PreCalc 2.3 2.3 Polynomial/Synthetic Division
DAY 2
Today you will learn: The Remainder Theorem and the Factor Theorem and its relevance when working with polynomials PreCalc 2.3 Remainder Theorem
If a polynomial f(x) is divided by x­k, the remainder is r = f(k)
Directions mean "find the remainder" Note: No work is needed for these
5. Use the remainder theorem to evaluate the following functions for the given values of x. Option #1: you can synthetically divide to find the remainder
­6 ­4 ­2
2 1 ­9
This option
gives both q(x)
and r(x), but you
only need r(x)
Option #2: you can replace the x value in the function and evaluate. ­9
This option gives
you only r(x)
Option #1: you can synthetically divide to find the remainder
­4 7
­7 ­1
Option #2: you can replace the x value in the function and evaluate. ­1
This option gives
you only r(x)
Calculator instructions
TI­84
This option
gives both q(x)
and r(x), but you
only need r(x)
Calculator instructions
TI­Nspire
TI­84
TI­Nspire
Next, type in the function
Next, type in the function
you should get the answer ­9
you should get the answer ­1
­9
­9
­9
­1
­1
­1
PreCalc 2.3 Note: Do NOT synthetically divide these problems to find the answer!!! It will be too long of a problem c) f(x)=x 30 ­3 if x=­1
f(­1)=(­1)30­3
=1­3
= ­2
d) f(x)=x 57 +5 if x=­1
f(­1)=(­1)57+5
= ­1 +5
=4
PreCalc 2.3 Factor Theorem
iff
A polynomial f(x) has a factor (x­k) if and only if f(k) =0
Show means "must be true"
NOTE: You can't just say f(2)=0 and f(­3)=0.
You have to show Synthetic Division
6) Show that a) (x­2) and b) (x+3) are factors of
f(x) = 2x4+7x3­4x2­27x­18
Answer (x­2) is a factor
since r(x)=0
Answer
or (x­2) is a factor
since r(2)=0
or
(x+3) is a factor
since r(x)=0
(x+3) is a factor
since r(­3)=0
7) Rewrite the function in the form f(x) = (x­k)q(x)+r for
the given value of k, and demonstrate that f(k)=r.
Function value of k
Function value of k
b) f(x) = 2x3­x2­4x+5 k=√2
a) f(x) = x3­19x­30 k=­3
Rewrite
Rewrite
Demonstrate f(k)=r
Demonstrate f(k)=r
PreCalc 2.3 7) Rewrite the function in the form f(x) = (x­k)q(x)+r for the given
value of k, and demonstrate that f(k)=r.
c) f(x) = x3­x2­13x+10 k=2­√3
Rewrite
Demonstrate f(k)=r
Try checking the answer in your calculator
store in your calculator
then evaluate STO
x Enter
= 6.46...
now type in f(x) and hit "enter"
x3­x2­13x+10 Enter
This is the same as you doing the following:
f( )= ( )3­( )2­13( )+10 = 3+2 3
= 6.46...
PreCalc 2.3 Homework p 159:
[
37­43 odd (like #7 notes)
day 1 45­48 all (use Remainder Thm)
for 45­48: Just plug the zero in to get an answer Example: f(2) = the answer
[
49, 50, 51, 53 ( like #8 notes)
day 2
55, 56 on given worksheet
2
Note: Assume x = 3
Your book will begin the answer as
The correct way that you will use is