Question

Decompose into partial fractions: $\frac{(2x^2+6x-2)}{(x^3-1)}$

• A. $\frac{2}{(x-1)}-\frac{4}{(x^2+x+1)}$
• B. $\frac{2}{(x-1)}+\frac{4}{(x^2+x+1)}$
• C. $-\frac{2}{(x-1)}+\frac{4}{(x^2+x+1)}$
• D. $\frac{2}{(x-1)}-\frac{3}{(x^2+x+1)}$

Answer 1

The correct option is B $\frac{2}{(x-1)}+\frac{4}{(x^2+x+1)}$

Let $\frac{2x^2+6x-2}{(x^3-1)}=\frac{(2x^2+6x-2)}{(x-1)(x^2+x+1)}=\frac{A}{(x-1)}+\frac{Bx+C}{(x^2+x+1)}$ ....(1)

Multiply $x^3-1$ on both the sides, we get

$2x^2+6x-2=A(x^2+x+1)+(Bx+C)(x-1)$

$=A{x}^2+Ax+A+B{x}^2-Bx+Cx-C$

$=x^2(A+B)+x(A-B+C)+(A-C)$

Comparing coefficients of $x^2,x$ and constant, we get

$A+B=2,A-B+C=6,A-C=-2$

Solving these equations, we get

$A=2,B=0,C=4$

Substituting the values of A, B and C in equation (1), we get

$\frac{2x^2+6x-2}{(x^3-1)}=\frac{2}{(x-1)}+\frac{4}{(x^2+x+1)}$