Alg_Complete

# We can then rewrite the original polynomial in terms

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ) 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...
View Full Document

## 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.

Ask a homework question - tutors are online