Resolve 5x-7/(x-1)(x-2)(x-3) into partial fraction using comparison coefficient method.pls i need ans asap with workings thanks in advance.?

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asked by poppy
  1. If two polynomials are identical, then all the coefficients of all the powers must be the same. You want to find A,B,C such that
    (5x-7)/(x-1)(x-2)(x-3) = A/(x-1) + B/(x-2) + C/(x-3)
    So, if you put all three terms on the right over the common denominator of (x-1)(x-2)(x-3), then all we have to do is compare the numerators. That is
    5x-7 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)
    5x-7 = A(x^2-5x+6) + B(x^2-4x+3) + C(x^2-3x+2)
    5x-7 = (A+B+C)x^2 - (5A+4B+3C)x + (6A+3B+2C)
    That means that
    A+B+C = 0
    5A+4B+3C = -5
    6A+3B+2C = -7
    Solve those equations, and your final solution is
    (5x-7)/(x-1)(x-2)(x-3) = -1/(x-1) - 3/(x-2) + 4(x-3)

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