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Lesson 2.3 Real Zeros of Polynomials. The Division Algorithm. Dividing by a polynomial Set up in long division. 2 terms in divisor (x + 1). How does this go into 1 st two terms in order to eliminate the 1 st term of the dividend. 2x. + 1. Multiply by the divisor

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Presentation Transcript
slide3

Dividing by a polynomial

Set up in long division

2 terms in divisor (x + 1).

How does this go into 1st

two terms in order to

eliminate the 1st term of

the dividend.

2x

+ 1

  • Multiply by the divisor
  • Write product under dividend
  • Subtract
  • Carry down next term
  • Repeat process

-

2x2 + 2x

-

x + 5

-

x + 1

-

4

Answer:

slide4

HINTS:

If a term is missing in the dividend – add a “0” term.

If there is a remainder, put it over the divisor and add it to the quotient (answer)

Example 1

(x4 – x2 + x) ÷ (x2 - x + 1)

slide5

Synthetic Division

  • Less writing
  • Uses addition
  • Setting Up
  • Divisor must be of the form: x – a
  • Use only “a” and coefficients of dividend
  • Write in “zero terms”

x – 2: a = 2

x + 3: a = -3

4 5 0 -2 5

slide7

Steps

  • Bring down
  • Multiply diagonally
  • Add
  • Remainder = last addition
  • Answer
    • Numbers at bottom are coefficients
    • Start with 1 degree less than dividend

REPEAT

slide10

The Remainder Theorem

If f(x) is divided by x – a , the remainder is

r = f(a)

The Factor Theorem

If f(x) has a factor (x – a) then f(a) = 0

slide12

Rational Zero Test

Every rational zero =

Factors of constant term

Factors of leading coefficient

=

slide13

Descartes’ Rule

Number of positivereal roots is:

► the number of variations in the signs, or

► less than that by a positive even integer

5x4 – 3x3 + 2x2 – 7x + 1

variations:

possible positive real roots:

slide15

Example 5

List possible zeros, verify with your calculator which are zeros, and check results with Descartes’ Rule

Problems Set 2.3