P x x r q x where q x is a polynomial with

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Unformatted text preview: 0 = 5 ( x + 1) ( x - 2 ) (b) Q ( x ) = x8 - 4 x 7 - 18 x 6 + 108 x5 - 135 x 4 = x 4 ( x - 3) (c) ( x + 5) 3 2 R ( x ) = x 7 + 10 x 6 + 27 x5 - 57 x3 - 30 x 2 + 29 x + 20 = ( x + 1) ( x - 1) ( x + 5 )( x - 4 ) 3 Solution In each of these the factoring has been done for us. Do not worry about factoring anything like this. You won’t be asked to do any factoring of this kind anywhere in this material. There are © 2007 Paul Dawkins 250 http://tutorial.math.lamar.edu/terms.aspx College Algebra only here to make the point that the zero factor property works here as well. We will also use these in a later example. (a) P ( x ) = 5 x5 - 20 x 4 + 5 x3 + 50 x 2 - 20 x - 40 = 5 ( x + 1) 2 ( x - 2) 3 Okay, in this case we do have a product of 3 terms however the first is a constant and will not make the polynomial zero. So, from the final two terms it looks like the polynomial will be zero for x = -1 and x = 2 . Therefore, the zeroes of this polynomial are, x = -1 and x = 2 (b) Q ( x ) = x8 - 4 x 7 - 18 x 6 + 108 x5 - 135 x 4 = x 4 ( x - 3) 3 ( x + 5) We...
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