1EPartialFracsT4


 * Partial Fractions**

Type 4 Quadratic in denominator can’t be factorised (Not in course)

math \text {Express } \dfrac {x+3}{(x+2)(x^2+1)} \text { in partial fractions} math

Because x 2 + 1 can't be factorised over R, use a polynomial of one degree less than the numerator.

math \\ \dfrac {x+3}{(x+2)(x^2+1)} = \dfrac {A}{x+2} + \dfrac {Bx + C}{x^2+1} \\. \\ \dfrac {x+3}{(x+2)(x^2+1)} = \dfrac {A(x^2+1)+(Bx+C)(x+2)}{(x+2)(x^2+1)} math

Equate numerators math . \qquad x+3 = A(x^2+1) + (Bx+C)(x+2) math

math \\ . \qquad \text {Let } x=-2 \\ . \qquad \Longrightarrow 1 = A(5) + (-2B + C)(0) \\ . \qquad \Longrightarrow A = \dfrac {1}{5} math

Substitute A = 1/5 and use x = 0 to eliminate B math \\ . \qquad \text {Let } x=0 \\ . \qquad \Longrightarrow 3 = \dfrac {1}{5}(1) + (C)(2) \\ . \qquad \Longrightarrow 2C = \dfrac {14}{5} \\ . \qquad \Longrightarrow C = \dfrac {7}{5} math

Substitute A = 1/5 and C = 7/5 and use any other x ( x = 1 is easy) math \\ . \qquad \text {Let } x=1 \\ . \qquad \Longrightarrow 4 = \dfrac {1}{5}(2) + (B +\dfrac {7}{5})(3) \\ . \qquad \Longrightarrow 3B = -\dfrac {3}{5} \\ . \qquad \Longrightarrow B = -\dfrac {1}{5} math

math \text {Thus } \dfrac {x+3}{(x+2)(x^2+1)} = \dfrac {\dfrac {1}{5}}{x+2} + \dfrac {-\dfrac {1}{5}x + \dfrac {7}{5}}{x^2+1} math

math \text {Or } \dfrac {x+3}{(x+2)(x^2+1)} = \left(\dfrac {1}{5} \right) \left( \dfrac {1}{x+2} \right) + \left( \dfrac {1}{5} \right) \left( \dfrac {-x + 7}{x^2+1} \right) math

math \text {Which gives } \dfrac {x+3}{(x+2)(x^2+1)} = \dfrac {1}{5(x+2)} + \dfrac {7 - x}{5(x^2+1)} math

Note: If there was a cubic that couldn’t be factorised in the denominator, you would use a quadratic as the numerator, etc

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