Answer: D.-6 and F.3
Step-by-step explanation:
The given rational number : [tex]\dfrac{(x-3)(x+6)}{(x+7)}[/tex]
To find : The value of x , where the rational expression becomes equal zero.
Let's check all the options :
A. 7
At x= 7 , [tex]\dfrac{(7-3)(7+6)}{(7+7)}=\dfrac{26}{7}\neq0[/tex]
B. -7
At x= 7 , [tex]\dfrac{(-7-3)(-7+6)}{(-7+7)}=\dfrac{10}{0}=\infty\neq0[/tex]
C. 6
At x= 6 , [tex]\dfrac{(6-3)(6+6)}{(6+7)}=\dfrac{36}{13}\neq0[/tex]
D. -6
At x= -6 , [tex]\dfrac{(-6-3)(-6+6)}{(-6+7)}=\dfrac{0}{1}=0[/tex]
E. -3
At x= -3 , [tex]\dfrac{(-3-3)(-3+6)}{(-3+7)}=\dfrac{-9}{2}\neq0[/tex]
F. 3
At x= 3 , [tex]\dfrac{(3-3)(3+6)}{(3+7)}=0[/tex]
Thus, at x= -6 and 3 , the rational expression equals to zero .
So , the correct options are : D.-6 and F.3
