Unformatted text preview: 4 x 2  2 x + 5 ) Note that we can always check our factoring by multiplying the terms back out to make sure we
get the original polynomial.
[Return to Problems] (b) x3 y 2 + 3 x 4 y + 5 x5 y 3
In this case we have both x’s and y’s in the terms but that doesn’t change how the process works.
Each term contains and x3 and a y so we can factor both of those out. Doing this gives, x3 y 2 + 3 x 4 y + 5 x5 y 3 = x3 y ( y + 3x + 5 x 2 y 2 ) [Return to Problems] (c) 3 x 6  9 x 2 + 3 x
In this case we can factor a 3x out of every term. Here is the work for this one. 3x 6  9 x 2 + 3 x = 3 x ( x5  3 x + 1) Notice the “+1” where the 3x originally was in the final term, since the final term was the term we
factored out we needed to remind ourselves that there was a term there originally. To do this we
need the “+1” and notice that it is “+1” instead of “1” because the term was originally a positive
term. If it had been a negative term originally we would have had to use “1”.
One of the more common mistak...
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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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