Question

$$m_{x-3}=\frac{x^{4}-81}{x^{2}-9}$$

Answer

$$m=x^3-(3*(27*x^2+2))/x$$

Solution


Add \(3\) to both sides.
\[mx={x}^{4}-81{x}^{2}-9+3\]
Simplify  \({x}^{4}-81{x}^{2}-9+3\)  to  \({x}^{4}-81{x}^{2}-6\).
\[mx={x}^{4}-81{x}^{2}-6\]
Divide both sides by \(x\).
\[m=\frac{{x}^{4}-81{x}^{2}-6}{x}\]
Simplify  \(\frac{{x}^{4}-81{x}^{2}-6}{x}\)  to  \({x}^{3}+\frac{-81{x}^{2}-6}{x}\).
\[m={x}^{3}+\frac{-81{x}^{2}-6}{x}\]
Factor out the common term \(3\).
\[m={x}^{3}+\frac{-3(27{x}^{2}+2)}{x}\]
Move the negative sign to the left.
\[m={x}^{3}-\frac{3(27{x}^{2}+2)}{x}\]