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Question: Find \(\dfrac{{2{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^3}}} = \) A. \(\dfrac{2}{{\left( {x + 1...

Find 2x2+x+1(x+1)3=\dfrac{{2{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^3}}} =
A. 2(x+1)+3(x+1)2−2(x+1)3\dfrac{2}{{\left( {x + 1} \right)}} + \dfrac{3}{{{{\left( {x + 1} \right)}^2}}} - \dfrac{2}{{{{\left( {x + 1} \right)}^3}}}
B. 2(x+1)−3(x+1)2+2(x+1)3\dfrac{2}{{\left( {x + 1} \right)}} - \dfrac{3}{{{{\left( {x + 1} \right)}^2}}} + \dfrac{2}{{{{\left( {x + 1} \right)}^3}}}
C. 2(x+1)+3(x+1)2+2(x+1)3\dfrac{2}{{\left( {x + 1} \right)}} + \dfrac{3}{{{{\left( {x + 1} \right)}^2}}} + \dfrac{2}{{{{\left( {x + 1} \right)}^3}}}
D. 2(x+1)−3(x+1)2−2(x+1)3\dfrac{2}{{\left( {x + 1} \right)}} - \dfrac{3}{{{{\left( {x + 1} \right)}^2}}} - \dfrac{2}{{{{\left( {x + 1} \right)}^3}}}

Explanation

Solution

Using the concept of partial fraction, we can write the given expression as2x2+x+1(x+1)3=A(x+1)+B(x+1)2+C(x+1)3\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
2x2+x+1(x+1)3=A(x+1)+B(x+1)2+C(x+1)3\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,
⇒2x2+x+1(x+1)3=A(x+1)2(x+1)3+B(x+1)(x+1)3+C(x+1)3\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,
⇒2x2+x+1(x+1)3=A(x2+2x+1)+B(x+1)+C(x+1)3\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,
⇒2x2+x+1=A(x2+2x+1)+B(x+1)+C\Rightarrow 2{x^2} + x + 1 = A\left( {{x^2} + 2x + 1} \right) + B\left( {x + 1} \right) + C
On expanding we get,
⇒2x2+x+1=Ax2+2Ax+A+Bx+B+C\Rightarrow 2{x^2} + x + 1 = A{x^2} + 2Ax + A + Bx + B + C
Taking like terms together we get,
⇒2x2+x+1=Ax2+(2A+B)x+(A+B+C)\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 x2{x^2}, we get,
A=2A = 2
On comparing coefficients of x2{x^2}, we get,
⇒2A+B=1\Rightarrow 2A + B = 1
On substituting the value of A, we get,
⇒2×2+B=1\Rightarrow 2 \times 2 + B = 1
⇒B=1−4=−3\Rightarrow B = 1 - 4 = - 3
On comparing constants, we get,
A+B+C=1A + B + C = 1
On substituting the value of A and B we get,
⇒2−3+C=1\Rightarrow 2 - 3 + C = 1
⇒C=1+1=2\Rightarrow C = 1 + 1 = 2
Now we can substitute the value of A, B and C in equation (1), we get,
2x2+x+1(x+1)3=2(x+1)−3(x+1)2+2(x+1)3\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 2(x+1)−3(x+1)2+2(x+1)3\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=0x = 0.
⇒2x2+x+1(x+1)3=2×02+0+1(0+1)3\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= 1
Now we can check each of the options giving the valuesx=0x = 0. Atx=0x = 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=32 + 3 - 2 = 3.
Option B becomes, 2−3+2=12 - 3 + 2 = 1
Option C becomes, 2+3+2=52 + 3 + 2 = 5
Option D becomes, 2−3−2=−32 - 3 - 2 = - 3
From the values of the options at x=0x = 0only option B has the same value as that of the expression.
Therefore, the correct answer is option B.