Return to problems d 9 x 2 2 x 7 12 x 2 x 7

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Unformatted text preview: es with these types of factoring problems is to forget this “1”. © 2007 Paul Dawkins 32 http://tutorial.math.lamar.edu/terms.aspx College Algebra Remember that we can always check by multiplying the two back out to make sure we get the original. To check that the “+1” is required, let’s drop it and then multiply out to see what we get. 3x ( x5 - 3x ) = 3x6 - 9 x 2 ¹ 3x6 - 9 x 2 + 3x So, without the “+1” we don’t get the original polynomial! Be careful with this. It is easy to get in a hurry and forget to add a “+1” or “-1” as required when factoring out a complete term. [Return to Problems] (d) 9 x 2 ( 2 x + 7 ) - 12 x ( 2 x + 7 ) This one looks a little odd in comparison to the others. However, it works the same way. There is a 3x in each term and there is also a 2 x + 7 in each term and so that can also be factored out. Doing the factoring for this problem gives, 9 x 2 ( 2 x + 7 ) - 12 x ( 2 x + 7 ) = 3x ( 2 x + 7 ) ( 3x - 4 ) [Return to Problems] Factoring By Grouping This is a method that isn’t used all that often, but when it can be used it can be somewhat useful. This method is best illustrated with an example...
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.

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