Solution:
Given:
[tex]\frac{2x^2-6x}{x^2-9}\div\frac{4x^3+28x^2}{x^2+8x+15}[/tex][tex]\begin{gathered} \frac{2x^2-6x}{x^2-9}\times\frac{x^2+8x+15}{4x^3+28x^2} \\ \\ To\text{ simplify the expression, we n}eed\text{ to factorize each expression and cancel them out.} \\ \text{Hence,} \\ \frac{2x(x-3)}{(x-3)(x+3)}\times\frac{(x+3)(x+5)}{4x^2(x+7)} \end{gathered}[/tex]
Canceling out the common factors,
[tex]\begin{gathered} \frac{2x(x-3)}{(x-3)(x+3)}\times\frac{(x+3)(x+5)}{4x^2(x+7)} \\ \frac{(x+5)}{2x(x+7)} \end{gathered}[/tex]
Thus,
[tex]\frac{2x^2-6x}{x^2-9}\div\frac{4x^3+28x^2}{x^2+8x+15}=\frac{(x+5)}{2x(x+7)}[/tex]
The restrictions will occur for the expression to be undefined.
The expression will be undefined when the denominator is zero.
[tex]\begin{gathered} \frac{(x+5)}{2x(x+7)} \\ \text{Hence,} \\ 2x(x+7)=0 \\ 2x=0\text{ OR x + 7 = 0} \\ x=\frac{0}{2}\text{ OR x = 0 - 7} \\ x=0\text{ OR x = -7} \\ \\ \text{Hence, there are restrictions at x = 0 and x = -7 because the expression will be undefined at these two points} \end{gathered}[/tex]
Therefore,
[tex]\begin{gathered} \frac{(x+5)}{2x(x+7)}\text{ will have the following restrictions,} \\ x\ne0\text{ and x }\ne-7 \end{gathered}[/tex]
