Transcript 2.3 - Polynomial and Synthetic Division

Warm up!!
•Pg. 130 #’s 16, 22, 36,
42, 52, 96
2.3 - Polynomial
and Synthetic Division
Long Division of Polynomials
1) Divide
2) Factor
3) Find Zeros
Now try this one yourself:
Divide 5x2 – 17x – 12 by x – 4
Answer: 5x + 3
Example:
Divide 6x3 – 19x2 + 16x – 4 by x – 2
Answer: 6x2 – 7x + 2
Which means that…
6x3 – 19x2 + 16x – 4 = (x – 2)(6x2 – 7x + 2)
6x3 – 19x2 + 16x – 4 = (x – 2)(2x – 1)(3x – 2)
Zeros: x = 2 x = ½ x = ⅔
Can check
zeros on
calculator
Rules of Long Division
 If the divisor goes into the equation evenly then it is
a factor and a zero.
 If the quotient has a remainder then it is not a factor.
 The proper form of writing the remainder is adding it
to the quotient and over the divisor.
Division Algorithm
Also written as:
Improper
F(x) = q(x) + r(x)
d(x)
d(x)
Proper
F(x)/d(x) is improper because f(x) ≥ d(x).
R(x)/d(x) is proper because r(x) ≤ d(x).
Using Division Algorithm
you can write it like this:
Division
Algorithm
x3 – 1 = x2 + x + 1
x–1
Example:
Divide x3-1 by x-1
– Write in descending powers
– Insert zeros where there are missing power
x2 + x + 1
x – 1 √(x3 + 0x2 + 0x – 1)
Because it equals 0,
x3 – x2
3 -1 is divisible by x-1
x
x2 + 0x
x2 – x
x–1
x–1
0
Synthetic Division
• Synthetic division is a shorthand, or shortcut, method of
polynomial division in the special case of dividing by a linear
factor.
– Used when finding zeroes of polynomials.
Example:
Divide x2 + 5x + 6 by x – 1
*Rational roots test determines that +/- 1, 2, 3, and 6 are
possible zeros.
• Synthetic provides the same quotient but in a quicker fashion.
Let’s divide:
x2 + 5x + 6 by x – 1
• If r = 0, (x – k) is a factor.
• If r = 0, (k, 0) is an x intercept of f. *k is a zero
• The remainder r gives the value of f at x = k, if
r = f(k) (Remainder Theorem)
Remainder Theorem
To evaluate polynomial function f(x) when x=k, divide f(x) by x-k
– Remainder will equal f(k)
Example 1:
Evaluate f(x) = 3x3 + 8x2 + 5x – 7 for x = -2
So f(-2) = -9, r = -9
Example 2:
Evaluate f(x) = x3 – x2 – 14x + 11 for x = 4
So f(4) = 3, r = 3
Repeated Division
Multi-step synthetic division…
Find all the zeros of the following polynomial function given (x – 2) and
(x + 3) are factors…
f (x) = 2x4 + 7x3 – 4x2 – 27x – 18.
Solution:
Using synthetic division with the factor (x – 2), you obtain the
following.
0 remainder, so f(2) = 0
and (x – 2) is a factor.
Repeated Division (Continued)
Take the result of this division and perform synthetic division again using the
factor (x + 3).
0 remainder, so f(-3) = 0
and (x +3 ) is a factor.
Because the resulting quadratic expression factors as
2x2 + 5x + 3 = (2x + 3)(x + 1)
the complete factorization of f (x) is
f (x) = (x – 2)(x + 3)(2x + 3)(x + 1).
Try #60 on
page 141
Factor Theorem/Repeated Division
Example 1:
Show that (x – 2) and (x + 3) are factors and find all the zeros of
f(x) = 2x4 + 7x3 – 4x2 – 27x – 18
x = 2, -3, -3/2, -1
Example 2:
Show that (x + 2) and (x – 1) are factors and find all the zeros of
f(x) = 2x3 + x2 – 5x + 2
x = ½, -2, 1
• Graph should confirm zeros
Homework
Page 140 #’s 5 – 35
(5’s), 47, 51, 55, 59,
61, 69, 73