Question

$\frac{2x^3+x^2-5}{x^4-25}=\frac{Ax+B}{x^2-5}+\frac{Cx+1}{x^2+5}\Rightarrow (A,B,C)=$

• A. (1, 1, 1)
• B. (1, 1, 0)
• C. (1, 0, 1)
• D. (1, 2, 1)

Answer 1

The correct option is C (1, 0, 1)

Taking RHS, we get

$\frac{Ax+B}{x^2-5}+\frac{Cx+1}{x^2+5}$
$=\frac{(A+C)x^3+(B+1)x^2+(5A-5C)x+5(B-1)}{x^4-25}$
$=\frac{2x^3+x^2-5}{x^4-25}$
Comparing the coefficients gives us
$A+C=2$,$B+1=1$, $5A-5C=0$ and $5(B-1)=-5$
Consider $A+C=2$ and $5A-5C=0$
Therefore
$5A-5C=0$
$\therefore A=C$
Substituting
$A+C=2$
$2A=2$
$A=1$
$\therefore C=1$
$B+1=1$
$\therefore B=0$
Hence $(A,B,C)=(1,0,1)$