Answer:
[tex]\frac{5x-4}{x(x^2+7)^2} = \frac{A}{x} + \frac{Bx+C}{x^2+7} + \frac{Dx+E}{(x^2+7)^2}[/tex]
Step-by-step explanation:
Given the expression [tex]\frac{5x-4}{x(x^2+7)^2}[/tex], we are to re-write the expression in form of a partial fraction.
Before we write in form of a partial fraction, we need to note the expression at the denominator. Since the expression in parenthesis is a quadratic equation, the equivalent numerator must be a linear expression.
Also the quadratic equation is a repeated form since it is squared. This means that we are to repeat the quadratic equation twice when writing as a partial fraction.
[tex]\frac{5x-4}{x(x^2+7)^2} = \frac{A}{x} + \frac{Bx+C}{x^2+7} + \frac{Dx+E}{(x^2+7)^2}[/tex]
From the above partial fraction, it can be seen that x² + 7 in parenthesis was repeated twice and their equivalent expressions at the numerator are both linear i.e Bx+E and Dx+ E where A, B, C, D and E are the unknown constant.
