Answer:
Explanation:
(8x²-18x+10)/(x²+5)(x-3)
express the expression as a partial fraction:
(8x²-18x+10)/[(x^2+5)(x-3)] =A/x-3 +bx+c/x²+5
both denominator are equal , so require only work with the nominator
(8x²-18x+10)=(x²+5)A+(x-3)(bx+c)
8x²-18x+10= x²A+5A+bx²+cx-3bx-3c
combine like terms:
x²(A+b)+x(-3b+c)+5A-3c
(8x²-18x+10)
looking at the equation
A+b=8
-3b+c=-18
5A-3c=10
solve for A,b and c (system of equation)
A=2 , B=6, and C=0
substitute in the value of A, b and c
(8x²-18x+10)/[(x^2+5)(x-3)] =A/x-3 +(bx+c)/x²+5
(8x²-18x+10)/[(x^2+5)(x-3)] = 2/x-3 + (6x+0)/(x²+5)
(8x²-18x+10)/[(x^2+5)(x-3)] =
2/(x-3)+6x/x²+5
(4x+2)/[(x²+4)(x-2)]
(4x+2)/[(x²+4)(x-2)]= A/(x-2) + bx+c/(x²-2)
(4x+2)=a(x²-2)+(bx+c)(x-2)
follow the same step in the previous answer:
the answer is :
(4x+2)/[(x²+4)(x-2)]= 5/4/(x-2) + (3/2 -5x/4)/(x²+4)