Question

If $\dfrac{{\left( {3x + 4} \right)}}{{{{\left( {x + 1} \right)}^2}\left( {x - 1} \right)}} = \dfrac{A}{{\left( {x - 1} \right)}} + \dfrac{B}{{\left( {x + 1} \right)}} + \dfrac{C}{{{{\left( {x + 1} \right)}^2}}}$ , then what is the value of $A$ ?
A. $\dfrac{{ - 1}}{2}$
B. $\dfrac{{15}}{4}$
C. $\dfrac{7}{4}$
D. $\dfrac{{ - 1}}{4}$

Answer 1

Hint: To begin, simplify the given equation and equalize the denominator on both sides. Then, using the numerators from both sides, create an equation. Then, in the equation, substitute the values for $x$ and calculate the required value of $A$.

Formula Used:
${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
$\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$

Complete step by step solution:
The given equation is $\dfrac{{\left( {3x + 4} \right)}}{{{{\left( {x + 1} \right)}^2}\left( {x - 1} \right)}} = \dfrac{A}{{\left( {x - 1} \right)}} + \dfrac{B}{{\left( {x + 1} \right)}} + \dfrac{C}{{{{\left( {x + 1} \right)}^2}}}$.
Let’s simplify the above equation by equalizing the denominator.
On the right-hand side of the equation, multiply the numerator and denominator of the first term by ${\left( {x + 1} \right)^2}$. Multiply the numerator and denominator of the second term by $\left( {x + 1} \right)\left( {x - 1} \right)$. And multiply the numerator and denominator of the third term by $\left( {x - 1} \right)$.

$\dfrac{{\left( {3x + 4} \right)}}{{{{\left( {x + 1} \right)}^2}\left( {x - 1} \right)}} = \dfrac{{A{{\left( {x + 1} \right)}^2}}}{{\left( {x - 1} \right){{\left( {x + 1} \right)}^2}}} + \dfrac{{B\left( {x + 1} \right)\left( {x - 1} \right)}}{{{{\left( {x + 1} \right)}^2}\left( {x - 1} \right)}} + \dfrac{{C\left( {x - 1} \right)}}{{{{\left( {x + 1} \right)}^2}\left( {x - 1} \right)}}$
$ \Rightarrow \dfrac{{\left( {3x + 4} \right)}}{{{{\left( {x + 1} \right)}^2}\left( {x - 1} \right)}} = \dfrac{{A{{\left( {x + 1} \right)}^2} + B\left( {x + 1} \right)\left( {x - 1} \right) + C\left( {x - 1} \right)}}{{\left( {x - 1} \right){{\left( {x + 1} \right)}^2}}}$

Now equate the numerators of the above equations we get,
$\left( {3x + 4} \right) = A{\left( {x + 1} \right)^2} + B\left( {x + 1} \right)\left( {x - 1} \right) + C\left( {x - 1} \right)$
Now simplify the above equation by using the formula $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$.
$3x + 4 = A{\left( {x + 1} \right)^2} + B\left( {{x^2} - 1} \right) + C\left( {x - 1} \right)$
Substitute $x = 1$ in the above equation.
$3\left( 1 \right) + 4 = A{\left( {1 + 1} \right)^2} + B\left( {{1^2} - 1} \right) + C\left( {1 - 1} \right)$
$ \Rightarrow 7 = A\left( 4 \right) + B\left( 0 \right) + C\left( 0 \right)$
$ \Rightarrow 7 = 4A$
Divide both sides by 4.
$A = \dfrac{7}{4}$

Option ‘C’ is correct

Note: Students often get confused about the concept of the equalization of the denominator.
While equalizing the denominator multiply the numerator and denominator with the common factors of the required denominator.