Question

The partial fraction of $\frac{3x^3-8x^2+10}{{\left(x-1\right)}^4}$ is

• A. $\frac{3}{\left(x-1\right)}+\frac{1}{{\left(x-1\right)}^2}+\frac{7}{{\left(x-1\right)}^3}+\frac{5}{{\left(x-1\right)}^4}$
• B. $\frac{3}{\left(x-1\right)}+\frac{1}{{\left(x-1\right)}^2}-\frac{7}{{\left(x-1\right)}^3}+\frac{5}{{\left(x-1\right)}^4}$
• C. $\frac{3}{\left(x-1\right)}+\frac{1}{{\left(x-1\right)}^2}-\frac{7}{{\left(x-1\right)}^3}-\frac{5}{{\left(x-1\right)}^4}$
• D. None of the above

Answer 1

The correct option is B

$\frac{3}{\left(x-1\right)}+\frac{1}{{\left(x-1\right)}^2}-\frac{7}{{\left(x-1\right)}^3}+\frac{5}{{\left(x-1\right)}^4}$

Explanation for the correct option:

Step 1: Write in the form of a partial equation.

The given fraction is $\frac{3x^3-8x^2+10}{{\left(x-1\right)}^4}$.

Let, the given fraction be resolved into a partial fraction as $\frac{A}{\left(x-1\right)}+\frac{B}{{\left(x-1\right)}^2}+\frac{C}{{\left(x-1\right)}^3}+\frac{D}{{\left(x-1\right)}^4}$.

So, $\frac{3x^3-8x^2+10}{{\left(x-1\right)}^4}=\frac{A}{\left(x-1\right)}+\frac{B}{{\left(x-1\right)}^2}+\frac{C}{{\left(x-1\right)}^3}+\frac{D}{{\left(x-1\right)}^4}$

$\Rightarrow$ $\frac{3x^3-8x^2+10}{{\left(x-1\right)}^4}=\frac{A{\left(x-1\right)}^3+B{\left(x-1\right)}^2+C\left(x-1\right)+D}{{\left(x-1\right)}^4}$

$\Rightarrow$ $3x^3-8x^2+10=A{\left(x-1\right)}^3+B{\left(x-1\right)}^2+C\left(x-1\right)+D$ $...\left(\mathrm{i}\right)$

Step 2: Form relations between $A,B,C$ and $D$.

Substituting $x=1$ in the equation $\left(\mathrm{i}\right)$, we get,

$3{\left(1\right)}^3-8{\left(1\right)}^2+10=A{\left(1-1\right)}^3+B{\left(1-1\right)}^2+C\left(1-1\right)+D$

$\Rightarrow$ $3-8+10=A{\left(0\right)}^3+B{\left(0\right)}^2+C\left(0\right)+D$

$\Rightarrow$ $5=D$

Substitute $D=5$ in the equation $\left(\mathrm{i}\right)$ and simplify it,

$$\begin{gathered} 3x^3-8x^2+10=A{\left(x-1\right)}^3+B{\left(x-1\right)}^2+C\left(x-1\right)+5 \\ \Rightarrow 3x^3-8x^2+10=A\left(x^3-1-3x^2+3x\right)+B\left(x^2+1-2x\right)+C\left(x-1\right)+5 \\ \Rightarrow 3x^3-8x^2+10=A{x}^3-A-3A{x}^2+3Ax+B{x}^2+B-2Bx+Cx-C+5 \\ \Rightarrow 3x^3-8x^2+10=A{x}^3-3A{x}^2+B{x}^2+3Ax-2Bx+Cx+B-C-A+5 \\ \Rightarrow 3x^3-8x^2+10=A{x}^3-\left(3A-B\right)x^2+\left(3A-2B+C\right)x+\left(B-C-A+5\right) \end{gathered}$$

On comparing the coefficients of $x^3,x^2,x$ and the constant terms of both sides on the above equation,

$A=3$

$3A-B=8$ $...\left(\mathrm{ii}\right)$

$3A-2B+C=0$ $...\left(\mathrm{iii}\right)$

And, $B-C-A+5=10$

Step 3: Calculate the values of $B$ and $C$.

Substituting $A=3$ in the equation $\left(\mathrm{ii}\right)$,

$\begin{array}{cr}\Rightarrow & 3A-B=8\\ \Rightarrow & 3\times 3-B=8\\ \Rightarrow & 9-B=8\\ \Rightarrow & B=1\end{array}$

Substituting $A=3$ and $B=1$in the equation $\left(\mathrm{iii}\right)$,

$3A-2B+C=0$

$\Rightarrow$ $3\times 3-2\times 1+C=0$

$\Rightarrow$ $9-2+C=0$

$\Rightarrow$ $C=-7$

Step 4: Determine the required partial fraction.

As we had supposed,

The required partial fraction $=\frac{A}{\left(x-1\right)}+\frac{B}{{\left(x-1\right)}^2}+\frac{C}{{\left(x-1\right)}^3}+\frac{D}{{\left(x-1\right)}^4}$

So, on substituting $A=3,B=1$ and $C=-7$ in the above we get,

The required partial fraction $=\frac{3}{\left(x-1\right)}+\frac{1}{{\left(x-1\right)}^2}+\frac{\left(-7\right)}{{\left(x-1\right)}^3}+\frac{5}{{\left(x-1\right)}^4}$

$\Rightarrow$ The required partial fraction $=\frac{3}{\left(x-1\right)}+\frac{1}{{\left(x-1\right)}^2}-\frac{7}{{\left(x-1\right)}^3}+\frac{5}{{\left(x-1\right)}^4}$

Hence, option $B$ is the correct option.