Solveeit Logo
Login

Question

Question: How do you express \(\dfrac{6{{x}^{2}}+1}{{{x}^{2}}{{(x-1)}^{2}}}\) in partial fractions?...

How do you express 6x2+1x2(x−1)2\dfrac{6{{x}^{2}}+1}{{{x}^{2}}{{(x-1)}^{2}}} in partial fractions?

Explanation

Solution

In this question we have the given expression in the form of a fraction which has polynomials in its numerator and denominator. We will first write the polynomial expression in the form of partial fractions with variables A,B,CA,B,C and DD. We will then substitute values such that individual values of the variables can be found directly or by comparison. We will then write values to get the required solution.

Complete step-by-step solution:
We have the given expression as:
⇒6x2+1x2(x−1)2\Rightarrow \dfrac{6{{x}^{2}}+1}{{{x}^{2}}{{(x-1)}^{2}}}
Now the expression can be written in the terms of partial fraction as:
⇒6x2+1x2(x−1)2=Ax+Bx2+C(x−1)+D(x−1)2→(1)\Rightarrow \dfrac{6{{x}^{2}}+1}{{{x}^{2}}{{(x-1)}^{2}}}=\dfrac{A}{x}+\dfrac{B}{{{x}^{2}}}+\dfrac{C}{\left( x-1 \right)}+\dfrac{D}{{{\left( x-1 \right)}^{2}}}\to (1)
On taking the lowest common multiple in the left-hand side, we get:
⇒6x2+1x2(x−1)2=Ax(x−1)2+B(x−1)2+Cx2(x−1)+Dx2x2(x−1)2\Rightarrow \dfrac{6{{x}^{2}}+1}{{{x}^{2}}{{(x-1)}^{2}}}=\dfrac{Ax{{\left( x-1 \right)}^{2}}+B{{\left( x-1 \right)}^{2}}+C{{x}^{2}}\left( x-1 \right)+D{{x}^{2}}}{{{x}^{2}}{{(x-1)}^{2}}}
Now on cancelling the denominator from both the sides, we get:
⇒6x2+1=Ax(x−1)2+B(x−1)2+Cx2(x−1)+Dx2→(2)\Rightarrow 6{{x}^{2}}+1=Ax{{\left( x-1 \right)}^{2}}+B{{\left( x-1 \right)}^{2}}+C{{x}^{2}}\left( x-1 \right)+D{{x}^{2}}\to (2)
Now on substituting x=0x=0, we get:
⇒6(0)2+1=A(0)(0−1)2+B(0−1)2+C(0)2(0−1)+D(0)2\Rightarrow 6{{\left( 0 \right)}^{2}}+1=A\left( 0 \right){{\left( 0-1 \right)}^{2}}+B{{\left( 0-1 \right)}^{2}}+C{{\left( 0 \right)}^{2}}\left( 0-1 \right)+D{{\left( 0 \right)}^{2}}
On simplifying, we get:
⇒1=0+B+0+0\Rightarrow 1=0+B+0+0
Hence, B=1B=1.
Now on substituting x=1x=1, we get:
⇒6(1)2+1=A(1)(1−1)2+B(1−1)2+C(1)2(1−1)+D(1)2\Rightarrow 6{{\left( 1 \right)}^{2}}+1=A\left( 1 \right){{\left( 1-1 \right)}^{2}}+B{{\left( 1-1 \right)}^{2}}+C{{\left( 1 \right)}^{2}}\left( 1-1 \right)+D{{\left( 1 \right)}^{2}}
On simplifying, we get:
⇒7=0+0+0+D\Rightarrow 7=0+0+0+D
Hence, D=7D=7.
Now on substituting B=1B=1 and D=7D=7 in equation (2)(2), we get:
⇒6x2+1=Ax(x−1)2+1(x−1)2+Cx2(x−1)+7x2\Rightarrow 6{{x}^{2}}+1=Ax{{\left( x-1 \right)}^{2}}+1{{\left( x-1 \right)}^{2}}+C{{x}^{2}}\left( x-1 \right)+7{{x}^{2}}
On expanding the terms on the right-hand side, we get:
⇒6x2+1=Ax(x2−2x+1)+(x2−2x+1)+Cx2(x−1)+7x2\Rightarrow 6{{x}^{2}}+1=Ax{{\left( {{x}^{2}}-2x+1 \right)}}+{{\left( {{x}^{2}}-2x+1 \right)}}+C{{x}^{2}}\left( x-1 \right)+7{{x}^{2}}
On simplifying, we get:
⇒6x2+1=A(x3−2x2+x)+(x2−2x+1)+C(x3−x2)+7x2\Rightarrow 6{{x}^{2}}+1=A{{\left( {{x}^{3}}-2{{x}^{2}}+x \right)}}+{{\left( {{x}^{2}}-2x+1 \right)}}+C\left( {{x}^{3}}-{{x}^{2}} \right)+7{{x}^{2}}
Now on grouping like terms, we get:
⇒6x2+1=(A+C)x3+(8−2A−C)x2+(A−2)x+1\Rightarrow 6{{x}^{2}}+1=\left( A+C \right){{x}^{3}}+\left( 8-2A-C \right){{x}^{2}}+\left( A-2 \right)x+1
On comparing the coefficients on both the sides, we get:
A+C=0→(3)A+C=0\to (3)
8−2A−C=6→(4)8-2A-C=6\to (4)
A−2=0→(5)A-2=0 \to (5)
From equation (5)(5), we get:
A=2A=2
Substituting A=2A=2 in equation (3)(3), we get:
2+C=02+C=0
Therefore, C=−2C=-2.
On substituting the values of all the coefficients in equation (1)(1), we get:
⇒6x2+1x2(x−1)2=2x+1x2+−2(x−1)+7(x−1)2\Rightarrow \dfrac{6{{x}^{2}}+1}{{{x}^{2}}{{(x-1)}^{2}}}=\dfrac{2}{x}+\dfrac{1}{{{x}^{2}}}+\dfrac{-2}{\left( x-1 \right)}+\dfrac{7}{{{\left( x-1 \right)}^{2}}}, which is the required solution.

Note: It is to be remembered that the partial fraction skeleton is different for different values of the polynomial expressions. The main application of using partial fractions is in integration when the integration has to be performed on complex polynomial fractions which cannot be directly integrated.