1EPartialFracsIntro


 * Introduction to Partial Fractions**

Recall that to add two fractions, first they must be changed to have a common denominator.

For example, math \\ . \qquad \dfrac {5}{x-2} + \dfrac {3}{x+4} = \dfrac {5(x+4)}{(x-2)(x+4)} + \dfrac {3(x-2)}{(x-2)(x+4)} \\. \\ . \qquad \qquad \qquad \qquad \quad = \dfrac {5(x+4)+3(x-2)}{(x-2)(x+4)} \\. \\ . \qquad \qquad \qquad \qquad \quad = \dfrac {8x+14}{(x-2)(x+4)} math

Aim: To do the reverse of that process ie split one fraction into two. The resulting two fractions are called __partial fractions__.

Example 1

math \text{Separate } \dfrac{11}{12} \text{ into partial fractions} math

Divide into two fractions with numerators A and B (unknown constants) math . \qquad \dfrac {11}{12} = \dfrac {A}{3} + \dfrac {B}{4} math

Add the right side fractions together by changing to a common denominator (which should be the same as the left side denominator) math . \qquad \dfrac {11}{12} = \dfrac {4A}{12} + \dfrac {3B}{12} = \dfrac {4A + 3B}{12} math

Since these two fractions are equal and have the same denominator, the numerators must be equal math . \qquad 11 = 4A + 3B math

Then (in algebra questions) we will use algebra to find A and B (in this case A = 2 and B = 1) math . \qquad \dfrac {11}{12} = \dfrac {2}{3} + \dfrac {1}{4} math

Example 2 math \text {Separate } \dfrac {8x+14}{(x-2)(x+4)} \text { into partial fractions.} math

Divide into two fractions with numerators A and B (unknown constants). math \text {Let } \dfrac {8x+14}{(x-2)(x+4)} = \dfrac {A}{x-2} + \dfrac {B}{x+4} math

Add the right side fractions together by changing to a common denominator (which should be the same as the left side denominator) math . \qquad \dfrac {8x+14}{(x-2)(x+4)} = \dfrac {A(x+4)}{(x-2)(x+4)} + \dfrac {B(x-2)}{(x-2)(x+4)} math

math . \qquad \dfrac {8x+14}{(x-2)(x+4)} = \dfrac {A(x+4) + B(x-2)}{(x-2)(x+4)} math

Since these two fractions are equal and have the same denominator, the numerators must be equal math . \qquad 8x + 14 = A(x + 4) + B(x - 2) \qquad \textbf{ [*] } math

We __could__ solve this by expanding the brackets and equating the coefficients and solving simultaneously: math \\ . \qquad 8x + 14 = Ax + 4A + Bx - 2A \\. \\ . \qquad 8x + 14 = (A + B)x + (4A - 2B) \\. \\ \text {Thus } \\ . \qquad 8 = A + B \text { and } 14 = 4A - 2B math etc


 * A simpler way is: **

math . \qquad 8x + 14 = A(x + 4) + B(x - 2) \qquad \textbf{ [*] } math

can be solved by substituting chosen values of x to eliminate either A or B.

math \\ . \qquad \text {Sub } x = -4 \text { into } \\ . \qquad 8x + 14 = A(x + 4) + B(x - 2) \\ \text {Gives } \\ . \qquad -18 = A(0) + B(-6) \\ \text {so} \\ . \qquad B = 3 math

and math \\ . \qquad \text {Sub } x = 2 \text { into } \\ . \qquad 8x + 14 = A(x + 4) + B(x - 2) \\ \text {Gives } \\ . \qquad 30 = A (6) + B(0) \\ \text {so} \\ . \qquad A = 5 math

Now put A = 5 and B = 3 back into the original expression:

math . \qquad \dfrac {8x+14}{(x-2)(x+4)} = \dfrac {5}{x-2} + \dfrac {3}{x+4} \: \text { as required} math

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