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Partial Fractions Marking Scheme
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[QUOTE="CAF123, post: 4412317, member: 419343"] You are writing the fraction as: $$\frac{15 - 17x}{(2+x)(1-3x)^2} = \frac{A(1-3x)^2 + B(2+x)(1-3x) + C(2-x)}{(2+x)(1-3x)^2}.$$For this to hold, the denominator must be the same on both sides. This is true by inspection. Similarly, the numerator on LHS must be the same as the numerator on the RHS. This gives: $$15 - 17x = A(1-3x)^2 + B(2+x)(1-3x) + C(2-x)$$ which has to be satisifed by all ##x##. The condition on x not being equal to 1/3 and -2 is so that we don't end up with f(x) being undefined. The exercise above is simply to determine A,B and C. The reason we choose 1/3 and -2 is because that simplies things greatly. [/QUOTE]
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