Answer:
which is the same as ![\frac{x^2+3x-4}{x^2+6x+8}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E2%2B3x-4%7D%7Bx%5E2%2B6x%2B8%7D)
If the math notation does not load properly, or the font size is too small, then it says ( (x+4)(x-1) )/( (x+4)(x+2) ) which is the same as (x^2+3x-4)/(x^2+6x+8)
==========================================================
Explanation:
We see that x cannot equal -2, as this prevents the denominator (x+2) from becoming zero. If x = -4, then x+4 = 0. So x not allowed to be -4 means (x+4) is prevented in being zero, and this is the missing factor in the denominator. We can never divide by zero.
Multiply top and bottom of this given fraction by (x+4) to get ![\frac{(x+4)(x-1)}{(x+4)(x+2)}](https://tex.z-dn.net/?f=%5Cfrac%7B%28x%2B4%29%28x-1%29%7D%7B%28x%2B4%29%28x%2B2%29%7D)
Note how the (x+4) terms divide and cancel to get back to the original fraction given. The portion
sticks around to make sure the domains line up properly.
----------
Optionally you can expand out each product in the numerator and denominator to get
(x+4)(x-1) = x^2-x+4x-4 = x^2+3x-4
(x+4)(x+2) = x^2+2x+4x+8 = x^2+6x+8
I used the FOIL rule
So, ![\frac{(x+4)(x-1)}{(x+4)(x+2)} = \frac{x^2+3x-4}{x^2+6x+8}](https://tex.z-dn.net/?f=%5Cfrac%7B%28x%2B4%29%28x-1%29%7D%7B%28x%2B4%29%28x%2B2%29%7D%20%3D%20%5Cfrac%7Bx%5E2%2B3x-4%7D%7Bx%5E2%2B6x%2B8%7D)
Set up the equation:
![(-4x - 2) \times 2x](https://tex.z-dn.net/?f=%28-4x%20-%202%29%20%5Ctimes%202x)
Distribute 2x to both of the terms in parentheses:
![-4x \times 2x = -8x^{2}](https://tex.z-dn.net/?f=-4x%20%5Ctimes%202x%20%3D%20-8x%5E%7B2%7D)
![-2 \times 2x = -4x](https://tex.z-dn.net/?f=-2%20%5Ctimes%202x%20%3D%20-4x)
![-8x^{2} - 4x](https://tex.z-dn.net/?f=-8x%5E%7B2%7D%20-%204x)
The answer is
B.
( Terminating Decimal) Hope this helps
Answer:
Idek
Step-by-step explanation:
ask someone or like wait for someone to Answer.Pacific them or