Then for each factor in the denominator we can use

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Unformatted text preview: r P ( x ) . x =1 x=2 x=3 ( multiplicity 2 ) ( multiplicity 1) ( multiplicity 1) Note that x = 1 has a multiplicity of 2 since it showed up twice in our work above. Before moving onto the next example let’s also note that we can now completely factor the polynomial P ( x ) = x 4 - 7 x3 + 17 x 2 - 17 x + 6 . We know that each zero will give a factor in the factored form and that the exponent on the factor will be the multiplicity of that zero. So, the factored form is, P ( x ) = x 4 - 7 x3 + 17 x 2 - 17 x + 6 = ( x - 1) ( x - 2 )( x - 3) 2 Let’s take a look at another example. Example 4 Find all the zeroes of P ( x ) = 2 x 4 + x 3 + 3 x 2 + 3 x - 9 . Solution From the second example we know that the list of all possible rational zeroes is, ±1 = ±1 ±1 ±3 = ±3 ±1 ±9 = ±9 ±1 ±1 1 =± ±2 2 ±3 3 =± ±2 2 ±9 9 =± ±2 2 The next step is to build up the synthetic division table. When we’ve got fractions it’s usually best to start with the integers and do those first. Also, this time we’ll start w...
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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.

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