Question

$\frac{1}{(x-1)(x+2)(2+3)}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{2x+3}$. Match the variables with their values.

• A. $\frac{1}{13}$
• B. $\frac{3}{13}$
• C. $-\frac{8}{13}$

Answer 1

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

$\Rightarrow \frac{A(x+2)(2x+3)+B(x-1)(2x+3)+C(x-1)(x+2)}{(x-1)(x+2)(2x+3)}=\frac{1}{(x-1)(x+2)(2x+3)}$

or

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

or

$\frac{x^2(2A+2B+C)+x(5A+B+C)+6A-3B-2C}{(x-1)(x+2)(2x+3)}=\frac{1}{(x-1)(x+2)(2x+3)}$

Now we'll compare the expression we got with the given expression. That is, comparing the numerator of the above expression with numerator of $\frac{1}{(x-1)(x+2)(2x+3)}$

So, 2A + 2B + C = 0 (coefficient of $x^2$ = 0)

5A + B + C = 0 (coefficient of x = 0)

6A - 3B - 2C = 1

We have three equations and three variables. On solving these equations we get A = $\frac{1}{13}$, B = $\frac{3}{13}$, C = $-\frac{8}{13}$