2.3 Real Zeros of Polynomial

8/22/2019 2.3 Real Zeros of Polynomial

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More Polynomial Solution

Methods (section 2.3) Synthetic Division

Remainder Theorem

Factor Theorem

Rational Zeros Test

Descartess Rule of Signs Upper and Lower Bounds Rules

1


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Division by a Monomial (pg 112)

Divide:5 4 3 2

2

15 21 6 2 3 6

3

x x x x x

x

Divide each term separately:

5 4 3 2

2 2 2 2 2 2

15 21 6 2 3 6

3 3 3 3 3 3

x x x x x

x x x x x x

3 2

2

2 1 25 7 2

3x x x

x x


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Division by a Binomial-Long Division

Divide:

4 3 23 1 3 2 4 7 5x x x x x

4 3 23 2 4 7 5 3 1x x x x x


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4 3 23 1 3 2 4 7 5x x x x x

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Division by a Binomial-Long Division34 3 2

4 3

3 2

3 1 3 2 4 7 5

3 -3 4

x

x x x x x

x x

x x

3 2

3x x2

3 7x x 2

3x x

6 5x 6 2x

3

Remainder

3 2 32

3 1x x x

x

Answer

2x x 2

See pg 114 for longer examples


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Synthetic Division (pg 115) Synthetic division: a method ofdividinga

polynomial by a binomial (x c).

c is a rootof the polynomial, iftheremainder R is zero.

3 23 2 3 2x x x x

rewrite 3 2 -3 -1 2

Leading coefficients


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Synthetic Division Steps3 2 -3 -1 2


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Synthetic Division Steps3 2 -3 -1 2 1. Bring down the 2

2

2. Multiply 2*3= 66

3. Add3 and 6= 33

9

4. Multiply 3*3 =9

8

5. Add1 and 9 =8

6. Multiply 8*3=24

24

7. Add 24 and 2 = 26 = R

26

Answer2 26

2 3 83

x xx


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Your Turn

Zeros in set-up?

8

Answer:

4 23 10 2 4x x x x

Place holders

-3 1 0 -10 -2 4

1

-3

-3

9

-1

3 -3

1 13 2 1

3 13

x x xx


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Your Turn #2

No remainder?

9

Answer:3 2

3 5 15x x x

4 3 22 11 5 30x x x x x

(x 2) is a factorx = 2 is a root

2 1 1 -11 -5 30

1

2

3

6

-5

-10

-15

-30

0


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The Remainder Theorem(pg 116)

If a polynomial f(x) is divided by (x c), where c isa constant, then the remainder R is f(c).

Dividend = Quotient * Divisor + Remainder

Proof (see Appendix C)

So, we can use Synthetic Division to: evaluatea function and

find pointson its graph


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Remainder Theorem Example Determine the remainder of the

polynomial for c=3. Using the earlier example:

3 23 2 3 2x x x x

2 262 3 8

3x x

x

p(3) = 2(3)3-3(3)2-3+2=54-27-3+2=26

p(3) = 26


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The Remainder Theorem What is the remainder when

f(x)=5x3 3x2 + x

is divided by x 2 ?

f(2) = 5(2) 3 3(2) 2 + 2

= 4012 + 2=30

So the remainder is 30


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Rational Roots (pg 118) There is no single method to

find all roots of a polynomial.

But there is a method to find ALLrationalroots of any polynomial.

Rational Root Test


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Rational Roots Iff(x)=anxn+an-1x

n-1+.+a2x2+a1x+a0

has integer coefficients, every rational zero off

has the form:

Rational zero = p/qwhere:

p and q have no common factors other than 1. P is a factor of a0 and

q is afactor ofan.


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Rational Roots

factors of constant termPossible rational zeros =factors of leading coefficient

p

q

f(x) = x4 -11x2 - 2x + 12p =12 q = 1

Factors of 12 are +1, + 2, + 3, + 4, + 6, + 12

Factors of 1= +1

1 2 3 4 6 12, , , , ,

1 1 1 1 1 1Possible rational zeros =


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Rational Roots Possible zeros are +1, + 2, + 3, + 4, + 6, + 12

We use synthetic division to determine which are

zeros. And we know from Descartes Rule that there are two

positive and 2 negative roots. To narrow down the math, if you graph it on the TI-83 first, you will find

that the graph crosses the x axis in four places between4 and 4.

Using the trace key provides rational zeros at3 and 1. Using the Factor Theorem if3 and 1 are zeros, then

f (-3) = 0 and f (1) = 0.


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Example #1 f(-3)=(-3)4-11(-3)2-2(-3)+12

= 8199 +6 +12=0

f(1)=(1)4-11(1)2-2(1)+12

= 1- 11 2 + 12=0

So3 and 1 are rational zeros.

From synthetic division:

f(x) = (x + 3)(x - 1)(x2 - 2x - 4)

The other two zeros are irrational numbers.


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Example #2 Find all the roots off(x) = 2x3 + x2 - 6x3

We know that since the index is 3 there will be a max of

three roots. The Rational Root theorem yields p =3 q = 2

Descartes Rule shows 1 positive and 2 negative roots.

We can use the upper and lower bounds rule to eliminate somepossibilities but

It is generally faster to use the TI-83 and graph it.

1 3 1 3, , ,

1 1 2 2Possible rational roots =


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Example #2 continued Graphing on the TI 83 and using the

trace key yields - as a rational root.

Knowing if x= - is a root then 2x+1 afactor. (2x+1 = 0 then x= -)

Use long division and divide by 2x+1 you

get on the next slide


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Example #2 continued

x2 3 is a factor:

x2 3 = 0 therefore

3 2

2

3 2

2x +x

0 - 6x

0 + 0

-6x - 3

-6x - 3

0

0 3

2 1 2 6 3

x

x x x x

3x

Continued on the next slide


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Example #2 continued Therefore all the roots, rational and

irrational in this case are

1, 3, 3

2

x


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Descartes Rule of Sign (pg 120) Let P(x) be a polynomial over the reals.

The number of positive roots of P(x)

Equals the number of variations in the sign of P(x) Or is fewer by an even number.

The number of negative roots of P(x)

Equals number of variations in the sign of P(-x)

Or is fewer is by an even number.

This is a useful technique for finding thezeros of a polynomial.


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Descartes Rule of Signs f(x)= +3x5+9x4+5x3-x2+2x-1

Look at the signs and note where the sign change

from positive to negative or vice versa. Count the sign changes

f(x)= +3x5+9x4+5x3-x2+2x-1

1 2 3


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Descartes Rule of Signs There are three changes.

This is the maximum number ofpositive

zeros. We might have only 1 (why not 2?).


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Descartes Rule of Signs We determine the negative roots by

replacingx in the original equation and

counting the sign changes. f(-x)= +3(-x)5+9(-x)4+5(-x)3-(-x)2+2(-x)-1

f(-x)= -3x5+9x4-5x3-x2-2x-1

Which means there are 2 or 0 negative zeros.

Notice that the total number of positive andnegative zeros equals the degree of the polynomial.

1 2


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Example Find the number of roots

f(x)= +2x5 - 3x4+ 4x3+ 5x2- 6x+7

Four changes, So 4, 2, or 0 positive roots.

Check for negative roots

f(-x)= +2(-x)5-3(-x)4+4(-x)3+5(-x)2-6(-x)+7

=-2x5-3x4-4x3+5x2+6x+7

One sign change. So 1 or 0 negative roots.

The maximum positive and negative roots equals 5.

1 2 3 4

1


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Bounds for Real Roots (pg 121)In solving higher order mathematicalequations or complex engineering problems it

is sometimes useful to determine the bounds(upper and lower) for which the roots(solutions) for a given function are between.


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Bounds The upper and lower bounds are

integer values of x between which the

roots (solution) of a given function lie. For instance, if the upper and lower

bound are x =3 and2 respectively,

then the roots or solutions to thefunction are between x=-3 and 2.


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Bounds for Real Roots Upper Bound

Let f(x) be a polynomial over R ( meaning allreal numbers) with a positive leading coefficient.

If f(x) is divided by x-c using synthetic division, ifc>0 and the numbers in the last row are positiveor zero, c is an upper bound.

Lower Bound If c < 0 and the numbers in the last row are

alternatively positive and negative, c is a lowerbound. (This is different from your book and a loteasier to remember)


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How to Locate the Bounds To find the bounds, use synthetic

division and start with any number

generally 1 for upper and1 for lower. Then continue with 2, 3, etc. until the

upper is found.

And2, -3, etc. until the lower bound isfound.


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Example #1 Find the upper and lower bounds

f(x)= 6x3 - 4x2+3x 2

Use synthetic division

1 6 -4 3 -2

6 2 56 2 5 3 All positive

So 1 is an upper bound


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Example #1 Lower bound f(x)= 6x3 - 4x2+3x 2

-1 6 -4 3 -2-6 10 -13

6 -10 13 -15

So1 is the lower bound

Alternating + an -


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Example #2 Find the upper and lower bounds of

f(x)=10x3-15x2-16x-12

Use synthetic division

1 10 -15 -16 12

10 -5 -21

10 -5 -21 -9

2 10 -15 -16 12

20 10 -12

10 5 -6 0

The 0 tells us that 2 is azero but not an upper bound3 10 -15 -16 12

30 45 87

10 15 29 99 All positive so 3 isthe upper bound


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Example #2 We found x-2 as the upper bound.

For the lower bound start with1.

-1 10 -15 -16 12

-10 25 -9

10 -25 9 3

-2 10 -15 -16 12

-20 70 -108

10 -35 54 -96

Alternating signs so2 is a lower bound.


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Review Find Roots

Synthetic Division

Remainder Theorem

Factoring

Narrow your search for roots

Rational Roots Test

Descartess Rule of Signs

Upper/Lower Bounds (Synth Div)

Documents

2.3 Real Zeros of Polynomial