2x-5÷(x+3)(x+1)^2 , change into partial fraction
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2x-5/(x+3)(x+1)^2
2x-5=A(x+1)^2+B(x+3)(x+1)+C(x+3)
2x-5=A(x^2+2x+1)+B(x^2+4x+3)+Cx+3C
2x-5=AX^2+2AX+A+BX^2+4BX+3B+3X+3C
A+B=0
2A+3B+C=2
A+3B+3C=-5
B=11/4 A=-11/4 C=23/2
2x-5=A(x+1)^2+B(x+3)(x+1)+C(x+3)
2x-5=A(x^2+2x+1)+B(x^2+4x+3)+Cx+3C
2x-5=AX^2+2AX+A+BX^2+4BX+3B+3X+3C
A+B=0
2A+3B+C=2
A+3B+3C=-5
B=11/4 A=-11/4 C=23/2
Bhaweshbhandari:
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