chapter 3 section 3.3 real zeros of polynomial functions

Chapter 3

Section 3.3

Real Zeros of Polynomial Functions

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Division of Polynomials Long Division (by hand):

Can we do this with polynomials ?

Writing the divisor and dividend as a base-10 polynomial we do the long division again

Polynomial Functions

245 ) 98154

98015

0 Quotient

Remainder9815245

= 40 +15245

We just did !

So = 40 +349

2(102) + 4(101) + 5 9(103) + 8(102) + 1(101) + 5 )

4(101)

8(103) + 16(102) + 20(101)

9(103) + 8(102) + 0(101)

1(101) + 5

+ 0 40

15

Adjust powers

8(103) + 18(102) + 0(101)

Quotient

Remainder

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Division of Polynomials

In general for polynomial functions f(x) and d(x) , d(x) ≠ 0

where Q(x) is the quotient, r(x) is the remainder ,

d(x) is the divisor and f(x) is the dividend Thus

where either r(x) = 0 or deg r(x) < deg d(x)


+f(x)d(x) = Q(x)

r(x)d(x)

f(x) = d(x) ∙ Q(x) + r(x)

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Division by Monomials Example


12x4 + 27x3 – 9x2 + 6x – 2 3x2

=12x4

3x2+

27x3

3x2

9x2

3x2– 6x

3x2+

23x2

–

= 4x2 + 9x – 3 2

3x2–

2x+

= 4x2 + 9x – 3 3x2

2–6x+

QuotientRemainder

OR

12x4 + 27x3 – 9x2 + 6x – 2 3x2)4x2

12x4

27x3

+ 9x

27x3

– 9x2

– 3

– 9x2

6x – 2 Remainder

Quotient

Note: deg r(x) = 1 < deg d(x) = 2

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Division by Linear Binomials Example


x4 + 3x3 – 4x + 1 x + 2 )x3

x4 + 2x3

x3

+ x2

x3 + 2x2

– 2x2

– 2x

– 2x2 – 4x 1

– 4x

Quotient

Remainder

Thusx4 + 3x3 – 4x + 1

x + 2+= x3 + x2 – 2x

x + 21

Question: Can we do this faster or more simply ?

Note: deg r(x) = 0 < deg d(x) = 1

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Synthetic Division Example

The arithmetic operations involve only the coefficients

So, using synthetic division we deal only with the coefficients


2x4 – 3x3 + 5x2 + 4x + 3 x + 2 )2x3

2x4 + 4x3

–7x3

– 7x2

–7x3 – 14x2

19x2

– 34

19x2 + 38x – 34x

+ 4x

+19x

+ 5x2

+ 3– 34x – 68

712 –3 5 4 32

2

4

–7

–14

19

38

–34

–68

71

Subtract

2 –3 5 4 3–2

2

–4

–7

14

19

–38

–34

68

71

Add

d(x) = x + k = x + 2

d(x) = x – k = x – (–2)

NOTE:deg r(x) = 0

< 1

= deg d(x)

Remainder

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Degree Facts For any polynomials A(x), B(x)

deg (A(x) B(x))

deg (A(x) + B(x))

Examples:

deg ( (x2 + 1) (3x3 – 4x2 + 5x + 7) )

deg ( (x – 1) (–2x5 + x – 5) )

deg ( (3x2 – 4) + (2x4 + 6x + 3) )

deg ( (4x3 – 7x2 – 2x +11) + (3x3 + x – 10) )


= deg A(x) + deg B(x)

= max { deg A(x), deg B(x) }

= 2 + 3 = 5

= 1 + 5 = 6

= max { 2, 4 } = 4

= max { 3, 3 } = 3

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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)


Note: This just says that

f(x)

d(x)Q(x)= +

r(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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Remainders and Functional Values In case the divisor d(x) is linear, i.e. d(x) = x – k , then

becomes

where deg r(x) < deg (x – k)

i.e. r(x) is just a constant r

At x = k this becomes

So the value of f at x = k, or f(k), is just the remainder

when f(x) is divided by x – k


f(x) = d(x) Q(x) + r(x)

f(x) = (x – k)Q(x) + r(x)

f(k) = (k – k)Q(k) + r

f(x) = (x – k)Q(x) + r

... which forces deg r(x) = 0

... and thus

… and thus …

f(k) = r

= 1

= r

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Remainders and Functional Values (continued) Example:

By synthetic division we found that dividing

yields

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


d(x) = (x + 2) = (x – (–2))f(x) = 2x4 – 3x3 + 5x2 + 4x + 3 by

+ 71

Question: Which method is easier?This leads to ...

f(x) = (2x3 – 7x2 – 34)+ 19x= (x + 2)Q(x)(x + 2) + r

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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)

and thus (x – k) is a factor of f(x)

The Factor Theorem

Polynomial f(x) has factor x – k if and only if f(k) = 0


= (x – k)Q(x)

… leading to …

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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

Note that f(x) = (x – k)Q1(x)


Question: What about the converse ?

WHY ?= xQ1(x) – kQ1(x)

= xQ1(x) – k(bn–1 xn–1 + … + b1x + b0)

= [ anxn + an–1xn–1 +…+ a1x ] + a0

= [ xQ1(x) – k(bn–1 xn–1 + b1x) ] – kb0

Since two polynomials are equal if and only if corresponding terms are equal, then

a0 = – kb0 Clearly, then, k is a factor of a0

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Factor Theorem Corollary (continued)

Is the converse true?

Example: f(x) = x4 + x3 – 19x2 + 11x + 30


1 1 –19 11 30

2

1

2

3

6

–13

–26

–15

–30

0

Thus (x – 2) is a factor of f(x)

Try k = 6

... and 2 is a factor of 30

1 1 –19 11 30

6

1

6

7

42

23

138

149

894

924

So (x – 6) is NOT a factor of f(x)

... BUT 6 IS a factor of 30

Try k = 2

Question:

What if k = ±1, –2, ±3, ±5, –6, ±10, ±15 ?

What about other factors of 30 ?

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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)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:


x

y

● ● ●(–7, 0) (2, 0) (11, 0)

Is f(x) = (x + 7)(x – 2)(x – 11) ? Not necessarily !

Note:

f(x) = (x + 7)Q1(x)

What is Q3(x) ? Is Q3(x) constant ?Does Q3(x) have factors x – k ?

= (x + 7)(x – 2)Q2(x) = (x + 7)(x – 2)(x – 1)Q3(x)

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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)f(–3) = f(5) = 0 so (x + 3) and (x – 5) are factors

Note: (x + 3)(x – 5) = x2 – 2x2 – 15

Question:


x

y

(–3, 0)

(5, 0) ● ●

Is f(x) equal to x2 – 2x – 15 ? Probably not !

Note: f(x) = (x + 3)Q1(x) = (x + 3)(x – 5)Q2(x)

Probably, but which ones and how many ?

Does Q2(x) have factors (x – k) ?

What does the graph of x2 – 2x – 15 look like ?

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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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Completely Factored PolynomialsExample: Factor

f(x) = 2x4 + 14x3 + 18x2 – 54x – 108

Now use synthetic division to check out zeros


= 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 –3

–6 (x + 3) is a factor

1

–3

–2

6

0

1 1 –6 –3

(x + 3) is a factor

(x – 2) is a factor

f(x) = 2(x + 3)3(x – 2)

Note:x = –3 is a repeated zero of multiplicity 3

Q1

Q2

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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 Degree of f(x) = n Number of real zeros m, m ≤ n If all zeros occur at x-intercepts then, counting multiplicities, total number of zeros is n

Example f(x) = 2(x + 3)3(x – 2)

one zero at 2 and one at –3 of multiplicity 3 total zeros, counting multiplicities, 1 + 3 = 4 = deg f(x)


, i.e. are real numbers,

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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 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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Even/Odd Multiplicities Examples


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

●

●

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Rational Zero Test Let f(x) = anxn + … + a2x2 + 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 f(x) = (x – k1)(x – k2)Q2(x)


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11

k = –1

Rational Zero Test Example Factor completely: f(x) = 3x4 – 12x3 – 24x2 + 36x + 45

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

–1

1 –4 –8 12 15

1

1

–3

–3

–11

–11

16 1

1

k = 1 3 1 –4 –8 12 15

1

3

–1

–3

–11

–33

–21

k = 3–63

–48

5 1 –4 –8 12 15 5

1

5

–3

–15

0

k = 5

–3

–15 1 1 –3 –3

0 3

0 –3

–1

0

Q1(x) Q2(x)

(x – 1) is not a factor of g(x)

(x – 3) is not a factor of g(x)

(x – 5) is a factor of g(x)

(x + 1) is a factor of Q1(x)

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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

and solve for x using the zero product property Thus

Note that the last two zeros are irrational


Q2(x) = x2 – 3 = (x – 3 )(x + 3 ) = 0

(x – 3 )(x + 3 )

f(x) = 3(x – 5)(x + 1)

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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:

1. x4 – 1 = 0

2. x3 = x

3. x4 – 5x2 + 4 = 0

4. x6 – 19x3 = 216


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Think about it !

chapter 3 section 3.3 real zeros of polynomial functions

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