Question

How do you express $\dfrac{{{x}^{2}}-3x+2}{4{{x}^{3}}+11{{x}^{2}}}$ in partial fractions?

Answer 1

To express $\dfrac{{{x}^{2}}-3x+2}{4{{x}^{3}}+11{{x}^{2}}}$ in partial fractions, we are going to take ${{x}^{2}}$ as common in the denominator of the above expression. Then we are going to equate this expression to $\dfrac{A}{{{x}^{2}}}+\dfrac{B}{x}+\dfrac{C}{4x+11}$. After equating, we are going to take the L.C.M of this addition and then will find the values of A, B and C.

Complete step by step solution:
The expression given in the above problem which we have to write in partial fraction form is as follows:
$\dfrac{{{x}^{2}}-3x+2}{4{{x}^{3}}+11{{x}^{2}}}$
Taking ${{x}^{2}}$ as common from the denominator of the above expression we get,
$\Rightarrow \dfrac{{{x}^{2}}-3x+2}{{{x}^{2}}\left( 4x+11 \right)}$ ………….. (1)
Now, to write the above fraction in partial fraction form, we are going to equate the above expression to:
$\dfrac{A}{{{x}^{2}}}+\dfrac{B}{x}+\dfrac{C}{4x+11}$
Equating the relation (1) to the above expression we get,
$\Rightarrow \dfrac{{{x}^{2}}-3x+2}{{{x}^{2}}\left( 4x+11 \right)}=\dfrac{A}{{{x}^{2}}}+\dfrac{B}{x}+\dfrac{C}{4x+11}$
Taking L.C.M in the denominator in the R.H.S of the above equation we get,
$\Rightarrow \dfrac{{{x}^{2}}-3x+2}{{{x}^{2}}\left( 4x+11 \right)}=\dfrac{A\left( 4x+11 \right)+Bx\left( 4x+11 \right)+C{{x}^{2}}}{{{x}^{2}}\left( 4x+11 \right)}$
Solving the numerator in the R.H.S of the above equation we get,
$\Rightarrow \dfrac{{{x}^{2}}-3x+2}{{{x}^{2}}\left( 4x+11 \right)}=\dfrac{4Ax+11A+4B{{x}^{2}}+11Bx+C{{x}^{2}}}{{{x}^{2}}\left( 4x+11 \right)}$
Now, the denominator is same in both the sides of the above equation so it will be cancelled out on both the sides and we get,
$\Rightarrow {{x}^{2}}-3x+2=4Ax+11A+4B{{x}^{2}}+11Bx+C{{x}^{2}}$
In the R.H.S of the above function, we are combining the terms which have same power of x and we get,
$\Rightarrow {{x}^{2}}-3x+2={{x}^{2}}\left( 4B+C \right)+\left( 4A+11B \right)x+11A$
Now, equating the coefficient of ${{x}^{2}},x$ and constant terms on both the sides we get,
$\begin{aligned} & 1=4B+C........(2) \\\ & -3=4A+11B.......(3) \\\ & 2=11A........(4) \\\ \end{aligned}$
Dividing 11 on both the sides of eq. (4) we get,
$\Rightarrow \dfrac{2}{11}=A$
Substituting the above value of A in eq. (3) we get,
$\begin{aligned} & \Rightarrow -3=4\left( \dfrac{2}{11} \right)+11B \\\ & \Rightarrow -3=\dfrac{8+121B}{11} \\\ \end{aligned}$
Cross multiplying the above equation we get,
$\begin{aligned} & -3\left( 11 \right)=8+121B \\\ & \Rightarrow -33=8+121B \\\ \end{aligned}$
Subtracting 8 on both the sides of the above equation we get,
$\begin{aligned} & \Rightarrow -33-8=121B \\\ & \Rightarrow -41=121B \\\ & \Rightarrow -\dfrac{41}{121}=B \\\ \end{aligned}$
Substituting the above value of B in eq. (2) we get,
$\begin{aligned} & 1=4\left( -\dfrac{41}{121} \right)+C \\\ & \Rightarrow 1=-\dfrac{164}{121}+C \\\ & \Rightarrow 1+\dfrac{164}{121}=C \\\ & \Rightarrow \dfrac{121+164}{121}=C \\\ & \Rightarrow \dfrac{285}{121}=C \\\ \end{aligned}$
From the above we got the following values of A, B and C:
$\Rightarrow \dfrac{2}{11}=A,B=-\dfrac{41}{121},C=\dfrac{285}{121}$
Now, substituting the above values of A, B and C in $\dfrac{A}{{{x}^{2}}}+\dfrac{B}{x}+\dfrac{C}{4x+11}$ and we get,
$\Rightarrow \dfrac{2}{11{{x}^{2}}}-\dfrac{41}{121x}+\dfrac{285}{121\left( 4x+11 \right)}$
Hence, we have written the given fraction into partial fraction form.

Note: The mistake which could be possible in the above problem is the calculation mistake and especially the positive or negative signs and the power of x too so make sure you will be alert while doing these calculations.