MTH5P1: Partial Fractions

This unit explains partial fractions

PARTIAL FRACTIONS

Partial Fractions.
The section on rational expressions included methods of addition and subtraction of algebraic fractions.


To find the values of A and B, we substitute convenient values for x to make each term on the top line
equal to zero.

It must be stressed that the substitution method and the cover-up rule can only be used exclusively
when we have distinct linear factors in the denominator.

In fact, when there are repeated or quadratic factors in the denominator, the cover-up method is ‘dodgy’ and should not be used (see later) ! Another method of finding A and B is to equate the coefficients.

gives A(x+1) + B(x-2) = x + 7.
From this, we can solve the simultaneous equations:
A + B = 1 (equating x-coefficients)
A – 2B = 7 (equating constants).
Subtracting the second equation from the first gives 3B = -6 and thus B = -2. Substituting in the first equation gives A – 2 = 1 and thus A = 3.

Most examination questions on partial fractions cover the following cases:
Two distinct linear factors in the denominator
Three distinct linear factors in the denominator
Three linear factors in the denominator, but one repeated
Two distinct linear factors in the denominator.
This case was covered in the previous example – here are two others.

(Method of equating coefficients)

Therefore A(x-3) + B(x-2) = 4x – 9.

Solving the simultaneous equations:
A + B = 4 (equating x-coefficients)
-3A – 2B = -9 (equating constants).
Hence 2A + 2B = 8 ; -3A – 2B = -9.
Adding the last two equations gives -A = -1 and thus A = 1.
Substituting in the first equation gives 1 + B = 4 and thus B = 3.

 

(Cover-up rule working)

(Method of equating coefficients)
The top line of the expanded expression, A(x+3) + B(x-4) = 4x + 5.
Solving the simultaneous equations:
A + B = 4 (equating x-coefficients)
3A – 4B = 5 (equating constants).
Hence 4A + 4B = 16 ; 3A – 4B = 5.
Adding the last two equations gives 7A = 21 and thus A = 3.
Substituting in the first equation gives 3 + B = 4 and thus B = 1.

Three distinct linear factors in the denominator.
The general method of solving partial fractions of this type is the same as the case with two distinct
linear factors, though there is more work involved.

(Long working)

(Cover-up rule working)

The method of equating coefficients can also be used when there are three linear factors in the denominator.

We need to expand all the quadratic products in the numerator:

Adding the two equations gives –5A = -5 and thus A = 1.
Substituting back into the original equation (1) gives C = 3.
Finally, substituting into the original equation (2) gives B = -2.

When we have three factors in the denominator, the method of equating coefficients can become increasingly difficult, but there are occasions when we do need to use it.

Note how equation (3) has turned out easy to solve – we can see at once that A = 3.
Substituting A = 3 into equations (1) and (2), we obtain

(Again we have multiplied top and bottom by 4 to get rid of the awkward fractions.)

The following simultaneous equations result:

Three linear factors in the denominator, but with one repeated pair.
The method here is again slightly different: here we have to use a combination of substitution and equating coefficients.

Note that the factor (x – 1) is repeated here. The required partial fraction will be of the form

The cover-up rule should not be used here, because it may lead the student to think that the solution
is of the form

(Substitution working)

Note that we have not been able to find B at this stage.
Final step – finding B by equating coefficients .

From the above, we can deduce B = 1 and (-4 – 1 + C) = 0 and thus C = 5.

(Substitution working)

From Equation (1) we can substitute –A for B in Equations 2 and 3.
3A + C = 4 (equating x coefficients) (2)
6A – C = 5 (equating constants) (3)
Adding gives 9A = 9 and hence A = 1 and B = -1.
Substituting in Eqn.2 gives 3 + C = 4 and hence C = 1.
From the above, we can deduce B = 1 and (-4 – 1 + C) = 0 and thus C = 5.

We still need to find B.

The equations are more messy than in the last one, but we can substitute 5 – 2B for A in Equations 2
and 3.

Adding the two equations gives 5C = 35 and hence C = 7.
Substituting in 20B + 8C = 136 gives 20B = 80 and hence B = 4.
Finally, recalling A = 5 – 2B gives A = -3.

ASSIGNMENT : Partial Fractions Assignment MARKS :   DURATION : EXPIRED

 

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