Answer: [tex]\frac{(x+4)(x-1)}{(x+4)(x+2)}[/tex] which is the same as [tex]\frac{x^2+3x-4}{x^2+6x+8}[/tex]
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)
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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 [tex]\frac{(x+4)(x-1)}{(x+4)(x+2)}[/tex]
Note how the (x+4) terms divide and cancel to get back to the original fraction given. The portion [tex]x \ne -4[/tex] sticks around to make sure the domains line up properly.
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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, [tex]\frac{(x+4)(x-1)}{(x+4)(x+2)} = \frac{x^2+3x-4}{x^2+6x+8}[/tex]
