Express the following in partial fractions: (x^2+4x+10)/(x+3)(x+4)(x+5)

(x^2+4x+10)/(x+3)(x+4)(x+1) = A/(x+3) + B/(x+4) + C/(x+1) [A(x+4)(x+1) + B(x+3)(x+1) + C(x+3)(x+4)]/(x+3)(x+4)(x+1) = (x^2+4x+10)/(x+3)(x+4)(x+1) [A(x+4)(x+1) + B(x+3)(x+1) + C(x+3)(x+4)] = (x^2+4x+10) A(x^2+5x+4) + B(x^2+4x+3) + C(x^2+7x+12) = (x^2+4x+10) A+B+C = 1 (1) 5A+4B+7C = 4 (2) 4A+3B+12C=10 (3) From (1): A = 1-B-C (4) Sub (4) into (2): 5(1-B-C)+4B+7C = 4 -B+2C = -1 (5) Sub (4) into (3):4(1-B-C)+3B+12C=10-B+8C=6 (6) (6)-(5): -B+8C=6-B+2C = -1 6C = 7 C = 7/6 From (5): -B + 2(7/6) = -1 -B = -1 - 14/6 B = 10/3 From (4): A = 1-(10/3)-(7/6) A = -7/2 Therefore: (x^2+4x+10)/(x+3)(x+4)(x+1) = -7/2(x+3) + 10/3(x+4) + 7/6(x+1)



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Answered by Alessandro S. Maths tutor

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