Question

PLEASE HELP!!!!!!!!!!!!!!

Answer 1

Answer:

Step-by-step 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)