Question

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

Answer 1

Hint: Using the concept of partial fraction, we can write the given expression as$\dfrac{{2{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^3}}} = \dfrac{A}{{\left( {x + 1} \right)}} + \dfrac{B}{{{{\left( {x + 1} \right)}^2}}} + \dfrac{C}{{{{\left( {x + 1} \right)}^3}}}$. Then we can take the sum of the RHS and solve for A B and C by equation the coefficients of each terms. Then we can substitute back the values of A B and C to get the required solution.

Complete step by step answer:

We can use the concept of partial fraction. So we can write the expression as
$\dfrac{{2{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^3}}} = \dfrac{A}{{\left( {x + 1} \right)}} + \dfrac{B}{{{{\left( {x + 1} \right)}^2}}} + \dfrac{C}{{{{\left( {x + 1} \right)}^3}}}$ .. (1)
We can make the denominators of the RHS equal by taking LCM. We get,
$ \Rightarrow \dfrac{{2{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^3}}} = \dfrac{{A{{\left( {x + 1} \right)}^2}}}{{{{\left( {x + 1} \right)}^3}}} + \dfrac{{B\left( {x + 1} \right)}}{{{{\left( {x + 1} \right)}^3}}} + \dfrac{C}{{{{\left( {x + 1} \right)}^3}}}$
We can take the sum of the RHS. We get,
$ \Rightarrow \dfrac{{2{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^3}}} = \dfrac{{A\left( {{x^2} + 2x + 1} \right) + B\left( {x + 1} \right) + C}}{{{{\left( {x + 1} \right)}^3}}}$
As the denominators are equal, their numerators will be also equal,
$ \Rightarrow 2{x^2} + x + 1 = A\left( {{x^2} + 2x + 1} \right) + B\left( {x + 1} \right) + C$
On expanding we get,
$ \Rightarrow 2{x^2} + x + 1 = A{x^2} + 2Ax + A + Bx + B + C$
Taking like terms together we get,
$ \Rightarrow 2{x^2} + x + 1 = A{x^2} + \left( {2A + B} \right)x + \left( {A + B + C} \right)$
Now we can compare the coefficients of the polynomial.
On comparing coefficients of ${x^2}$, we get,
$A = 2$
On comparing coefficients of ${x^2}$, we get,
$ \Rightarrow 2A + B = 1$
On substituting the value of A, we get,
$ \Rightarrow 2 \times 2 + B = 1$
$ \Rightarrow B = 1 - 4 = - 3$
On comparing constants, we get,
$A + B + C = 1$
On substituting the value of A and B we get,
$ \Rightarrow 2 - 3 + C = 1$
$ \Rightarrow C = 1 + 1 = 2$
Now we can substitute the value of A, B and C in equation (1), we get,
$\dfrac{{2{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^3}}} = \dfrac{2}{{\left( {x + 1} \right)}} - \dfrac{3}{{{{\left( {x + 1} \right)}^2}}} + \dfrac{2}{{{{\left( {x + 1} \right)}^3}}}$
So the required expression is $\dfrac{2}{{\left( {x + 1} \right)}} - \dfrac{3}{{{{\left( {x + 1} \right)}^2}}} + \dfrac{2}{{{{\left( {x + 1} \right)}^3}}}$
Therefore, the correct answer is option B.

Note: Alternate approach to this problem is:
We can take the value of the expression at $x = 0$.
$ \Rightarrow \dfrac{{2{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^3}}} = \dfrac{{2 \times {0^2} + 0 + 1}}{{{{\left( {0 + 1} \right)}^3}}}$
$ = 1$
Now we can check each of the options giving the values$x = 0$. At$x = 0$, the denominators of all the option will become 1. So we need to just add the numerators.
Thus option A becomes, $2 + 3 - 2 = 3$.
Option B becomes, $2 - 3 + 2 = 1$
Option C becomes, $2 + 3 + 2 = 5$
Option D becomes, $2 - 3 - 2 = - 3$
From the values of the options at $x = 0$only option B has the same value as that of the expression.
Therefore, the correct answer is option B.