Two linear factors are given. The example below shows how to break two linear factors into its partial fractions.
Case 2: One linear factor and one repeated factor are given. The example below shows how to break into its partial fractions.
Case 3: One linear factor and one quadratic factor are given. The example below shows how to break into its partial fractions.
**Note that the denominator of a given fraction can have various combinations of linear factor, quadratic factor and/or repeated factor. But, the basic way to deal with each type of factors is as given in the above cases.
Step 4: Solve for the unknown constants A, B, C etc. One can use substitution of values of x on both sides of the equation or comparing coefficients of like terms on both sides of equation (e.g. coefficient of x with x and x² with x²). <See following videos on how these two ways are being used. Notice that substituition of values of x is used more often in the following videos as this is an easy and faster way than comparing coefficients of like terms. However, sometimes substituition of values of x cannot be used to completely find all the unknown constants, so comparing coefficients must be used together with substituition of values of x. One should know all ways of doing things but always choose the easiest and fastest way to get things done. Sometimes, combining two different ways into doing the same question may yield the best result. Always think flexibly and efficiently!>
Step 4: Solve for the unknown constants A, B, C etc. One can use substitution of values of x on both sides of the equation or comparing coefficients of like terms on both sides of equation (e.g. coefficient of x with x and x² with x²). <See following videos on how these two ways are being used. Notice that substituition of values of x is used more often in the following videos as this is an easy and faster way than comparing coefficients of like terms. However, sometimes substituition of values of x cannot be used to completely find all the unknown constants, so comparing coefficients must be used together with substituition of values of x. One should know all ways of doing things but always choose the easiest and fastest way to get things done. Sometimes, combining two different ways into doing the same question may yield the best result. Always think flexibly and efficiently!>
Example showing long division is needed before breaking a fraction into its partial fractions
Partial fractions (Case 1: two linear factors in denominator)
Partial fractions (Case 2: One repeated factor and one linear factor in denominator)
Partial fractions (Case 3: One linear factor and one quadratic factor in denominator)
Note that in this video, after breaking the initial fraction into its partial fractions, the way of comparing coeffients of like terms of x², x and constants on both sides of the equation is used to solve for the unknowns A, B and C. In this video, three simultaneous equations involving A, B and C are formed to solve for A, B and C. Though this can be done, I personally think that one can substitute x = -1/2 into the equation to find A first and then substituting x = 0 to find C next and lastly substituting x = any value to find B since A and C values are found already. Try it for yourself to see if my proposed way of entirely using substitution works and whether it is easier and faster this way instead of solving three simultaneous equations.
Kindly ignore the second part of this video from time 9:50 onwards which is not relevant to partial fractions.
Kindly ignore the second part of this video from time 9:50 onwards which is not relevant to partial fractions.