Question 656859
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Presuming you meant:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x^2\ -\ 16}{3x\ -\ 12}]


rather than


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ \frac{16}{3x}\ -\ 12]


or one of the other two interpretations of your ambiguous rendering, then:


Step 1: Factor the difference of two squares in the numerator.


Step 2: Factor a 3 out of the denominator binomial.


Step 3: Eliminate the factor that is now common to both numerator and denominator.


Next time use parentheses to group your numerator and denominator quantities, thus, in this case:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ](x^2 - 16)/(3x - 12)


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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