Answer:
[tex]\frac{4a-2}{2a+1}[/tex]
Step-by-step explanation:
Factorise the numerator and denominator
8a² - 2 ← factor out 2 from each term
= 2(4a² - 1) ← 4a² - 1 is a difference of squares
= 2(2a - 1)(2a + 1)
4a² + 4a + 1 ← is a perfect square
= (2a + 1)²
Thus
[tex]\frac{8a^2-2}{4a^2+4a+1}[/tex]
= [tex]\frac{2(2a-1)(2a+1)}{(2a+1)(2a+1)}[/tex] ← cancel (2a + 1) on numerator/ denominator
= [tex]\frac{2(2a-1)}{2a+1}[/tex]
= [tex]\frac{4a-2}{2a+1}[/tex]
