(2x-5) / (x+3) (x+1)² into partial fractions
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Step-by-step explanation:
Answered by
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Step-by-step explanation:
2x-5/(x+3)(x+1)²
2x-5 = A(x+1)² + B(x+1) (x-3)+ C (x+3)
put x=-1
2(-1)-5 = A(-1+1)² + B (-1+1) + C(-1+3)
-2-5 = C (+2)
-7 = 2C
C = -7/2
2(-3)-5 = A(-3+1)²
-6-5 = A(-2)²
-11 = 4A
A = -11/4
comparing constants
-5 = A + 3B + 3C
-5 = -11/4 + 3B+ 3(-7/2)
3B = -5 + 21/2 + 11 /4
3B = -20+42+11/4
3B = 33/4
B = 11/4
Hence ,
2x-5/(x+3)(x+1)² = -11/4(x+3) + 11/4/(x+1) + -7/2/(x+1)²
= -11/4x+3 + 11/4(x+1) - 7/(x+1)²
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