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Unformatted text preview: = 3 So, this second degree polynomial has two zeroes or roots.
Now, let’s find the zeroes for P ( x ) = x 2 - 14 x + 49 . That will mean solving, x 2 - 14 x + 49 = ( x - 7 ) = 0
2 Þ x=7 So, this second degree polynomial has a single zero or root. Also, recall that when we first
looked at these we called a root like this a double root.
We solved each of these by first factoring the polynomial and then using the zero factor property
on the factored form. When we first looked at the zero factor property we saw that it said that if
the product of two terms was zero then one of the terms had to be zero to start off with.
The zero factor property can be extended out to as many terms as we need. In other words, if
we’ve got a product of n terms that is equal to zero, then at least one of them had to be zero to
start off with. So, if we could factor higher degree polynomials we could then solve these as
Let’s take a look at a couple of these. Example 1 Find the zeroes of each of the following polynomials.
(a) P ( x ) = 5 x5 - 20 x 4 + 5 x3 + 50 x 2 - 20 x - 4...
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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