Download 2.3 - Villa Walsh Academy

SFM Productions Presents:
Another saga in your continuing
Pre-Calculus experience!
2.3
Polynomial and Synthetic Division
Homework for section 2.3
p156 #13-45 eoo, 55-73 odd
f ( x )  6 x 3  19 x 2  16 x  4
5
Y
And because 2 is a zero, (x-2) is a
factor…
4
3
2
f ( x )  x  2
1

 q( x )
X
-5
-4
-3
-2
-1
0
-1
-2
1
2
3
4
5
q( x ) 
f (x)
x  2
-3
-4
-5
Presto: Long Division

x 2
6 x 3  19 x 2  16 x  4
6 x 3  19 x 2  16 x  4   x  2
 2 x
 1 3 x  2
(product of linear factors)

Try:
x
x 1
2
 3x  5

  x  1
x2  3x  5
x2  3x  5
 x  1 x  2

3
OR
x2  3x  5
x 1
 x  2

3

x 1
The Division Algorithm
f ( x )  d ( x )  q( x )  r ( x )
dividend
divisor
remainder
quotient
What type of f raction is it w hen the
degree of f ( x )  the degree of d ( x )
f
d
?
x 
x 
What type of f raction is it w hen the
degree of f ( x )  the degree of d ( x )
?
x
Try:
x 1
3
 1   x  1
x3 
0 x2  0 x  1
1


x 3  1   x  1  x 2  x  1
Product of linear and quadratic factors
x
3

 1  i 3   
 1  i 3  
 1   x  1   x  
   x  
 




2
2







Product of linear factors…
Synthetic Division:
only works if dividing by something that has the
form of x- k
You have to be dividing by a linear term.
In Synthetic Division, all you use is the coefficients.
Try:
-3
x 4  10 x 2  2 x  4  x  3
1
0
-10
-2
4
-3
9
3
1
-3
-1
1
1 • -3
-3 • -3
-1 • -3
1 • -3
-3
1
x 4  10 x 2  2 x  4  x  3
-3
1
1
0
-10
-3
9
3
-3
-1
1
Depressed equation
(quotient)
-2
4
Original equation
(dividend)
-3
1

divisor
remainder
f ( x )  d ( x )  q( x )  r ( x )
x 4  10 x 2  2 x  4
1
3
2
 x  3x  x  1
x 3
x 3
D o me the f avor of f inding the f ollow ing:
f ( 3 )  x 4  10 x 2  2 x  4
The Remainder Theorem
f ( k)  r
An easy way to do find the f of something is to do the synthetic
division in your head…
Go! Do!