Polynomial and Synthetic Division, p' 289294

1
Polynomial and Synthetic Division, p. 289-294

• OBJECTIVES
• How to use long division to divide polynomials by
  other polynomials
• How to use synthetic division to divide
  polynomials by binomials of the form (x k)
• How to use the Remainder Theorem and Factor
  Theorem

2
p. 285 42

• Find all real zeroes of f (x) x3 4x2 25x
  100
• Real zeroes are x intercepts, thus let f (x)
  0 .
• 0 x2 ( x 4 ) 25 ( x 4 )
• 0 ( x2 25 ) ( x 4 )
• 0 ( x 5 ) ( x 5 ) ( x 4 )
• ( x 5 ) 0 , ( x 5 ) 0 , ( x 4 ) 0
• x 5 , x 5 , x 4 are the three
  zeroes.

3
Notice, if f (x) x3 4x2 25x 100 , then
f (4) (4)3 4(4)2 25(4) 100 f (4) 64
64 100 100 0 and x 4 is a zero. This is
an example of the following theorem

• Remainder Theorem p. 293
• If a polynomial f (x) is divided by x k, then
  the remainder is r f (k).

4
Since f (4) 0 , x 4 is a factor of f
(x) x3 4x2 25x 100 , and x 4 is a
zero. This is an example of the following
theorem

• The Factor Theorem p. 293
• A polynomial f (x) has a factor (x k) if and
  only if f (k) 0.

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x3 4x2 25x 100 ( x 4 ) ( x 2 25 ) 0
Dividend Divisor
QuotientRemainder
6
x3 4x2 25x 100 ( x2 25 )( x 4 )
x2 25
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The Division Algorithm p. 290
Let f(x) and d(x) are polynomials such that d(x)
0. If the degree of d(x) is less than or
equal to the degree of f(x), there exist unique
polynomials q(x) and r(x) such that f(x)
d(x)q(x) r(x) where r(x) 0 or the degree of
r(x) is less than the degree of d(x). If the
remainder r(x) is zero, d(x) divides evenly into
f(x).
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Synthetic Division p. 292

• Consider f (x) x3 4x2 25x 100 and x
  4.
• 1 4 25 100

4
0
100
4
1
0
25
0
Thus, the quotient is x2 0x 25 or x2
25.
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Uses of the Remainder in Synthetic Division p.
294

• The remainder r, obtained in the synthetic
  division of f (x) by x k, provides the
  following information
• 1. The remainder r gives the value of f at x
  k. That is, r f(k).
• 2. If r 0, (x k) is a factor of f(x).
• 3. If r 0, (k,0) is an x-intercept of the graph
  of f.

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p. 295 58 List all real zeroes
f (x) 3x3 2x2 19x 6 and x 3 .

• 3 2 19 6

-3
6
9
21
0
3
7
2
0 3x2 7x 2 0 ( 3x 1 ) ( x 2 ) , x
1/3 or x 2
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Homework

• P. 290-292 1-64 alt odd ( 57-64 skip part e)
• Synthesis 75-86
• Review 87-96
• Read p. 298-307
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