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Unformatted text preview: he division algorithm and see what we get,
© 2007 Paul Dawkins 248 http://tutorial.math.lamar.edu/terms.aspx College Algebra P (r ) = (r  r )Q (r ) + R
= ( 0) Q ( r ) + R
=R
Now, that’s convenient. The remainder of the division algorithm is also the value of the
polynomial evaluated at r. So, from our previous examples we now know the following function
evaluations. If P ( x ) = 5 x3  x 2 + 6 then P ( 4 ) = 310
If P ( x ) = 2 x 3  3 x  5 then P ( 2 ) = 15
If P ( x ) = 4 x 4  10 x 2 + 1 then P ( 6 ) = 4825
This is a very quick method for evaluating polynomials. For polynomials with only a few terms
and/or polynomials with “small” degree this may not be much quicker that evaluating them
directly. However, if there are many terms in the polynomial and they have large degrees this can
be much quicker and much less prone to mistakes than computing them directly.
As noted, we will be using this fact in a later section to greatly reduce the amount of work we’ll
need to do in those problems. © 2007 Paul Dawkins 249...
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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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