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Does this help?
John, Here is a similar problem that shows you how it is done. Starting with 3r^{3}  9r, we can see that all the terms have an r, so we can factor that out:
And 3 goes into both terms as well, so
In the parentheses, we have two terms: a squared term and a constant and they are separated by a minus sign. We could use the "difference of squares" factoring to finish this off. A difference of squares factors two positive terms separated by a minus sign. If you have a^{2}  b^{2}, the factors are (ab)(a+b). In your case, the a^{2} is actually r^{2} and the 3 can be thought of as (the square root of 3) squared. This is a little weird to think of the first time, but I am sure you can see that it is true. So we now have
Since each term is linear, we have finished our factoring now. Hope this example helps,
Hi John, Begin by factoring out the common factor of 2p: 2p (p^{2}  17) The factor of (p^{2}  17) may be considered unfactorable (prime) if you are asked to factor the polynomial over the set of natural numbers. However, if you were asked to factor over the set of REAL numbers, it could be treated as a difference of squares and factored as: Hope this helps,  


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