polynomial factors polynomial factor theorem. 7/23/2013 polynomial factors 2 the remainder theorem...
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Polynomial Factors
Polynomial Factor Theorem
Polynomial Factors 27/23/2013
The Remainder Theorem
Polynomial f(x) divided by x – k
yields a remainder of f(k)
Question:
Polynomial Factor Theorem
= (x – k)Q(x)
… leading to …
What if the remainder is 0 ?
Then we know that f(x) = (x – k)Q(x) + r(x)
and thus (x – k) is a factor of f(x)
Polynomial Factors 37/23/2013
The Factor Theorem
Polynomial f(x) has factor x – k
if and only if
f(k) = 0 Factor Theorem Corollary
If (x – k) is a factor of f(x) = anxn + an–1xn–1 + • • • + a1x + a0
then k is a factor of a0
Polynomial Factor Theorem
Question: What about the converse ?
Polynomial Factors 47/23/2013
Example: f(x) = x4 + x3 – 19x2 + 11x + 30
6 42 138 894
Polynomial Factor Theorem
1 1 –19 11 30 2
1
2
3
6
–13
–26
–15
–30
0
(x – 2) is a factor of f(x)
Try factor k = 6
... and 2 is a factor of 30
1 1 –19 11 30
1 7 23 149 924
(x – 6) NOT a factor of f(x)
... BUT 6 IS a factor of 30
Try factor k = 2
Question:
What if k = ±1, –2, ±3, ±5, –6, ±10, ±15 ?
What about other factors of 30 ?
≠ 0
6
Polynomial Factors 57/23/2013
Factor Theorem Example Given the graph of
polynomial f(x)
Estimate the degree of f(x) Even or odd degree ?
Polynomial Factor Theorem
x
y
● ● ●(–7, 0) (2, 0) (11, 0)
Odd degree ! WHY ?
Polynomial Factors 67/23/2013
Factor Theorem Example Given the graph of polynomial f(x) Find the factors of f(x)
Since f(–7) = f(2) = f(11) = 0
… then factors are
(x + 7), (x – 2), (x – 11)
Note:
(x + 7)(x – 2)(x – 11) = x3 – 6x2 – 69x + 154
Polynomial Factor Theorem
x
y
● ● ●(–7, 0) (2, 0) (11, 0)
Polynomial Factors 77/23/2013
Factor Theorem Example The graph of polynomial f(x)
Question:
Is f(x) = (x + 7)(x – 2)(x – 11) ?
Polynomial Factor Theorem
x
y
● ● ●(–7, 0) (2, 0) (11, 0)
Not necessarily !f(x) = (x + 7)Q1(x) = (x + 7)(x – 2)Q2(x)
= (x + 7)(x – 2)(x – 1)Q3(x)
Polynomial Factors 87/23/2013
Factor Theorem Example The graph of polynomial f(x)
Question:
Polynomial Factor Theorem
x
y
● ● ●(–7, 0) (2, 0) (11, 0)
f(x) = (x + 7)(x – 2)(x – 1)Q3(x)
What is Q3(x) ?
Is Q3(x) constant ?
Does Q3(x) have factors x – k ?
Polynomial Factors 97/23/2013
Factor Theorem Example Given the graph of polynomial f(x)
estimate the degree of f(x) Even or odd degree ?
Find the factors of f(x)
Polynomial Factor Theorem
x
y
(–3, 0) (5, 0) ● ●
Note: (x + 3)(x – 5) = x2 – 2x2 – 15
f(–3) = f(5) = 0so (x + 3) and (x – 5) are factors
WHY !
Polynomial Factors 107/23/2013
Factor Theorem Example Given the graph of f(x)
Polynomial Factor Theorem
x
y
(–3, 0) (5, 0) ● ●
Is f(x) equal to x2 – 2x – 15 ? Probably not !
Question:
(x + 3)(x – 5) = x2 – 2x2 – 15
What is the graph of x2 – 2x – 15 ?Note: f(x) = (x + 3)Q1(x) = (x + 3)(x – 5)Q2(x)
Which ones and how many ?
Does Q2(x) have factors (x – k) ?
Polynomial Factors 117/23/2013
These things are related
If f(k) = 0, then point (k, 0) on the graph of f is an x-intercept
The number k is a zero for f(x), i.e. f(k) = 0
(x – k) is a factor of f(x)
The number k is a factor of the constant term of f(x)
Intercepts, Zeros and Factors
WHY ?
Polynomial Factors 127/23/2013
Example: Completely Factor
f(x) = 2x4 + 14x3 + 18x2 – 54x – 108
Now use synthetic division to check out zeros
Completely Factored Polynomials
= 2(x4 + 7x3 + 9x2 – 27x – 54)
Polynomial Factors 137/23/2013
Completely Factored Polynomials
f(x) = 2(x4 + 7x3 + 9x2 – 27x – 54)
1
–3
4
–12 9
–18
54
0
1 7 9 –27 –54 –3
–3
x – k = x – (–3)
(x + 3) is a factor
1–3
1 –3 18
0
1 4 –3 –18
–6 (x + 3) is a second factor
Q1 –3
Polynomial Factors 147/23/2013
Completely Factored Polynomials
f(x) = 2(x4 + 7x3 + 9x2 – 27x – 54)1 7 9 –27 –54
0 1 4 –18 –3 (x + 3) is a factor
1 1 0 –6 (x + 3) is a second factor
1–3 –2
6 0
1 1 –6
(x + 3) is a third factor
(x – 2) is a fourth factor
2(x + 3)3(x – 2)
Note: x = –3 is a repeated zero of multiplicity 3
Q2 –3
Polynomial Factors 157/23/2013
Completely Factored Polynomials
f(x) = 2(x4 + 7x3 + 9x2 – 27x – 54)
1
–3
4
–12 9
–18
54
0
1 7 9 –27 –54 –3
–3
x – k = x – (–3)
(x + 3) is a factor
1–3
1 –3 18
0
1 4 –3 –18
–6 (x + 3) is a second factor
1–3 –2
6 0
1 1 –6
(x + 3) is a third factor
(x – 2) is a fourth factor
2(x + 3)3(x – 2)
Note:x = –3 is a repeated zero of multiplicity 3
Q1
Q2
–3
–3
Polynomial Factors 167/23/2013
Complete Factoring with Multiple Zeros General Form
f(x) = anxn + an–1 xn–1 + ... + a1x + a0
= an(x – kn)(x – kn–1 ) ... (x – k1)
Some of the zeros ki may be repeated
Polynomial Functions
Polynomial Factors 177/23/2013
Complete Factoring with Multiple Zeros
f(x) = an(x – kn)(x – kn–1 ) ... (x – k1)
Degree of f(x) = n
Number of zeros is m, m ≤ n
Real zeros occur at x-intercepts
Counting multiplicities, total number of zeros is n
Polynomial Functions
Polynomial Factors 187/23/2013
Complete Factoring with Multiple Zeros Example
f(x) = 2(x + 3)3(x – 2)
deg f(x) = 4
One zero at 2, one at –3 of multiplicity 3
Total zeros, counting multiplicities,
1 + 3 = 4 = deg f(x)
Polynomial Functions
Polynomial Factors 197/23/2013
Complete Factoring Examples 1. f(x) = 5x3 – 10x2 – 15x
= 5x(x2 – 2x – 3)
= 5x(x – 3)(x + 1)
= 5(x + 1)(x – 0)(x – 3)
Polynomial Functions
Zeros of f(x) are –1 , 0 , 3
Polynomial Factors 207/23/2013
Complete Factoring Examples 2. Given:
f(x) is a quadratic polynomial lead coefficient is 7 f(–3) = 0 and f(2) = 0
Write f(x) in completely factored form
Polynomial Functions
Polynomial Factors 217/23/2013
Complete Factoring Examples Write f(x) in completely factored form
Note that –3 and 2 are zeros of f(x) From the Factor Theorem
x –(–3) and x – 2 are factors of f(x) Thus
f(x) = 7(x + 3)(x – 2)
Polynomial Functions
Polynomial Factors 227/23/2013
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 Factors 237/23/2013
Think about it !