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**On a blank piece of paper please complete PG 328 complete #1-12 on the Prerequisite Skills**

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**Chapter 5: Polynomials & Polynomial Functions**

BIG IDEAS: Performing operations with polynomials Solving polynomial equations and finding zeros

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**Lesson 1: Use Properties of Exponents**

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**How do you simplify algebraic expressions with exponents?**

Essential question How do you simplify algebraic expressions with exponents?

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VOCABULARY Scientific notation: The representat5ion of a number in the form c x 10n where 1≤c<10 and n is an integer.

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**Power of a product property**

EXAMPLE 1 Evaluate numerical expressions a. (– )2 = (– 4)2 (25)2 Power of a product property = Power of a power property = = 16,384 Simplify and evaluate power. b. 115 118 –1 118 115 = Negative exponent property = 118 – 5 Quotient of powers property = 113 = 1331 Simplify and evaluate power.

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**Product of powers property**

EXAMPLE 3 Simplify expressions a. b–4b6b7 = b– = b9 Product of powers property b. r–2 –3 s3 ( r –2 )–3 ( s3 )–3 = Power of a quotient property = r 6 s–9 Power of a power property = r6s9 Negative exponent property c m4n –5 2n–5 = 8m4n –5 – (–5) Quotient of powers property = 8m4n0= 8m4 Zero exponent property

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**Power of a product property**

EXAMPLE 4 Standardized Test Practice SOLUTION (x–3y3)2 x5y6 = (x–3)2(y3)2 x5y6 Power of a product property x –6y6 x5y6 = Power of a power property

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GUIDED PRACTICE for Examples 3, 4, and 5 Simplify the expression. Tell which properties of exponents you used. 5. x–6x5 x3 ANSWER x2 ; Product of powers property (7y2z5)(y–4z–1) 7z4 y2 ; Product of powers property, Negative exponent property ANSWER

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GUIDED PRACTICE for Examples 3, 4, and 5 7. s 3 2 t–4 s6t8 ANSWER ; Power of a power property, Negative exponent property 8. x4y– x3y6 x3 y24 ; Quotient of powers property, Power of a Quotient property, Negative exponent property ANSWER

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**How do you simplify algebraic expressions with exponents.**

Essential question How do you simplify algebraic expressions with exponents. Use the properties of exponents to rewrite it with only positive exponents

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**Simplify the expression (-3x3) (5x)**

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**Lesson 3: Add, Subtract and Multiply Polynomials**

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**What are the special product patterns?**

Essential question What are the special product patterns?

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VOCABULARY Like Terms: Terms that have the same variable parts. Constant terms are also like terms.

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EXAMPLE 1 Add polynomials vertically and horizontally a. Add 2x3 – 5x2 + 3x – 9 and x3 + 6x in a vertical format. SOLUTION a x3 – 5x2 + 3x – 9 + x3 + 6x 3x3 + x2 + 3x + 2

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EXAMPLE 1 Add polynomials vertically and horizontally b. Add 3y3 – 2y2 – 7y and –4y2 + 2y – 5 in a horizontal format. (3y3 – 2y2 – 7y) + (–4y2 + 2y – 5) = 3y3 – 2y2 – 4y2 – 7y + 2y – 5 = 3y3 – 6y2 – 5y – 5

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EXAMPLE 2 Subtract polynomials vertically and horizontally a. Subtract 3x3 + 2x2 – x + 7 from 8x3 – x2 – 5x + 1 in a vertical format. SOLUTION a. Align like terms, then add the opposite of the subtracted polynomial. 8x3 – x2 – 5x + 1 – (3x3 + 2x2 – x + 7) 8x3 – x2 – 5x + 1 + – 3x3 – 2x2 + x – 7 5x3 – 3x2 – 4x – 6

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EXAMPLE 2 Subtract polynomials vertically and horizontally b. Subtract 5z2 – z + 3 from 4z2 + 9z – 12 in a horizontal format. Write the opposite of the subtracted polynomial, then add like terms. (4z2 + 9z – 12) – (5z2 – z + 3) = 4z2 + 9z – 12 – 5z2 + z – 3 = 4z2 – 5z2 + 9z + z – 12 – 3 = –z2 + 10z – 15

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GUIDED PRACTICE for Examples 1 and 2 Find the sum or difference. 1. (t2 – 6t + 2) + (5t2 – t – 8) ANSWER 6t2 – 7t – 6 2. (8d – 3 + 9d3) – (d3 – 13d2 – 4) ANSWER 8d3 + 13d2 + 8d + 1

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**Multiply –2y2 + 3y – 6 by –2 . Multiply –2y2 + 3y – 6 by y**

EXAMPLE 3 Multiply polynomials vertically and horizontally a. Multiply –2y2 + 3y – 6 and y – 2 in a vertical format. b. Multiply x + 3 and 3x2 – 2x + 4 in a horizontal format. SOLUTION a –2y2+ 3y – 6 y – 2 4y2 – 6y + 12 Multiply –2y2 + 3y – 6 by –2 . –2y3 + 3y2 – 6y Multiply –2y2 + 3y – 6 by y –2y3 + 7y2 –12y + 12 Combine like terms.

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EXAMPLE 3 Multiply polynomials vertically and horizontally b. (x + 3)(3x2 – 2x + 4) = (x + 3)3x2 – (x + 3)2x + (x + 3)4 = 3x3 + 9x2 – 2x2 – 6x + 4x + 12 = 3x3 + 7x2 – 2x + 12

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EXAMPLE 4 Multiply three binomials Multiply x – 5, x + 1, and x + 3 in a horizontal format. (x – 5)(x + 1)(x + 3) = (x2 – 4x – 5)(x + 3) = (x2 – 4x – 5)x + (x2 – 4x – 5)3 = x3 – 4x2 – 5x + 3x2 – 12x – 15 = x3 – x2 – 17x – 15

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**Sum and difference Square of a binomial Cube of a binomial EXAMPLE 5**

Use special product patterns a. (3t + 4)(3t – 4) = (3t)2 – 42 Sum and difference = 9t2 – 16 b. (8x – 3)2 = (8x)2 – 2(8x)(3) + 32 Square of a binomial = 64x2 – 48x + 9 c. (pq + 5)3 Cube of a binomial = (pq)3 + 3(pq)2(5) + 3(pq)(5)2 + 53 = p3q3 + 15p2q2 + 75pq + 125

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GUIDED PRACTICE for Examples 3, 4 and 5 Find the product. 3. (x + 2)(3x2 – x – 5) ANSWER 3x3 + 5x2 – 7x – 10 4. (a – 5)(a + 2)(a + 6) (a3 + 3a2 – 28a – 60) ANSWER 5. (xy – 4)3 ANSWER x3y3 – 12x2y2 + 48xy – 64

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**What are the special product patterns?**

Essential question What are the special product patterns? (a+b)(a-b) = a2 – b2 (a+b)2= a2 + 2ab + b2 (a-b)2= a2 - 2ab + b2 (a+b)3= a3 + 3a2b + 3ab2 + b3 (a-b)3= a3 - 3a2b + 3ab2 - b3

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**Multiply the polynomials: (x+2) (x+3) (2x – 1) (2x + 1)**

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**Lesson 4: Factor and Solve Polynomial Equations**

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**How can you solve a higher-degree polynomial equation?**

Essential question How can you solve a higher-degree polynomial equation?

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VOCABULARY Factored Completely: A factorable polynomial with integer coefficients is factored completely if it is written as a product of un-factorable polynomials with integer coefficients Factor by grouping: To factor a polynomial with four terms by grouping, factor common monomials from pairs of terms, then look for a common binomial factor Quadratic Form: The form au2 + bu + c, where u is any expression in x.

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**Perfect square trinomial**

EXAMPLE 1 Find a common monomial factor Factor the polynomial completely. a. x3 + 2x2 – 15x = x(x2 + 2x – 15) Factor common monomial. = x(x + 5)(x – 3) Factor trinomial. b. 2y5 – 18y3 = 2y3(y2 – 9) Factor common monomial. = 2y3(y + 3)(y – 3) Difference of two squares c. 4z4 – 16z3 + 16z2 = 4z2(z2 – 4z + 4) Factor common monomial. = 4z2(z – 2)2 Perfect square trinomial

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**Factor common monomial.**

EXAMPLE 2 Factor the sum or difference of two cubes Factor the polynomial completely. a. x3 + 64 = x3 + 43 Sum of two cubes = (x + 4)(x2 – 4x + 16) b z5 – 250z2 = 2z2(8z3 – 125) Factor common monomial. = 2z2 (2z)3 – 53 Difference of two cubes = 2z2(2z – 5)(4z2 + 10z + 25)

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GUIDED PRACTICE for Examples 1 and 2 Factor the polynomial completely. x3 – 7x2 + 10x ANSWER x( x – 5 )( x – 2 ) y5 – 75y3 ANSWER 3y3(y – 5)(y + 5 ) b b2 ANSWER 2b2(2b + 7)(4b2 –14b + 49) 4. w3 – 27 ANSWER (w – 3)(w2 + 3w + 9)

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**Distributive property**

EXAMPLE 3 Factor by grouping Factor the polynomial x3 – 3x2 – 16x + 48 completely. x3 – 3x2 – 16x + 48 = x2(x – 3) – 16(x – 3) Factor by grouping. = (x2 – 16)(x – 3) Distributive property = (x + 4)(x – 4)(x – 3) Difference of two squares

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**Write as difference of two squares. Difference of two squares**

EXAMPLE 4 Factor polynomials in quadratic form Factor completely: (a) 16x4 – 81 and (b) 2p8 + 10p5 + 12p2. a x4 – 81 = (4x2)2 – 92 Write as difference of two squares. = (4x2 + 9)(4x2 – 9) Difference of two squares = (4x2 + 9)(2x + 3)(2x – 3) Difference of two squares Factor common monomial. b. 2p8 + 10p5 + 12p2 = 2p2(p6 + 5p3 + 6) Factor trinomial in quadratic form. = 2p2(p3 + 3)(p3 + 2)

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GUIDED PRACTICE for Examples 3 and 4 Factor the polynomial completely. x3 + 7x2 – 9x – 63 ANSWER (x + 3)(x – 3)(x + 7) g4 – 625 (4g2 + 25)(2g + 5)(2g – 5) ANSWER t6 – 20t4 + 24t2 4t2(t2 – 3)(t2 – 2 ) ANSWER

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**How can you solve a higher-degree polynomial equation?**

Essential question How can you solve a higher-degree polynomial equation? Factor the polynomial completely and use the zero product property

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**A company’s income is modeled by the function P = 22x2 – 571 x**

A company’s income is modeled by the function P = 22x2 – 571 x. What is the value of P when x = 200?

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**Lesson 5: Apply the Remainder and Factor Theorem**

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Essential question If you know one zero of a polynomial function, how can you determine another zero?

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VOCABULARY Polynomial long division: A method used to divide polynomials similar to the way you divide numbers Synthetic division: A method used to divide a polynomial by a divisor of the form x - k

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EXAMPLE 1 Use polynomial long division Divide f (x) = 3x4 – 5x3 + 4x – 6 by x2 – 3x + 5. SOLUTION Write polynomial division in the same format you use when dividing numbers. Include a “0” as the coefficient of x2 in the dividend. 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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**) quotient Multiply divisor by 3x4/x2 = 3x2**

EXAMPLE 1 Use polynomial long division 3x2 + 4x – 3 quotient x2 – 3x + 5 3x4 – 5x3 + 0x2 + 4x – 6 ) Multiply divisor by 3x4/x2 = 3x2 3x4 – 9x3 + 15x2 Subtract. Bring down next term. 4x3 – 15x x 4x3 – 12x2 + 20x Multiply divisor by 4x3/x2 = 4x Subtract. Bring down next term. –3x2 – 16x – 6 –3x2 + 9x – 15 Multiply divisor by – 3x2/x2 = – 3 –25x + 9 remainder

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EXAMPLE 1 Use polynomial long division 3x4 – 5x3 + 4x – 6 x2 – 3x + 5 = 3x2 + 4x – 3 + –25x + 9 ANSWER CHECK You can check the result of a division problem by multiplying the quotient by the divisor and adding the remainder. The result should be the dividend. (3x2 + 4x – 3)(x2 – 3x + 5) + (–25x + 9) = 3x2(x2 – 3x + 5) + 4x(x2 – 3x + 5) – 3(x2 – 3x + 5) – 25x + 9 = 3x4 – 9x3 + 15x2 + 4x3 – 12x2 + 20x – 3x2 + 9x – 15 – 25x + 9 = 3x4 – 5x3 + 4x – 6

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**) quotient Multiply divisor by x3/x = x2. Subtract.**

EXAMPLE 2 Use polynomial long division with a linear divisor Divide f(x) = x3 + 5x2 – 7x + 2 by x – 2. x2 + 7x quotient x – 2 x3 + 5x2 – 7x ) x3 – 2x2 Multiply divisor by x3/x = x2. 7x2 – 7x Subtract. 7x2 – 14x Multiply divisor by 7x2/x = 7x. 7x + 2 Subtract. 7x – 14 Multiply divisor by 7x/x = 7. 16 remainder ANSWER x3 + 5x2 – 7x +2 x – 2 = x2 + 7x + 7 + 16

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GUIDED PRACTICE for Examples 1 and 2 Divide using polynomial long division. (2x4 + x3 + x – 1) (x2 + 2x – 1) (2x2 – 3x + 8) + –18x + 7 x2 + 2x – 1 ANSWER (x3 – x2 + 4x – 10) (x + 2) (x2 – 3x + 10) + –30 x + 2 ANSWER

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EXAMPLE 3 Use synthetic division Divide f (x)= 2x3 + x2 – 8x + 5 by x + 3 using synthetic division. SOLUTION – – – –21 2 – –16 2x3 + x2 – 8x + 5 x + 3 = 2x2 – 5x + 7 – 16 ANSWER

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EXAMPLE 4 Factor a polynomial Factor f (x) = 3x3 – 4x2 – 28x – 16 completely given that x + 2 is a factor. SOLUTION Because x + 2 is a factor of f (x), you know that f (–2) = 0. Use synthetic division to find the other factors. – –4 – –16 – 3 – –

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**Write original polynomial.**

EXAMPLE 4 Factor a polynomial Use the result to write f (x) as a product of two factors and then factor completely. f (x) = 3x3 – 4x2 – 28x – 16 Write original polynomial. = (x + 2)(3x2 – 10x – 8) Write as a product of two factors. = (x + 2)(3x + 2)(x – 4) Factor trinomial.

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GUIDED PRACTICE for Examples 3 and 4 Divide using synthetic division. 3. (x3 + 4x2 – x – 1) (x + 3) x2 + x – 4 + 11 x + 3 ANSWER 4. (4x3 + x2 – 3x + 7) (x – 1) 4x2 + 5x + 2 + 9 x – 1 ANSWER

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GUIDED PRACTICE for Examples 3 and 4 Factor the polynomial completely given that x – 4 is a factor. f (x) = x3 – 6x2 + 5x + 12 ANSWER (x – 4)(x –3)(x + 1) f (x) = x3 – x2 – 22x + 40 ANSWER (x – 4)(x –2)(x +5)

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**Use synthetic division and factor the results**

Essential question If you know one zero of a polynomial function, how can you determine another zero? Use synthetic division and factor the results to find the other zeros.

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