Transcript 1314 College Algebra

Chapter 3
Section 3.3
Real Zeros of Polynomial
Functions
Polynomial Functions

Division of Polynomials

40
Long Division (by hand): 245 ) 9815
980
9815
15
3
So
15
= 40 +
= 40 +
245
49
245
Quotient
Remainder

Can we do this with polynomials ? We just did !

Writing the divisor and dividend as a base-10
polynomial we do the long division again
4(101) + 0
2(102) + 4(101) + 5
Quotient
) 9(103)
+ 8(102) + 1(101) + 5
8(103) + 16(102) + 20(101) Adjust powers
8(103) + 18(102) + 0(101)
9(103) + 8(102) + 0(101)
1(101) + 5
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Remainder
5
Polynomial Functions

Division of Polynomials

In general for polynomial functions f(x) and d(x) , d(x) ≠ 0
f(x)
r(x)
Q(x) +
=
d(x)
d(x)
where Q(x) is the quotient, r(x) is the remainder ,
d(x) is the divisor and f(x) is the dividend

Thus
f(x) = d(x) ∙ Q(x) + r(x)
where either r(x) = 0 or deg r(x) < deg d(x)
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Polynomial Functions

Division by Monomials

Example
12x4
12x4 + 27x3 – 9x2 + 6x – 2
27x3
9x2
6x
2
–
–
+
+
=
3x2
3x2
3x2
3x2 3x2
3x2
2
2
–
= 4x2 + 9x – 3 +
x
3x2
Quotient
Remainder
6x – 2
= 4x2 + 9x – 3 +
3x2
OR
4x2 + 9x – 3
Quotient
3x2 ) 12x4 + 27x3 – 9x2 + 6x – 2
12x4
27x3
27x3
– 9x2
– 9x2
6x – 2
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Note: deg r(x) = 1 < deg d(x) = 2
Remainder
7
Polynomial Functions

Division by Linear Binomials

Example
x3 + x2 – 2x
x+2
) x4 + 3x3
Quotient
– 4x + 1
x4 + 2x3
x3
x3 + 2x2
– 2x2 – 4x
– 2x2 – 4x
1
Thus
x4 + 3x3 – 4x + 1
x+2
= x3 + x2 – 2x +
Remainder
1
x+2
Note: deg r(x) = 0 < deg d(x) = 1
Question: Can we do this faster or more simply ?
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Polynomial Functions

2x3 – 7x2 +19x – 34
Synthetic Division

x+2
Example
) 2x4 – 3x3
+ 5x2 + 4x + 3
2x4 + 4x3
 The arithmetic operations
–7x3 + 5x2
–7x3 – 14x2
involve only the coefficients
19x2 + 4x
 So, using synthetic division we
19x2 + 38x
deal only with the coefficients
– 34x + 3
– 34x – 68
d(x) = x + k = x + 2
71
5
4
3
2 2 –3
Subtract
4 –14 38 –68
2 –7 19 –34 71
Remainder
d(x) = x – k = x – (–2)
–2
NOTE:
deg r(x) = 0
<1
= deg d(x)
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Add
2
–3
–4
–7
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4
14 –38
19 –34
3
68
71
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Polynomial Functions

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Degree Facts

For any polynomials A(x), B(x)
deg (A(x)  B(x)) = deg A(x) + deg B(x)
deg (A(x) + B(x)) = max { deg A(x), deg B(x) }

Examples:

deg ( (x2 + 1)  (3x3 – 4x2 + 5x + 7) ) = 2 + 3 = 5

deg ( (x – 1)  (–2x5 + x – 5) ) = 1 + 5 = 6

deg ( (3x2 – 4) + (2x4 + 6x + 3) ) = max { 2, 4 } = 4

deg ( (4x3 – 7x2 – 2x +11) + (3x3 + x – 10) ) = max { 3, 3 } = 3
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Polynomial Functions

Division Algorithm for Polynomials
If f(x) and d(x) are polynomial functions with
0 < deg d(x) < deg f(x)
then there exist unique polynomial functions
Q(x) and r(x) such that
f(x) = d(x) ∙ Q(x) + r(x)
where either r(x) = 0 or deg r(x) < deg d(x)
f(x)
r(x)
Note: This just says that
= Q(x) +
d(x)
d(x)
Example
x2 – 1
x–1
= x+1
deg f(x) = 2 , deg d(x) = 1 , r(x) = 0
Thus f(x) = d(x) ∙ Q(x) + r(x) becomes x2 – 1 = (x – 1) ∙ (x + 1) + 0
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Polynomial Functions

Remainders and Functional Values

In case the divisor d(x) is linear, i.e. d(x) = x – k , then
f(x) = d(x)  Q(x) + r(x)
becomes
f(x) = (x – k)Q(x) + r(x)
where deg r(x) < deg (x – k) = 1 ... which forces deg r(x) = 0
i.e. r(x) is just a constant r ... and thus
f(x) = (x – k)Q(x) + r

At x = k this becomes
f(k) = (k – k)Q(k) + r = r
So the value of f at x = k, or f(k), is just the remainder
when f(x) is divided by x – k … and thus …
f(k) = r
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Polynomial Functions

Remainders and Functional Values (continued)

Example:
By synthetic division we found that dividing
f(x) = 2x4 – 3x3 + 5x2 + 4x + 3 by d(x) = (x + 2) = (x – (–2))
yields
f(x) = Q(x)(x + 2) + r = (2x3 – 7x2 + 19x – 34)(x + 2) + 71
Thus f(–2) = 71

To evaluate this by the usual “brute force” method
f(–2) = 2(–2)4 – 3(–2)3 + 5(–2)2 + 4(–2) + 3
= 2(16) – 3(–8) + 5(4) – 8 + 3
= 32 + 24 + 20 – 8 + 3
= 71
Question: Which method is easier?
This leads to ...
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Polynomial Functions

The Remainder Theorem
Polynomial f(x) divided by x – k yields a remainder of f(k)

Question:


What if the remainder is 0 ?
Then we know that
f(x) = (x – k)Q(x) + r(x) = (x – k)Q(x)
and thus (x – k) is a factor of f(x) … leading to …
The Factor Theorem
Polynomial f(x) has factor x – k if and only if f(k) = 0
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Polynomial Functions

Factor Theorem Corollary
If (x – k) is a factor of f(x) = an xn + an–1xn – 1 + ... + a1x + a0
then k is a factor of a0
WHY ?

Note that
f(x) = (x – k)Q1(x) = xQ1(x) – kQ1(x)
= xQ1(x) – k(bn–1 xn–1 + … + b1x + b0)
= [ xQ1(x) – k(bn–1 xn–1 + b1x) ] – kb0
= [ anxn + an–1xn–1 +…+ a1x ] + a0
Since two polynomials are equal if and only if
corresponding terms are equal, then
a0 = – kb0
Clearly, then, k is a factor of a0
Question: What about the converse ?
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Polynomial Functions

Factor Theorem Corollary (continued)

Is the converse true?

Example: f(x) = x4 + x3 – 19x2 + 11x + 30
Try k = 6
Try k = 2
2 1
1
1 –19 11 30
2
6 –26 –30
3 –13 –15
0
6 1
1
Thus (x – 2) is a factor of f(x)
... and 2 is a factor of 30
1 –19 11 30
6 42 138 894
7 23 149 924
So (x – 6) is NOT a factor of f(x)
... BUT 6 IS a factor of 30
Question: What about other factors of 30 ?
What if k = ±1, –2, ±3, ±5, –6, ±10, ±15 ?
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Polynomial Functions

Factor Theorem Example
y

Given the graph of polynomial f(x)

Estimate the degree of f(x)


Even or odd degree ?
(–7, 0)
●
(2, 0)
●
●
(11, 0)
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

Question: Is f(x) = (x + 7)(x – 2)(x – 11) ? Not necessarily !
Note:
f(x) = (x + 7)Q1(x) = (x + 7)(x – 2)Q2(x) = (x + 7)(x – 2)(x – 1)Q3(x)
What is Q3(x) ? Does Q3(x) have factors x – k ? Is Q3(x) constant ?
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Polynomial Functions

Factor Theorem Example

Given the graph of polynomial f(x)
estimate the degree of f(x)


y
Even or odd degree ?
●
(–3, 0)
●
(5, 0)
x
Find the factors of f(x)
f(–3) = f(5) = 0 so (x + 3) and (x – 5) are factors
Note: (x + 3)(x – 5) = x2 – 2x2 – 15

Question: Is f(x) equal to x2 – 2x – 15 ?
Probably not !
What does the graph of x2 – 2x – 15 look like ?
Note: f(x) = (x + 3)Q1(x) = (x + 3)(x – 5)Q2(x)
Does Q2(x) have factors (x – k) ?
Probably, but which ones and how many ?
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Polynomial Functions

Intercepts, Zeros and Factors

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)
WHY ?
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Polynomial Functions

Completely Factored Polynomials
Example: Factor
f(x) = 2x4 + 14x3 + 18x2 – 54x – 108 = 2(x4 + 7x3 + 9x2 – 27x – 54)
Now use synthetic division to check out zeros
x – k = x – (–3)
–3 1
7
9 –27 –54
–3 –12
9 54
0
1 4 –3 –18
–3 1
f(x) = 2(x +
3)3(x
– 2)
Note:
x = –3 is a repeated
zero of multiplicity 3
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–3
1 1
–3
–3
–6
–3 1
1 –6
–3
6
1 –2 0
–18
18
0
(x + 3) is a factor
Q1
(x + 3) is a factor
Q2
(x + 3) is a factor
(x – 2) is a factor
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Polynomial Functions

Complete Factoring with Multiple Zeros

General Form
f(x) = anxn + an–1 xn–1 + ... + a1x + a0
= an(x – kn)(x – kn–1 ) ... (x – k1)
where some of the zeros ki may be repeated




Example
f(x) = 2(x + 3)3(x – 2)


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Degree of f(x) = n
Number of real zeros m, m ≤ n
If all zeros occur at x-intercepts , i.e. are real numbers,
then, counting multiplicities, total number of zeros is n
one zero at 2 and one at –3 of multiplicity 3
total zeros, counting multiplicities, 1 + 3 = 4 = deg f(x)
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Polynomial Functions

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)
2. Given:



f(x) is a quadratic polynomial
lead coefficient is 7
f(–3) = 0 and f(2) = 0
Write f(x) in complete factored form



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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)
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Polynomial Functions
Even/Odd Multiplicities
Examples

y(x)
y=x+3
●
y(x)
y(x)
●
●
x
y = (x – 3)2
y(x)
●
x
y = (x – 3)5
●
x
y = (x – 3)4
y = (x – 3)3
y(x)
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●
x
y(x)
●
x
y = (x + 3)3(x – 3)
Section 3.3 v5.0
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
●
x
y = (x + 2)3(x – 3)2
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Polynomial Functions

Rational Zero Test
n
2
 Let f(x) = anx + … + a2x + a1x + a0 where an ≠ 0 and all
coefficients are integers
Then all rational zeros of f(x) are of form p/q , in lowest
terms, where p is a factor of a0 and q is a factor of an

NOTE:




This works only if the coefficients are integers
It does NOT say all zeros are rational numbers
It does NOT include any irrational zeros of f(x)
FACT:
If two polynomials are equal they have the same factors
If f(x) = (x – k1)Q1(x) and if Q1(x) = (x – k2)Q2(x) then
we have
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f(x) = (x – k1)(x – k2)Q2(x)
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Polynomial Functions

Rational Zero Test Example

Factor completely: f(x) = 3x4 – 12x3 – 24x2 + 36x + 45





k=1
1
f(x) = 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 rational zeros p/q are: 1, 3, 5, 15
Check zeros of g(x) with synthetic division:
1
–4
1
–8 12
–3 –11
1 –3 –11
k=5
5
1
–4
5
1
1
15
1
1 16
–8 12 15
5 –15 –15
–3 –3
0
k=3
(x – 1) is not a
factor of g(x)
1
1
–4
3
–8 12 15
–3 –33 –63
–1 –11 –21 –48
k = –1 –1
1
1
–1
–3
0
–3
3
(x – 5) is a
factor of g(x)
1
0
–3
0
Q2(x)
Q1(x)
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Section 3.3 v5.0
(x – 3) is not a
factor of g(x)
(x + 1) is a
factor of Q1(x)
26
Polynomial Functions

Rational Zero Test Example (continued)




Factor completely: f(x) = 3x4 – 12x3 – 24x2 + 36x + 45
We now have:
f(x) = 3g(x) = 3(x – 5)(x + 1)(x2 – 3)
Synthetic division on Q2(x) = x2 – 3 shows that none of
the possible rational zeros (1, 3) are zeros of Q2(x)
To find the zeros of Q2(x) use difference of squares
Q2(x) = x2 – 3 = (x –  3 )(x +  3 ) = 0
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
and solve for x using the zero product property
Thus
f(x) = 3(x – 5)(x + 1)(x –  3 )(x +  3 )

Note that the last two zeros are irrational
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Polynomial Functions

Equations


Each new function f(x) we define leads to a new type of
equation to solve when we set f(x) = 0 and find the zeros
Examples
Find all real solutions of:
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
1. x4 – 1 = 0

2. x3 = x

3. x4 – 5x2 + 4 = 0

4. x6 – 19x3 = 216
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Think about it !
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