Factors of 3x²±1÷3y² are

Question

asked Mar 25, 2020 40.7k views

1 Answer

Lellansin · Answer 1 · 2020-03-31T15:24:10+0000

Answer:

$(1)/(3) ,(1)/(y) , (1)/(y^(2) ) \textrm { and } (3x^(2) + 1)$

$(1)/(3) ,(1)/(y) , (1)/(y^(2) ), (√(3) x + 1) \textrm { and } (√(3) x - 1)$

Explanation:

We have to find the factors of the expression (3x² ± 1) ÷ 3y²

Now, (3x² ± 1) ÷ 3y² = (3x² + 1) ÷ 3y² and (3x² - 1) ÷ 3y²

Now, factors of (3x² + 1) ÷ 3y² are
$(1)/(3) ,(1)/(y) , (1)/(y^(2) ) \textrm { and } (3x^(2) + 1)$

Again the factors of (3x² - 1) ÷ 3y² are
$(1)/(3) ,(1)/(y) , (1)/(y^(2) ), (√(3) x + 1) \textrm { and } (√(3) x - 1)$ (Answer)