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Unformatted text preview: exists? The answer comes
from our old friend, polynomial division. Dividing x3 + 4x2 − 5x − 14 by x − 2 gives
x2 + 6x + 7
x− 2 x3 + 4x2 − 5x − 14
− x3 − 2x2
6x2 − 5x
− 6x2 − 12x)
7x − 14
− (7x − 14)
As you may recall, this means x3 + 4x2 − 5x − 14 = (x − 2) x2 + 6x + 7 , and so to ﬁnd the zeros
of f , we now can solve (x − 2) x2 + 6x + 7 = 0. We get x − 2 = 0 (which gives us our known
zero, x = 2) as well as x2 + 6x + 7 = 0. The latter doesn’t factor nicely, so we apply the Quadratic
Formula to get x = −3 ± 2. The point of this section is to generalize the technique applied here.
First up is a friendly reminder of what we can expect when we divide polynomials.
Theorem 3.4. Polynomial Division: Suppose d(x) and p(x) are nonzero polynomials where
the degree of p is greater than or equal to the degree of d. There exist two unique polynomials,
q (x) and r(x), such that p(x) = d(x) q (x) + r(x), where either r(x) = 0 or the degree of r is
strictly less than the degree of d.
2 and probably forgot
pun intended 198 Polynomial Functions As you may recall, a...
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