Question
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.?
Answers
oobleck
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)
(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)