`(4x^2+3)/(x-5)^3`

To decompose this fraction, consider the factor in the denominator. The factor is repeated three times. So the partial fraction of this will have denominator in increasing exponent of the factor.

`A/(x-5)` , `B/(x-5)^2` , and `C/(x-5)^3`

Add these three fractions and set it equal to the given fraction.

`(4x^2+3)/(x-5)^3=A/(x-5) + B/(x-5)^2+C/(x-5)^3`

To solve for the values of A, B and C, eliminate the fractions in the equation. So, multiply both sides by the LCD.

`(x-5)^3*(4x^2+3)/(x-5)^3= (A/(x-5)+B/(x-5)^2+C/(x-5)^3)*(x-5)^3`

`4x^2+3=A(x-5)^2+B(x-5)+C`

`4x^2+3=A(x^2-10x+25)+B(x-5)+C`

Then, group together the terms with x^2, the terms with x and the constants.

`4x^2+3=Ax^2 - 10Ax + 25A +Bx - 5B+C`

`4x^2+3=Ax^2 + (-10A + B)x + (25A-5B+C)`

Set the coefficient of x^2 at the left side equal to the coefficient of x^2 at the right side.

`4=A ` This is the value of A.

Then, set the coefficient of x at the left side equal to the coefficient of x at the right side.

` ``0=-10A+B` (Let this be EQ1.)

And, set the constant at the left side equal to the constant at the right side.

`3=25A - 5B + C ` (Let this be EQ2.)

To get the value of B, plug-in A=4 to EQ1.

`0=-10A+B`

`0=-10(4) + B`

`0=-40+B`

`40=B`

And to get the value of C, plug-in A=4 and B=40 to EQ2.

`3=25A - 5B + C`

`3=25(4)-5(40)+C`

`3=100-200+C`

`3=-100+C`

`3+100=C`

`103=C`

So the partial fraction decomposition of the given rational expression is:

`4/(x-5) + 40/(x-5)^2 + 103/(x-5)^3`

To check, express the fractions with same denominators.

`4/(x-5)+40/(x-5)^2+103/(x-5)^3`

`=4/(x-5)*(x-5)^2/(x-5)^2 + 40/(x-5)^2*(x-5)/(x-5)+103/(x-5)^3`

`=(4(x^2-10x+25))/(x-5)^3 + (40(x-5))/(x-5)^3)+103/(x-5)^3`

Now that they have same denominators, proceed to add them.

`= (4(x^2-10x+25) + 40(x-5) + 103)/(x-5)^3`

`=(4x^2-40x+100+40x-200+103)/(x-5)^3`

`=(4x^2+3)/(x-5)^3`

**Therefore, `(4x^2+3)/(x-5)^3=4/(x-5)+40/(x-5)^2+103/(x-5)^3` .**

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