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2 2 So, we can use the third special form from above. x 4 - 25 = ( x 2 + 5 ) ( x 2 - 5 ) Neither of these can be further factored and so we are done. Note however, that often we will
need to do some further factoring at this stage.
[Return to Problems] (c) x 4 + x 2 - 20
Let’s start this off by working a factoring a different polynomial. u 2 + u - 20 = ( u - 4 ) ( u + 5 ) We used a different variable here since we’d already used x’s for the original polynomial. © 2007 Paul Dawkins 39 http://tutorial.math.lamar.edu/terms.aspx College Algebra () So, why did we work this? Well notice that if we let u = x 2 then u 2 = x 2 2 = x 4 . We can then rewrite the original polynomial in terms of u’s as follows, x 4 + x 2 - 20 = u 2 + u - 20
and we know how to factor this! So factor the polynomial in u’s then back substitute using the
fact that we know u = x 2 . x 4 + x 2 - 20 = u 2 + u - 20
= ( u - 4 )( u + 5 ) = ( x 2 - 4 )( x 2 + 5 )
Finally, notice that the first term will also factor since it is th...
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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