This is a general question
Let's consider the general second degree (quadratic) equation ax^{2 }+ bx + c = 0
If a quadratic expression on the left is to have factors, those factors will be of the form
(qx + p) and (sx + r) where p, q, r and s are all real numbers. So we now have the equation (qx + p) (sx + r) = 0
Either qx + p is 0 or sx + r is 0
or x = -p/q or -r/s which means x is rational. But we know from quadratic formula that x can be solved by
x = (-b + √(b^{2} - 4ac))/2a or (-b - √(b2 - 4ac))/2a
In order for x to be rational, clearly b^{2} - 4ac must be a perfect square
So from the above analysis, in order for a quadratic expression to have factors, b2 - 4ac must be a perfect square.
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