Question 426540
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Hi
(8x^2 + 1)/(2x^2-5x-3)   |cannot divide by ZERO
 2x^2 -5x - 3 = 0
factoring
(2x +1)(x-3) = 0
(2x +1)=0   x = -1/2
(x-3) = 0   x = 3
{x: x &#8800; 3 and x &#8800; -1/2}  the set of all real numbers except 3 and -1/2