SOLUTION: i have worked on this problem and have several different answerw 5sqrt96a^12b^8 i came up with 2a^2b^3 5sqrt24

Question 35659: i have worked on this problem and have several different answerw
5sqrt96a^12b^8
i came up with 2a^2b^3 5sqrt24
Answer by rapaljer(4671) (Show Source): You can put this solution on YOUR website!
There is NO SUCH THING as a "5sqrt" of a quantity. Do you mean or do you mean . This could be your problem. If indeed you mean , then I can do that for you.

HOWEVER, I continue to insist that the problem PROBABLY was this:

If it is a 5th root problem, then you must find perfect 5th powers that divide into this quantity. The ONLY perfect 5th power that is small enough to be used in most problems is . It just happens (a coincidence???) that 32 is a factor of 96 and it divides into 96 exactly 3 times. Also, when you have powers involved, like or , you need to find a power that is DIVISIBLE by 5, which would be like , , , etc. Now, sort this out into TWO 5th root radicals, and place all the perfect 5th powers up front in the first radical sign, and place all the leftover factors in the second radical. In this case, you will have to use and as the perfect powers of "a" and "b" respectively. When you do a 5th root of a quantity that is raised to a power, you DIVIDE exponents. For example and or .

The first radical is a perfect 5th power, so you can actually DO this one. The second radical, containing all the left-overs, will have to stay in the radical sign.

If this problem was too deep for you, or if you need additional help, check out my Lesson Plans in algebra.com for Radicals (under Square Roots) at Intermediate or College Algebra level, or see my own website under "Math in Living Color" and look for those particular topics at those levels of math.

R^2 at SCC

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