Question 985153
<font face="Times New Roman" size="+2">


In order to be a perfect square, all prime factors of a number must be represented an even number of times.


The prime factorization of 48 is *[tex \Large 2\ \times\ 2\ \times\ 2\ \times\ 2\ \times\ 3].  You have an even number of factors of 2, but only 1 factor of 3.  If you include another factor of 3, then there will be four 2s and two 3s, which results in 48 times 3 or 144.  144 is indeed a perfect square.


Do the other problems the same way.  By the way, next time read and follow instructions:  One problem per post.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \