Question

Hello, I need some assistance with this precalculus homework question, please?HW Q16

Answer 1

The expression is given below as

`(x^2)/((x-2)^2(x+3))`

Concept:

We will create a template using the denominator

`\begin{gathered} (x^2)/(\left(x-2\right)^2\left(x+3\right))=(A)/((x-2))+(B)/((x-2)^2)+(C)/((x+3)) \\ \end{gathered}`

By cross multiplying, we will have

`\begin{gathered} (x^(2))/((x-2)^(2)(x+3))=(A)/((x-2))+(B)/((x-2)^(2))+(C)/((x+3)) \\ x^2=A(x-2)(x+3)+B(x+3)+C(x-2)^2 \end{gathered}`

Step 1:

Put x=2

`\begin{gathered} x^2=A(x-2)(x+3)+B(x+3)+C(x-2)^2 \\ 2^2=A(2-2)(2+3)+B(2+3)+C(2-2)^2 \\ 4=5B \\ divide\text{ both sides by 5} \\ (5B)/(5)=(4)/(5) \\ B=(4)/(5) \end{gathered}`

Step 2:

Put x=-3

`\begin{gathered} x^2=A(x-2)(x+3)+B(x+3)+C(x-2)^2 \\ (-3)^2=A(-3-2)(-3+3)+B(-3+3)+C(-3-2)^2 \\ 9=25C \\ divide\text{ both sides by 25} \\ (25C)/(25)=(9)/(25) \\ C=(9)/(25) \end{gathered}`

Step 3:

Expand the brackets

`\begin{gathered} x^(2)=A(x-2)(x+3)+B(x+3)+C(x-2)^(2) \\ x^2=A(x^2+3x-2x-6)+Bx+3B+C(x^2-4x+4) \\ x^2=Ax^2+3Ax-6A+Bx+3B+Cx^2-4Cx+4C \\ By\text{ comparing coefficient, we will have} \\ A+C=1 \\ A+(9)/(25)=1 \\ A=1-(9)/(25) \\ A=(25-9)/(25) \\ A=(16)/(25) \end{gathered}`

Hence,

The partial fraction decomposition will be

`\Rightarrow(16)/(25(x-2))+(4)/(5(x-2)^2)+(9)/(25(x+3))`