When you factor this, the easiest way to do it is to set the expression equal to 0 and then factor out anything common to get it into a simpler form. 3 goes into both 12 and 21. [tex] 3(4x^2-7)=0 [/tex]. Now of course, [tex] 3\neq 0 [/tex], so that means that [tex] 4x^2-7=0 [/tex]. We solve for x to get [tex] 4x^2=7 [/tex] and [tex] x^2=\frac{7}{4} [/tex] and finally that [tex] x=\sqrt{\frac{7}{4}} [/tex]. 4 is a perfect square that can be simplified, so that leaves us with [tex] x=+/-\frac{\sqrt{7}}{2} [/tex]. If we expand that back out we have the 2 factors as [tex] (4x-\sqrt{7})(4x+\sqrt{7}) [/tex]. Obviously, that is not one of your choices, so your choice is the last one: prime.
