Answer:
see explanation
Step-by-step explanation:
given
[tex]\frac{x8x^2-8}{4x+4}[/tex] , x ≠ - 1
The x ≠ - 1 means that x cannot take the value of - 1 as this would make the denominator equal to zero which would make the fraction undefined.
Factor the numerator/ denominator
8x² - 8 = 8(x² - 1) ← factor out 8 from both terms
x² - 1 ← is a difference of squares, hence
8x² - 8 = 8(x - 1)(x + 1)
4x + 4 = 4(x + 1) ← factor out 4 from both terms
Thus the fraction now simplifies to
[tex]\frac{8(x-1)(x+1)}{4(x+1)}[/tex]
Cancel the 8 and 4 by 4 and cancel the factor (x + 1), leaving
= 2(x - 1) = 2x - 2 ← in the form ax + b
