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Algebra 2 Chapter 6 Notes Polynomials and Polynomial Functions Algebra 2 Chapter 6 Notes Polynomials and Polynomial Functions

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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 = 15 0 = 1 Quotient of Powers Property a m = a m − n a n 5 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 ⁽⁾ ⁽⁾

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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 s 10 (7 b −3 ) 2 b 5 b= 7 2 b −6 b 5 b= 49 b −6 + 5 + 1 = 49 b 0 = 49 ( x y 2 ) 2 x 3 y −1 = x 2 y 4 x 3 y −1 = x 2−3 y 4 − (−1) = x −1 y 5 = y 5 x 1 ⁽⁾ ⁽ ⁾

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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 polynomial Polynomial Function in standard form: Descending order of exponents from left to right. DegreeTypeStandard form 0Constantf(x) = a 0 1Linearf(x) = a 1 x + a 0 2Quadraticf(x) = a 2 x 2 + a 1 x + a 0 3Cubicf(x) = a 3 x 3 + a 2 x 2 + a 1 x + a 0 4Quarticf(x) = a 4 x 4 + a 3 x 3 + a 2 x 2 + a 1 x + a 0

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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 1620− 7215− 7 177 − 7998 f (x) = 98

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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 20− 85− 7 61830105 2 (multiply by 3) 6 (multiply by 3) 10 (multiply by 3) 35 (multiply by 3) 98 f (x) = 98

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NOTES: Page 73b, Section 6.2 Graphing Polynomial Functions f ( x ) = x 3 + x 2 – 4 x – 1 f ( – 3 ) = (– 3 ) 3 + (– 3 ) 2 – 4 (– 3 ) – 1 f ( – 3 ) = – 27 + 9 + 12 – 1 f ( – 3 ) = – 7 x– 3–2– 10123 f ( x )– 733– 1– 3323

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NOTES: Page 74, Section 6.3 Adding, Subtracting Polynomials ADD a)3 x 3 + 2 x 2 − x− 7b)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+ 19 x 3 + 5 x 2 + 10 x− 3 SUBTRACT a)8 x 3 − 3 x 2 − 2 x+ 98 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

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

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NOTES: Page 75, Section 6.3 Multiplying Polynomials SPECIAL PRODUCT PATTERNS EXAMPLES 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 ( 3t 2 – 2 ) 2 = ( 3t 2 – 2 ) (3t 2 – 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) = a 3 + 3a 2 b + 3ab 2 + b 3 a 3 + 3a 2 b + 3ab 2 + b 3 (x + 1) 3 = (x + 1) (x + 1) (x + 1) = (x 2 + 2 x + 1) (x + 1) = x 3 + 3x 2 + 3x + 1 (a – b) 3 = (a – b ) (a – b ) (a – b ) (a 2 – 2 a b + b 2 ) (a – b ) = a 3 – 3a 2 b + 3ab 2 – b 3 (p – 2) 3 = (p – 2 ) (p – 2 ) (p – 2 ) (p 2 – 4 p + 4 ) (p – 2 ) = p 3 – 6p 2 + 12p – 8

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NOTES: Page 75, Section 6.3 Multiplying Polynomials SPECIAL PRODUCT PATTERNS EXAMPLES 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 ( 3t 2 – 2 ) 2 = 9 t 4 – 12 t 2 + 4 CUBE OF A BINOMIAL (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3 a 3 + 3a 2 b + 3ab 2 + b 3 (x + 1) 3 = x 3 + 3x 2 + 3x + 1 (a – b) 3 = a 3 – 3a 2 b + 3ab 2 – b 3 (p– 2) 3 = p 3 – 6p 2 + 12p – 8

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NOTES: Page 75a, Section 6.3 Multiplying Polynomials

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NOTES: Page 76, Section 6.4 Factoring and Solving Polynomial Equations

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NOTES: Page 76a, Section 6.4 Factoring and Solving Polynomial Equations

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NOTES: Page 76a, Section 6.4 Factoring by Grouping & Factor in Quadratic Form Factoring by Grouping x 3 – 2 x 2 – 9 x + 18 x 2 ( x – 2 ) – 9 ( x – 2 ) ( x 2 – 9) ( x – 2 ) ( x – 3 ) ( x + 3 ) ( x – 2 ) Factor in Quadratic Form 81 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 )

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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 = 0 2 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

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A Review of Long Division 32+1 312 99853 1 2 − ( 93 6 ) 625 − ( 62 4 ) 1 QUOTIENT+REMINDER DIVISORDIVIDENDDIVISOR

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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.

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NOTES: Page 78a, Section 6.5 x 3 + 2 x 2 – 6 x – 9 x – 2 x 3 x 2 x 1 212– 6– 9 +2 8 4 Quotient 14 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 ]

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NOTES: Page 78b, Section 6.5 x 3 + 2 x 2 – 6 x – 9 x + 3 x 3 x 2 x 1 – 312– 6– 9 +– 3 3 9 Quotient 1 – 1 – 30 = 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 ]

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Remainder Theorem: If polynomial f (x) is divided by x – k, then the remainder is r = f (k) 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 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) x3x3 2 x 2 – 6 x– 9 – 312– 6– 9 – 339 1– 1– 30 Quotientx2x2 – 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 – 9 f (– 3)=0 f (x)=(x) 3 2(x) 2 – 6(x)– 9 2 f (2)= 2(2)32(2)3 2 2(2) 2 2 – 6(2)– 9 2 f ( 2)=88– 12– 9 2 f (2)=– 5 k2 Where the remainder, r = f (k) x – ( k ) = x – ( 2) x3x3 2 x 2 – 6 x– 9212– 6– 9 284 142– 5 Quotientx2x2 + 4 x+ 2– 5 2 x – 2

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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 − 321118 9 +– 6– 15 – 9 Quotient 25 30 2 x 3 + 11 x 2 + 18 x + 9= ( x + 3 ) ( 2 x 2 + 5 x + 3 ) = ( x + 3 ) ( 2 x + 3 ) ( x + 1 )

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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 21− 2− 9 18 + 2 0 – 18 Quotient 10 − 90 f ( x ) = ( x − 2 ) ( x 2 − 9 ) = 0 = ( x − 2 ) ( x + 3 ) ( x − 3 ) = 0 By the Factor Theorem x = 2, − 3, 3

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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 Theorem If 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: The Rational Zero Theorem If 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,±2121,±3131,±4141,±6161,±12 1 – 112– 11– 12 – 1 12 11– 120 So, f(x) = ( x + 1 ) ( x 2 + x – 12 ) = ( x + 1 ) ( x – 3 ) ( x + 4 )

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Turning Points of Polynomial Functions The 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. Turning Points of Polynomial Functions The 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. ● ● Local maximum Local minimum

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