suppose f(x)= 3x2- 3x + 8/x(x2+4) then f(x)=A/x + Bx +C/x2+4 where is A = where is B = where is C =

Question

suppose

f(x)= 3x²- 3x + 8/x(x²+4)

then

f(x)=A/x + Bx +C/x²+4

where is A =

where is B =

where is C =

Answer 1

axl, recall I answereed a partial fraction question you had where a denominator term was squared and unsquared (what is call repeated), and said the only way to solve is with simultaneous equations. Here simultaneous equations also can be used but I want to show you a technique that can be useful in future math learning.

call your problem f(x)/[g(x)h(x)j(x)], f(x)=3x²-3x+8, g(x)=x, h(x)=x+2i, j(x)=x-2i

wanted is f(x)/[g(x)h(x)j(x)]=[A/g(x)]+[B/h(x)]+[C/j(x)]

in algebra I can prove A=value of f(x) at x where g(x)=0 divided by h(x)*j(x) at x where g(x)=0

A=f(0)/(h(0)*g(0))=8/4=2

B=value of f(x) at x where h(x)=0 divided by g(x)*j(x) at x where h(x)=0

B=f(-2i)/(g(-2i)*j(-2i))=(2-3i)/4

C=value of f(x) at x where j(x)=0 divided by g(x)*h(x) at x where j(x)=0

C=f(2i)/(g(2i)*h(2i))=(2+3i)/4

that last two terms are (2-3i)/(4*(x+2i))+(2+3i)/(4*(x-2i))=(x-3)/(x²+4)

so in conclusion (3x²-3x+8)/(x*(x²+4))=(2/x)+(x-3)/(x²+4)

The A for your answer is 2

The B for your answer is 1

The C for your answer is -3