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Unformatted text preview: or two. Example 2 Factor by grouping each of the following.
(a) 3 x 2  2 x + 12 x  8 [Solution]
(b) x5 + x  2 x 4  2 [Solution]
(c) x5  3 x3  2 x 2 + 6 [Solution]
Solution
(a) 3 x 2  2 x + 12 x  8
In this case we group the first two terms and the final two terms as shown here, ( 3x 2  2 x ) + (12 x  8 ) Now, notice that we can factor an x out of the first grouping and a 4 out of the second grouping.
Doing this gives, 3 x 2  2 x + 12 x  8 = x ( 3 x  2 ) + 4 ( 3 x  2 )
We can now see that we can factor out a common factor of 3 x  2 so let’s do that to the final
factored form. 3x 2  2 x + 12 x  8 = ( 3 x  2 ) ( x + 4 ) And we’re done. That’s all that there is to factoring by grouping. Note again that this will not
always work and sometimes the only way to know if it will work or not is to try it and see what
you get.
[Return to Problems] (b) x5 + x  2 x 4  2
In this case we will do the same initial step, but this time notice that both of the final two terms
are negative so we’ll factor out a “” as well when we group them. Doing this gives, (x 5 + x )  ( 2 x4 + 2)...
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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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