Algebra 2

Chapter 6 Notes

Polynomials and

Polynomial Functions

NOTES: Page 72, Section 6.1

Using Properties of Exponents

Product of Powers Property a m • a n = a m + n 5 2 • 5 3 = 5 2 + 3 = 5 5

Power of a Power Property ( a m ) n = a m n ( 5 2 ) 3 = 5 6

Power of a Product Property ( a b ) m = a m b m ( 5 • 4 ) 2 = 5 2 4 2

Negative Exponent Property

a − m = 1 a m

5 − 2 = 1 5 2

Zero Exponent Property a 0 = 1 5 0 = 1

Quotient of Powers Property

a m = a m − n

a n5 3 = 5 3 − 2 = 5 1

5 2

Power of a Quotient Property

a m = a m

b b m 5 3 = 5 3

4 4 3

, a ≠ 0

, a ≠ 0

, a ≠ 0

, a ≠ 0

⁽ ⁾ ⁽ ⁾

NOTES: Page 72a, Section 6.1

Using Properties of Exponents

Evaluating Numerical Expressions

( 2 3 ) 4 = 2 3 • 4 = 2 12 = 4096

3 2

4 = 3 2

4 2 = 9 16

( −5 ) −6 ( −5 ) 4 = ( −5 ) −6 + 4 = ( −5 ) −2 = 1 ( −5 ) 2

= 1 25

Simplifying Algebraic Expressions

r 2

s−5 = ( r ) 2

( s−5 )2 = r 2

s−10 = r 2 • s10

(7 b −3 )2 b5 b = 7 2 b −6 b5 b = 49 b −6 + 5 + 1 = 49 b 0 = 49

( x y 2 )2

x3 y −1= x 2 y 4

x3 y −1 = x 2−3 y 4 − (−1) = x −1 y 5 = y 5 x 1

⁽ ⁾

⁽ ⁾

NOTES: Page 73, Section 6.2

Evaluating and Graphing Polynomial Functions

f(x) = a n x n + a n −1 x n − 1 + …+ a 1 x + a 0

Leading coefficient Constant term

Degree of polynomialPolynomial Function in standard form:

Descending order of exponents from left to right.

Degree Type Standard form

0 Constant f(x) = a 0

1 Linear f(x) = a 1 x + a 0

2 Quadratic f(x) = a 2 x 2+ a 1 x + a 0

3 Cubic f(x) = a 3 x 3 + a 2 x 2+ a 1 x + a 0

4 Quartic f(x) = a 4 x 4 + a 3 x 3 + a 2 x 2+ a 1 x + a 0

NOTES: Page 73a, Section 6.2

Direct Substitution

f(x) = 2 x 4 − 8 x 2 + 5 x − 7 when x = 3

x = 3

2 (x) 4 0 (x) 3 − 8 (x) 2 5 (x) − 7

2 (3) 4 0 (3) 3 − 8 (3) 2 5 (3) − 7

2 (81) 0 (27) − 8 (9) 5 (3) − 7

162 0 − 72 15 − 7

177 − 79 98

f (x) = 98

NOTES: Page 73a, Section 6.2

Synthetic Substitution

f(x) = 2 x 4 − 8 x 2 + 5 x − 7 when x = 3

x = 3

2 x 4 0 x 3 − 8 x 2 5 x − 7

2 0 − 8 5 − 7

6 18 30 105

2 (multiply by 3)

6 (multiply by 3)

10(multiply by 3)

35(multiply by 3) 98

f (x) = 98

NOTES: Page 73b, Section 6.2

Graphing Polynomial Functions

f ( x ) = x3 + x2 – 4 x – 1

f ( – 3 ) = (– 3 ) 3 + (– 3 ) 2 – 4 (– 3 ) – 1f ( – 3 ) = – 27 + 9 + 12 – 1f ( – 3 ) = – 7

x – 3 –2 – 1 0 1 2 3

f ( x ) – 7 3 3 – 1 – 3 3 23

•

•

NOTES: Page 74, Section 6.3

Adding, Subtracting Polynomials

ADD

a) 3 x 3 + 2 x 2 − x − 7 b) 9 x 3 − 2 x + 1

+ x 3 − 10 x 2 + 8 + + 5 x 2 + 12 x − 4

4 x 3 − 8 x 2 − x + 1 9 x 3 + 5 x 2 + 10 x − 3

SUBTRACT

a) 8 x 3 − 3 x 2 − 2 x + 9 8 x 3 − 3 x 2 − 2 x +9

− ( 2 x 3 + 6 x 2 − x + 1 ) − 2 x 3 − 6 x 2 + x − 1

6 x 3 − 9 x 2 − x + 8

b) ( 2 x 2 + 3 x ) − ( 3 x 2 + x − 4 ) = 2 x 2 + 3 x − 3 x 2 − x + 4 = − x 2 + 2 x + 4

NOTES: Page 74b, Section 6.3

Multiplying Polynomials

Multiply

a) − x 2 + 2 x + 4 b) ( x – 3 ) ( 3 x 2 – 2 x – 4 ) =

x – 3 3 x 2 ( x – 3 ) – 2 x( x – 3 ) – 4 ( x – 3 ) =

+ 3 x 2 – 6 x – 12 3 x 3 – 9 x 2 – 2 x 2 + 6 x – 4 x + 12 =

− x 3 + 2 x 2 + 4 x 3 x 3 – 11 x 2 + 2 x + 12

− x 3 + 5 x 2 – 2 x – 12

c) ( x – 1 ) ( x + 4 ) ( x + 3 ) = (x2 + 3x – 4 ) ( x + 3 )

= x2 ( x + 3 ) + 3 x ( x + 3 ) – 4 ( x + 3 )

= x 3 + 3 x 2 + 3 x 2 + 9 x – 4 x – 12

= x 3 + 6 x 2 + 5 x – 12

NOTES: Page 75, Section 6.3

Multiplying Polynomials

SPECIAL PRODUCT PATTERNS EXAMPLES

SUM & DIFFERENCE

( a + b ) ( a – b ) = a 2 + ab – ab – b 2 = a 2 – b 2

( x + 3 ) ( x – 3 ) = x 2 + 3x – 3x – 3 2 = x 2 – 9

SQUARE OF A BINOMIAL

( a + b ) 2 = ( a + b ) ( a + b ) =a 2 + 2 a b + b 2

(y + 4 ) 2 = ( y + 4 ) ( y + 4 ) = y 2 + 8 y + 16

( a – b ) 2 = ( a – b ) ( a – b ) =a 2 – 2 a b + b 2

( 3t2 – 2 ) 2 = ( 3t2 – 2 ) (3t2 – 2 ) =9 t 4 – 12 t 2 + 4

CUBE OF A BINOMIAL

(a + b)3 = (a + b) (a + b) (a + b) = (a 2 + 2 a b + b 2) (a + b) = a3 + 3a2b + 3ab2 + b3

(x + 1)3 = (x + 1) (x + 1) (x + 1) = (x 2 + 2 x + 1) (x + 1) = x3 + 3x2 + 3x + 1

(a – b)3 = (a – b ) (a – b ) (a – b ) (a 2 – 2 a b + b 2 ) (a – b ) =a3 – 3a2b + 3ab2 – b3

(p – 2)3 = (p – 2 ) (p – 2 ) (p – 2 ) (p 2 – 4 p + 4 ) (p – 2 ) =p3 – 6p2 + 12p – 8

NOTES: Page 75, Section 6.3

Multiplying Polynomials

SPECIAL PRODUCT PATTERNS EXAMPLES

SUM & DIFFERENCE

( a + b ) ( a – b ) = a 2 – b 2

( x + 3 ) ( x – 3 ) = x 2 – 9

SQUARE OF A BINOMIAL

( a + b ) 2 =a 2 + 2 a b + b 2

( y + 4 ) 2 = y 2 + 8 y + 16

( a – b ) 2 = a 2 – 2 a b + b 2

( 3t2 – 2 ) 2 =9 t 4 – 12 t 2 + 4

CUBE OF A BINOMIAL

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(x + 1)3 = x3 + 3x2 + 3x + 1

(a – b)3 =a3 – 3a2b + 3ab2 – b3

(p– 2)3 = p3 – 6p2 + 12p – 8

NOTES: Page 75a, Section 6.3

Multiplying Polynomials

SUM & DIFFERENCE ( 4 n + 5 ) ( 4 n – 5 ) = ( 4 n ) 2 – 5 2 = 16 n 2 - 25

SQUARE OF A BINOMIAL

( 9 y – x2 ) 2 = ( 9 y ) 2 – 2 ( 9 y ) ( x ) 2 + ( x 2 ) 2

= 81 y 2 – 18 y x 2 + x 4

CUBE OF A BINOMIAL

( a b + 2 ) 3 = ( a b )3 + 3 ( a b ) 2 ( 2 ) + 3 ( a b ) ( 2 ) 2 + 2 3

= a 3 b 3 + 6 a 2 b 2 + 12 a b + 8

NOTES: Page 76, Section 6.4

Factoring and Solving Polynomial Equations

Types of Factoring

General Trinomial 2 x 2 – 5 x – 12 = ( 2 x + 3 ) ( x – 4 )

Perfect Square Trinomial x 2 + 10 x + 25 = ( x + 5 ) 2

Difference of Two Squares 4 x 2 – 9 = ( 2 x + 3 ) ( 2 x – 3 )

Common Monomial Factors 6 x 2 + 15 x = 3 x ( 2 x + 5 )

Sum and Difference of two cubes

Sum of 2 Cubes a 3 + b 3 = ( a + b ) ( a 2 – a b + b 2 )

Ex 1 of sum of 2 cubes x 3 + 8 = ( x + 2 ) ( x 2 – 2 x + 4 )

Ex 2 of sum of 2 cubes x 3 + 27 = ( x + 3 ) ( x 2 – 3 x + 9 )

Difference of 2 cubes a 3 − b 3 = ( a − b ) ( a 2 + a b + b 2 )

Ex 1 of Difference of 2 cubes 8 x 3 − 1 = ( 2 x − 1 ) ( 4 x 2 + 2 x + 1)

Ex 2 of Difference of 2 cubes 16 u 5 − 250 u 2 = 2 u 2 ( 8 u 3 − 125 )

= 2 u 2 ( 2 u − 5 ) ( 4 u 2 + 10 u + 25)

NOTES: Page 76a, Section 6.4

Factoring and Solving Polynomial Equations

Sum and Difference of two cubes

Sum of 2 Cubes a 3 + b 3 = ( a + b ) ( a 2 – a b + b 2 ) = a3 – a2b + ab2 + a2b – ab2 + b3

Ex 1 of sum of 2 cubes x 3 + 8 = ( x + 2 ) ( x 2 – 2 x + 4 ) = x3 – 2x2 + 4x + 2x2 – 4x + 8

Ex 2 of sum of 2 cubes x 3 + 27 = ( x + 3 ) ( x 2 – 3 x + 9 ) = x3 – 3x2 + 9x + 3x2 – 9x + 27

Difference of 2 cubes a 3 − b 3 = ( a − b ) ( a 2 + a b + b 2 ) = a3 + a2b + ab2 – a2b – ab2 – b3

Ex 1 of Difference of 2 cubes

8 x 3 − 1 = ( 2 x − 1 ) ( 4 x 2 + 2 x + 1) = 8x3 + 4x2 + 2x – 4x2 – 2x – 1

Ex 2 of Difference of 2 cubes

16 u 5 − 250 u 2 = 2 u 2 ( 8 u 3 − 125 )

= 2 u 2 ( 2 u − 5 ) ( 4 u 2 + 10 u + 25) = 8 u3 + 20 u2 + 50 u –20 u2 – 50 u – 125

Types of Factoring

General Trinomial 2 x 2 – 5 x – 12 = ( 2 x + 3 ) ( x – 4 )

Perfect Square Trinomial x 2 + 10 x + 25 = ( x + 5 ) 2

Difference of Two Squares 4 x 2 – 9 = ( 2 x + 3 ) ( 2 x – 3 )

Common Monomial Factors 6 x 2 + 15 x = 3 x ( 2 x + 5 )

Factoring and Solving Polynomial Functions

SUM of 2 CUBES a 3 + b 3 = ( a + b ) ( a 2 – a b + b 2 )x 3 + 8 = ( x + 2 ) ( x 2 – 2 x + 4 )

DIFFERENCE OF 2 CUBES a 3 − b 3 = ( a − b ) ( a 2 + a b + b 2 )8 x 3 − 1 = ( 2 x − 1 ) ( 4 x 2 + 2 x + 1)

SUM & DIFFERENCE of a Binomial ( a + b ) ( a – b ) = a 2 – b 2 ( x + 3 ) ( x – 3 ) = x 2 – 9

SQUARE of a Binomial (positive) ( a + b ) 2 = a 2 + 2 a b + b 2

( y + 4 ) 2 = y 2 + 8 y + 16

SQUARE of a Binomial (negative) ( a – b ) 2 = a 2 – 2 a b + b 2

( 3t2 – 2 ) 2 = 9 t 4 – 12 t 2 + 4

CUBE of a Binomial (positive) (a + b)3 = a3 + 3a2b + 3ab2 + b3

(x + 2)3 = x3 + 6x2 + 12x + 8

CUBE of a Binomial (negative) (a – b)3 = a3 – 3a2b + 3ab2 – b3

(p– 2)3 = p3 – 6p2 + 12p – 8

NOTES: Page 76a, Section 6.4

Factoring by Grouping & Factor in Quadratic Form

Factoring by Grouping x 3 – 2 x 2 – 9 x + 18x 2 ( x – 2 ) – 9 ( x – 2 ) ( x 2 – 9) ( x – 2 ) ( x – 3 ) ( x + 3 ) ( x – 2 )

Factor in Quadratic Form81 x 4 – 16 == ( 9 x 2 ) 2 – ( 4 ) 2

= ( 9 x 2 – 4 ) ( 9 x 2 + 4 )= ( 3 x – 2 ) ( 3 x + 2 ) ( 9 x 2 + 4 )

4 x 6 – 20 x 4 + 24 x 2 4 x 2 ( x 4 – 5 x 2 + 6 ) 4 x 2 ( x 2 – 3) ( x 2 – 2 )

NOTES: Page 77, Section 6.4

Factoring & Solving Polynomial Equations

2 x 5 + 24 x = 14 x 3 2 x 5 – 14 x 3 + 24 x = 02 x ( x 4 – 7 x 2 + 12 ) = 0 2 x ( x 2 – 3) ( x 2 – 4) 2 x ( x 2 – 3) ( x – 2 ) ( x + 2 ) = 0

x = 0 , ± 3 , 2, – 2

A Review of Long Division

3 2 + 1

3 1 2 9 9 8 5 3 1 2

− ( 9 3 6 )

6 2 5

− ( 6 2 4 )

1

QUOTIENT + REMINDER

DIVISOR DIVIDEND DIVISOR

NOTES: Page 78, Section 6.5

The Remainder and Factor Theorems

Polynomial Long Division,

divide 2 x 4 + 3 x 3 + 5 x – 1 by x 2 – 2 x + 2

2 x 2 + 7 x + 10 + 11 x – 21

x 2 – 2 x + 2 2 x 4 + 3 x 3 + 0 x 2 + 5 x – 1 x 2 – 2 x + 2– ( 2 x 4 – 4 x 3 + 4 x 2 )

7 x 3 – 4 x 2 + 5 x

– ( 7 x 3 – 14 x 2 + 14 x )

10 x 2 – 9 x – 1

– ( 10 x 2 – 20 x + 20)

11 x – 21 [ remainder ]

At each stage divide the term with the highest power in what is left of the dividend by the first term of the divisor. This gives the next term of the quotient.

NOTES: Page 78a, Section 6.5

x 3 + 2 x 2 – 6 x – 9x – 2

x 3 x 2 x 1

2 1 2 – 6 – 9

+ 2 8 4

Quotient 1 4 2 – 5

= x 2 + 4 x + 2 – 5 x – 2

x 2 + 4 x + 2 + – 5

x – 2 x 3 + 2 x 2 – 6 x – 9 x – 2– ( x 3 – 2 x 2 )

4 x 2 – 6 x

– ( 4 x 2 – 8 x )

2 x – 9

– ( 2 x – 4)

– 5

Example 1: Long Division versus Synthetic Division [ form: x – h ]

NOTES: Page 78b, Section 6.5

x 3 + 2 x 2 – 6 x – 9x + 3

x 3 x 2 x 1

– 3 1 2 – 6 – 9

+ – 3 3 9

Quotient 1 – 1 – 3 0

= x 2 – x – 3

x 2 – x – 3

x + 3 x 3 + 2 x 2 – 6 x – 9

– ( x 3 + 3 x 2 )

– x 2 – 6 x

– ( – x 2 – 3 x )

– 3 x – 9

– (– 3 x – 9 )

0

Example 2: Long Division versus Synthetic Division [ form: x – h ]

Remainder Theorem:

If polynomial f (x) is divided by x – k , then the remainder is r = f (k)

Factor Theorem:

A polynomial f (x) has a factor x – k , if and only if f (k) = 0

To be a factor, there should be NO remainder, so f (k) = 0 x – ( k ) = x – ( – 3)

x3 2 x2 – 6 x – 9

– 3 1 2 – 6 – 9

– 3 3 9

1 – 1 – 3 0

Quotient x2 – x – 3

f (x) = (x)3 2(x)2 – 6(x) – 9

f (– 3) = (– 3)3 2(– 3)2 – 6(– 3) – 9

f (– 3) = – 27 + 18 + 18 – 9

f (– 3) = 0

f (x) = (x)3 2(x)2 – 6(x) – 9

f (2) = (2)3 2(2)2 – 6(2) – 9

f ( 2) = 8 8 – 12 – 9

f (2) = – 5

Where the remainder, r = f (k) x – ( k ) = x – ( 2)

x3 2 x2 – 6 x – 9

2 1 2 – 6 – 9

2 8 4

1 4 2 – 5

Quotient x2 + 4 x + 2 – 5

x – 2

NOTES: Page 79, Section 6.5

Factor Theorem: A polynomial f (x) has a factor ( x – h ) if and only if f ( k ) = 0

Where k is called a “ZERO of the Function” because f ( k ) = 0

Example of Factoring a Polynomial

f ( x ) = 2 x 3 + 11 x 2 + 18 x + 9 given f ( − 3 ) = 0

… because f ( − 3 ) = 0, the x − ( − 3 ) or x + 3 is factor of f ( x )

x 3 x 2 x 1

− 3 2 11 18 9

+ – 6 – 15 – 9

Quotient 2 5 3 0

2 x 3 + 11 x 2 + 18 x + 9 = ( x + 3 ) ( 2 x 2 + 5 x + 3 )

= ( x + 3 ) ( 2 x + 3 ) ( x + 1 )

NOTES: Page 79a, Section 6.5

Example of Findng Zeros of a Polynomial Function

One zero of f ( x ) = x 3 − 2 x 2 − 9 x + 18 is x = 2

Find the other zeros of the function.

x 3 x 2 x 1

2 1 − 2 − 9 18

+ 2 0 – 18

Quotient 1 0 − 9 0

f ( x ) = ( x − 2 ) ( x 2 − 9 ) = 0

= ( x − 2 ) ( x + 3 ) ( x − 3 ) = 0

By the Factor Theorem x = 2, − 3, 3

NOTES: Page 80, Section 6.6

Using the Rational Zero Theorem

This polynomial function f ( x ) = 64 x 3 + 120 x 2 − 34 x − 105 has

as its zeros. Note the numerators ( – 3 , − 5 , and 7 ) are factors of the constant term , − 105, while the

denominators, (2, 4,and 8) are factors of leading coefficient, 64

– 3 – 5 7

2 4 8

The Rational Zero TheoremIf f (x) = a n x n + . . . + a 1 x n-1 + a 0 has integer coefficients, then every rational zero of f has the following form:

p=

factor of constant term, a 0

q factor of leading coefficient a n

Ex 1. Find the rational zeros of f ( x ) = 1 x 3 + 2 x 2 − 11 x − 12 x = ± 1

1, ± 2

1, ± 3

1, ± 4

1, ± 6

1, ± 12

1

– 1 1 2 – 11 – 12

– 1 – 1 12

1 1 – 12 0

So, f(x) = ( x + 1 ) ( x2 + x – 12 ) = ( x + 1 ) ( x – 3 ) ( x + 4 )

Turning Points of Polynomial FunctionsThe graph of every polynomial function of degree n has at most n – 1 turning points. Moreover, if a polynomial function has n distinct real zeros then the graph has exactly n – 1 turning points.

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Local maximum

Local minimum