Question

If $\frac{1}{(x-1)(x+2)(2x+3)}$ can be expressed as $\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{2x+3}$ then what will be the respective values of A, B and C?

• A. $\frac{1}{15},\frac{1}{3},\frac{4}{5}$
• B. $\frac{1}{15},\frac{1}{3},\frac{4}{-5}$
• C. $\frac{4}{5},\frac{1}{3},\frac{1}{15}$
• D. $\frac{4}{-5},\frac{1}{3},\frac{1}{15}$

Answer 1

The correct option is B

$\frac{1}{15},\frac{1}{3},\frac{4}{-5}$

The approach to solve these kind of questions should be by cross checking. Let's take LCM of $\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{2x+3}$ and rewrite this-

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

or

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

or

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

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 = $\frac{1}{15}$, B = $\frac{1}{3}$, C = $\frac{4}{-5}$

We can also put the values given in the options and check which option satisfies these 3 equations.