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Unformatted text preview: Again, we can always distribute the “” back through the parenthesis to make sure we get the
© 2007 Paul Dawkins 33 http://tutorial.math.lamar.edu/terms.aspx College Algebra original polynomial.
At this point we can see that we can factor an x out of the first term and a 2 out of the second
term. This gives, x5 + x  2 x 4  2 = x ( x 4 + 1)  2 ( x 4 + 1) We now have a common factor that we can factor out to complete the problem. x5 + x  2 x 4  2 = ( x 4 + 1) ( x  2 ) [Return to Problems] (c) x5  3 x3  2 x 2 + 6
This one also has a “” in front of the third term as we saw in the last part. However, this time the
fourth term has a “+” in front of it unlike the last part. We will still factor a “” out when we
group however to make sure that we don’t lose track of it. When we factor the “” out notice that
we needed to change the “+” on the fourth term to a “”. Again, you can always check that this
was done correctly by multiplying the “” back through the parenthsis. (x 5  3x3 )  ( 2 x 2  6 ) Now that we’ve do...
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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.
 Spring '12
 MrVinh

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