Answer: First Option
[tex]\frac{x(x-4)(x-1)}{2(x+4)}[/tex]
Step-by-step explanation:
We have the following expression
[tex]\frac{x^2-16}{2x+8}*\frac{x^3-2x^2+x}{x^2+3x-4}[/tex]
First factorize the denominators:
[tex]\frac{x^2-16}{2(x+4)}*\frac{x^3-2x^2+x}{(x+4)(x-1)}[/tex]
Now we factor the numerators
[tex]\frac{(x-4)(x+4)}{2(x+4)}*\frac{x(x^2-2x+1)}{(x+4)(x-1)}[/tex]
[tex]\frac{(x-4)(x+4)}{2(x+4)}*\frac{x(x-1)^2}{(x+4)(x-1)}[/tex]
now we simplify the expression
[tex]\frac{(x-4)}{2}*\frac{x(x-1)}{(x+4)}[/tex]
[tex]\frac{x(x-4)(x-1)}{2(x+4)}[/tex]
The answer is the first option
