Question

$\frac{3x-1}{(1-x+x^2)(2+x)}=$

• A. $\frac{x}{x^2-x+1}-\frac{1}{x+2}$
• B. $\frac{x}{x^2-x+1}+\frac{1}{x+2}$
• C. $\frac{x}{x^2+x+1}+\frac{2}{x+2}$
• D. $\frac{x}{-x+1}-\frac{2}{x+2}$

Answer 1

Answer: (A)

$\frac{x}{x^2-x+1}-\frac{1}{x+2}$

$\frac{3x-1}{(1-x+x^2)(2+X)}=\frac{A}{(2+X)}+\frac{Bx+C}{(x^2-x+1)}$

finding values of A, B, C

$3x-1=A(x^2-x+1)+(Bx+C)(2+x)$

Comparing the co-efficients of x A, B, C we get

Comparing $x^2$ co-efficient we get

$A+B=O--\left(1\right)$

$x$ co-efficient

$3=-A+2B+C--\left(2\right)$

1 co-efficient

$-1=A+2C--\left(3\right)$

solving the equations

we get

$A=-1;B=1$

$C=0$

we get $\frac{-1}{(2+X)}+\frac{x}{(x^2-x+1)}$