Question 530528
4x + 3
---------------
20x² + 23x + 6


4x + 3
----------------- {factored bottom into two binomials}
(4x + 3)(5x + 2)


1
------ {cancelled 4x + 3 on top and bottom}
5x + 2


Restrictions on x, would be those that would make the denominator equal to 0.


5x + 2 = 0  {set denominator equal to 0}
5x = -2 {subtracted 2 from both sides}
x = -2/5 {divided both sides by 5}
Therefore x cannot be -2/5
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