7.5.1 Zeros of Polynomial 7.5.1 Zeros of Polynomial FunctionsFunctions

7.5.1 Zeros of Polynomial 7.5.1 Zeros of Polynomial FunctionsFunctions

Objectives: •Use the Rational Root Theorem to find the zeros of a polynomial function

Rational Root Theorem

• p is a factor of the constant term of f(x)

Let f(x) be a polynomial function with integer coefficients in standard form. If (in lowest terms) is a root of f(x) = 0, then

pq

• q is a factor of the leading coefficient of f(x)

Example 1Find all rational roots of 8x3 + 10x2 – 11x + 2 = 0.

Step 1: Make an organized list of all possible roots.

factors of 2: 1, 2

factors of 8: 1, 2, 4, 8

1 1 1 1, , ,

1 2 4 8

2 2 2 2, , ,

1 2 4 8

List: ±1, ±½, ±¼, ± ⅛ , ±2

Example 1Find all rational roots of 8x3 + 10x2 – 11x + 2 = 0.

Step 2: Use substitution or synthetic division to test possible roots, until you find one that works.

List: ±1, ±½, ±¼, ± ⅛ , ±2

Example 1Find all rational roots of 8x3 + 10x2 – 11x + 2 = 0.

Step 3: Use factoring or quadratic formula to find the other two.

Found: ½ Resulting Equation: 8x2 + 14x – 4 = 0

Example 1Find all rational roots of 8x3 + 10x2 – 11x + 2 = 0.

Step 2: Use a graphing calculator to identify possible roots.

possible roots: -21

,4

1,2

1 1 1 1, , ,

1 2 4 8

2 2 2 2, , ,

1 2 4 8

Example 1Find all rational roots of 8x3 + 10x2 – 11x + 2 = 0.

Step 3: Use substitution or synthetic division to test all possible roots.

possible roots: -21

,4

1,2

8

-16

-6

121

-2

0

-2 8 10 -11 2

roots: -2

Example 1Find all rational roots of 8x3 + 10x2 – 11x + 2 = 0.

Step 3: Use substitution or synthetic division to test all possible roots.

possible roots: -21

,4

1,2

8 10 -11 2

roots -2

14

8

2

12

3

-8

-2

0

1,

4

Example 1Find all rational roots of 8x3 + 10x2 – 11x + 2 = 0.

Step 3: Use substitution or synthetic division to test all possible roots.

possible roots: -21

,4

1,2

8 10 -11 2

roots: -2

12

8

4

14

7

-4

-2

0

1,

41

,2

Example 2Find all of the zeros of Q(x) = x3 + 4x2 – 6x - 12.First, use the Rational Root Theorem and a graph of the polynomial function to determine some possibilities.Then use synthetic division to test your choices.

1

2

6

12

6

12

0

2 1 4 -6 -12

Since the remainder is 0, x – 2 is a factor of x3 + 4x2 – 6x - 12.

Example 2Find all of the zeros of Q(x) = x3 + 4x2 – 6x - 12.

Since the remainder is 0, x – 2 is a factor of x3 + 4x2 – 6x - 12.

x3 + 4x2 – 6x – 12 = 0

(x – 2)(x2 + 6x + 6) = 0x = 2

or

26 6 4(1)(6)

x2(1)

6 12x

2

6 2 3x

2

x 3 3

Homework

p.463 #11-21 Odd

7.5.2 Zeros of Polynomial 7.5.2 Zeros of Polynomial FunctionsFunctions

7.5.2 Zeros of Polynomial 7.5.2 Zeros of Polynomial FunctionsFunctions

Objectives: •Use the Complex Conjugate Root Theorem to find the zeros of a polynomial function•Use the Fundamental Theorem to write a polynomial function given sufficient information about its zeros

Example 1Find all of the zeros of P(x) = -4x3 + 2x2 – x + 3.First, use the Rational Root Theorem and a graph of the polynomial function to determine some possibilities.Then use synthetic division to test your choices.

-4

-4

-2

-2

-3

-3

0

1 -4 2 -1 3

Since the remainder is 0, x – 1 is a factor of -4x3 + 2x2 – x + 3.

-4x3 + 2x2 – x + 3 = 0

(x – 1)(-4x2 - 2x - 3) = 0

x = 1

or

2( 2) ( 2) 4( 4)( 3)x

2( 4)

2 44

x8

2 2 11

x8

1 i 11

x4

Example 1Find all of the zeros of P(x) = -4x3 + 2x2 – x + 3.

Since the remainder is 0, x – 1 is a factor of -4x3 + 2x2 – x + 3.

Complex Conjugate Root Theorem

If P is a polynomial function with real-number coefficients and a + bi (where b = 0) is a root of P(x) = 0, then a – bi is also a root of P(x) = 0.

Graph R(x) = 2x3 – x2 – 4x. How many real zeros does R have?

Graph S(x) = 2x3 – x2 – 4x + 3. How many real zeros does S have?

Graph T(x) = 2x3 – x2 – 4x + 6. How many real zeros does T have?

Homework

Page 464 #23-33 Odd

Example 2Write a polynomial function, P, given that P has a degree of 3, P(0) = 120, and its zeros are -3, 2, and 4.

x = -3 and

P(x) = a(x + 3)(x – 2)(x – 4)

Since the zeros are -3, 2, and 4, x = 4

x + 3= 0

x – 4= 0and

x = 2

x - 2 = 0

P(x) = a(x2 + x - 6)(x – 4)

P(x) = a(x3 - 3x2 – 10x + 24)

120 = a((0)3 - 3(0)2 – 10(0) + 24)120 = 24 a a = 5

P(x) = 5(x3 - 3x2 – 10x + 24)

P(x) = 5x3 - 15x2 – 50 x + 120

Example 3Write a polynomial function, P, given that P has a degree of 2, P(0) = 18, and its zeros are -3 + 3i and -3 – 3i.

x = -3 + 3i

and

x + 3 - 3i( )( ) = 0

x2 + 3x + 3xi x2 + 6x + 9 – 9(-1)

= 0

Since the zeros are -3 + 3i and -3 – 3i,

x = -3 - 3i

x – (-3 + 3i) = 0

x – (-3 - 3i) = 0

and

x + 3 - 3i = 0

x + 3 + 3i = 0

and

x + 3 + 3i

+ 3x + 9 + 9i

– 3xi – 9i – 9i2 = 0

x2 + 6x + 18 = 0

Homework

Page 464 #41-49 Odd