polynomial zeros real zeros of polynomial functions
TRANSCRIPT
Polynomial Zeros
Real Zeros of
Polynomial Functions
Polynomial Zeros 27/23/2013
Even/Odd Multiplicity Examples
Polynomial Functions
x
y(x)
x
y(x)
x
y(x)
x
y(x)
x
y(x)
●y = (x – 3)2
●y = x + 3
●y = (x – 3)3
●y = (x – 3)4
●y = (x – 3)5
●●
y = (x + 3)3(x – 3)
x
y(x)
y = (x + 2)3(x – 3)2
● ●
Polynomial Zeros 37/23/2013
Rational Zero Test Let
f(x) = anxn + … + a2x2 + a1x + a0
for integer coefficients with an ≠ 0
Then all rational zeros of f(x) are of form
p/q p is a factor of a0 and q is a factor of an
p and q have no common factors
Polynomial Functions
Polynomial Zeros 47/23/2013
Rational Zero Test
f(x) = anxn + … + a2x2 + a1x + a0
All rational zeros of f(x) of form p/q , with p is a factor of a0 and q is a factor of an
NOTE: This works only for integer coefficients NOT all zeros are rational numbers NO irrational zeros of f(x) are included
Polynomial Functions
Polynomial Zeros 57/23/2013
Zeros and Factors FACT:
If two polynomials are equal then
they have the same factors
If f(x) = (x – k1)Q1(x) and if
Q1(x) = (x – k2)Q2(x) then we have
f(x) = (x – k1)(x – k2)Q2(x)
Polynomial Functions
Polynomial Zeros 67/23/2013
Rational Zero Test Example Factor completely:
f(x) = 3x4 – 12x3 – 24x2 + 36x + 45
= 3(x4 – 4x3 – 8x2 + 12x + 15)
= 3g(x) Here an = a4 = 1 and a0 = 15 Factors p of 15 are: ±1, ±3, ±5, ±15 ; Factors q of 1 are: ±1 Possible p/q values are: ±1, ±3, ±5, ±15
Polynomial Functions
Polynomial Zeros 77/23/2013
Rational Zero Test Example g(x) = x4 – 4x3 – 8x2 + 12x + 15
Possible p/q values: ±1, ±3, ±5, ±15 Check zeros of g(x)
Polynomial Functions
1 1 –4 –8 12 15
1
1
–3
–3
–11
–11
16 1
1k = 1
(x – 1) is NOT a factor of g(x)
1 –4 –8 12 15
1
3
–1
–3
–11
–33
–21
k = 3–63
–48 (x – 3) is NOT a factor of g(x)
3
Polynomial Zeros 87/23/2013
Rational Zero Test Example Possible p/q values: ±1, ±3, ±5, ±15
Polynomial Functions
5 1 –4 –8 12 15
1
5
1
5
–3
–15
0 –3
–15 k = 5
(x – 5) IS a factor of g(x)
1 1 –3 –3
1
–1
0
0
–3
3
0
k = –1
(x + 1) IS a factor of Q1(x)
–1
Q1(x)
Q2(x)
Polynomial Zeros 97/23/2013
Rational Zero Test Example
Polynomial Functions
Thus farf(x) = 3g(x) = 3(x – 5)Q1(x)
= 3(x – 5)(x + 1)Q2(x)= 3(x – 5)(x + 1)(x2 – 3)
Q2(x) = x2 – 3
with possible rational zeros of ±1 and ±3
Synthetic division shows none of these are zeros of Q2(x)
Polynomial Zeros 107/23/2013
Rational Zero Test Example
Polynomial Functions
Q2(x) = x2 – 3
= 3 x2 – ( ) 2
= 3 (x – ) 3 (x + )
f(x) = 3(x – 5)(x + 1)Q2(x)Thus
3 (x – ) 3 (x + ) = 3(x – 5)(x + 1)
Two rational zeros and two irrational zeros
3 x = 5, –1, ±
Polynomial Zeros 117/23/2013
Equations New functions f(x) lead to new types
of equations to solve Set f(x) = 0 and find the zeros
Examples
Find all real solutions of: 1. x4 – 1 = 0
Polynomial Functions
Polynomial Zeros 127/23/2013
Equations Examples
Find all real solutions of:
2. x3 = x
3. x4 – 5x2 + 4 = 0
4. x6 – 19x3 = 216
Polynomial Functions
Polynomial Zeros 137/23/2013
Problem Solving Find the width W of the rectangle from its length and area A. Also determine W when L = 5 inches.
A = WL
= W(x2 + 1)
= 3x3 + 3x – 5x2 – 5
= 3x(x2 + 1) – 5(x2 + 1)
= (3x – 5)(x2 + 1)
Polynomial Functions
W A = 3x3 – 5x2 + 3x – 5
L = x2 + 1
A
Polynomial Zeros 147/23/2013
Problem SolvingA = (3x – 5)(x2 + 1)
Polynomial Functions
W A = 3x3 – 5x2 + 3x – 5
L = x2 + 1
W =x2 + 1
A
=x2 + 1
(3x – 5)(x2 + 1)
= 3x – 5
At x = 5 inches, W = 3(5) – 5 = 10 inches
Note: To find W we could have used conventional long division
Polynomial Zeros 157/23/2013
Think about it !