3-1 © 2011 Pearson Prentice Hall. All rights reserved Chapter 6 Exponents, Polynomials, and Polynomial Functions Active Learning Questions

3-2 © 2011 Pearson Prentice Hall. All rights reserved Section 6.1 Exponents and Scientific Notation Simplify: b.) c.) a.)

3-3 © 2011 Pearson Prentice Hall. All rights reserved Section 6.1 Exponents and Scientific Notation Simplify: b.) c.) a.)

3-4 © 2011 Pearson Prentice Hall. All rights reserved Section 6.1 Exponents and Scientific Notation Find and correct the error in the following: a.) b.) c.)

3-5 © 2011 Pearson Prentice Hall. All rights reserved Section 6.1 Exponents and Scientific Notation Find and correct the error in the following: a.) b.) c.)

3-6 © 2011 Pearson Prentice Hall. All rights reserved Section 6.1 Exponents and Scientific Notation Which of the following numbers have values that are less than 1? a.) 3.5  10 5 b.) 3.5  10 – 5 c.) – 3.5  10 5

3-7 © 2011 Pearson Prentice Hall. All rights reserved Section 6.1 Exponents and Scientific Notation Which of the following numbers have values that are less than 1? a.) 3.5  10 5 b.) 3.5  10 – 5 c.) – 3.5  10 5

3-8 © 2011 Pearson Prentice Hall. All rights reserved Section 6.2 More Work with Exponents and Scientific Notation simplifies to: a.) – 6 b.) – 9 c.) 9

3-9 © 2011 Pearson Prentice Hall. All rights reserved Section 6.2 More Work with Exponents and Scientific Notation simplifies to: a.) – 6 b.) – 9 c.) 9

3-10 © 2011 Pearson Prentice Hall. All rights reserved Section 6.3 Polynomials and Polynomial Functions Simplify: (2x 2 + x + xy 3 ) – (x 2 + 5xy 3 ) a.) x 2 + x – 4xy 3 b.) x 2 + x + 6xy 3 c.) 2 + x – 4xy 3

3-11 © 2011 Pearson Prentice Hall. All rights reserved Section 6.3 Polynomials and Polynomial Functions Simplify: (2x 2 + x + xy 3 ) – (x 2 + 5xy 3 ) a.) x 2 + x – 4xy 3 b.) x 2 + x + 6xy 3 c.) 2 + x – 4xy 3

3-12 © 2011 Pearson Prentice Hall. All rights reserved Section 6.3 Polynomials and Polynomial Functions Which polynomial is the opposite of 16x 3 – 5x + 7? a.) – 16x 3 – 5x + 7 b.) 16x 3 + 5x + 7 c.) – 16x 3 + 5x – 7

3-13 © 2011 Pearson Prentice Hall. All rights reserved Section 6.3 Polynomials and Polynomial Functions Which polynomial is the opposite of 16x 3 – 5x + 7? a.) – 16x 3 – 5x + 7 b.) 16x 3 + 5x + 7 c.) – 16x 3 + 5x – 7

3-14 © 2011 Pearson Prentice Hall. All rights reserved Section 6.4 Multiplying Polynomials Find the product: (x + 1)(x – 1)(x 2 – 1) a.) x 4 – 1 b.) x 4 – 2x c.) x 2 – 1

3-15 © 2011 Pearson Prentice Hall. All rights reserved Section 6.4 Multiplying Polynomials Find the product: (x + 1)(x – 1)(x 2 – 1) a.) x 4 – 1 b.) x 4 – 2x c.) x 2 – 1

3-16 © 2011 Pearson Prentice Hall. All rights reserved Section 6.4 Multiplying Polynomials Find the error. What should the correct answer be? a.) 4x(–x) + 4x(5) + 2x b.) 4x(x) + 4x(– 5) + 2x c.) 4x(x) + (– 5) + 2x 4x(x – 5) + 2x = 4x(x) + 4x(– 5) + 4x(2x)

3-17 © 2011 Pearson Prentice Hall. All rights reserved Section 6.4 Multiplying Polynomials Find the error. What should the correct answer be? a.) 4x(–x) + 4x(5) + 2x b.) 4x(x) + 4x(– 5) + 2x c.) 4x(x) + (– 5) + 2x 4x(x – 5) + 2x = 4x(x) + 4x(– 5) + 4x(2x)

3-18 © 2011 Pearson Prentice Hall. All rights reserved Which factorization of 12x 2 + 9x – 3 is correct? a.) 3(4x 2 + 3x + 1) b.) 3(4x 2 + 3x – 1) c.) 3(4x 2 + 3x) Section 6.5 The Greatest Common Factor and Factoring by Grouping

3-19 © 2011 Pearson Prentice Hall. All rights reserved Which factorization of 12x 2 + 9x – 3 is correct? a.) 3(4x 2 + 3x + 1) b.) 3(4x 2 + 3x – 1) c.) 3(4x 2 + 3x) Section 6.5 The Greatest Common Factor and Factoring by Grouping

3-20 © 2011 Pearson Prentice Hall. All rights reserved Section 6.6 Factoring Trinomials Substitute x for y 2 in 9y y 2 – 6. a.) 9x x – 6 b.) 3x x – 6 c.) 9x x 2 – 6

3-21 © 2011 Pearson Prentice Hall. All rights reserved Section 6.6 Factoring Trinomials Substitute x for y 2 in 9y y 2 – 6. a.) 9x x – 6 b.) 3x x – 6 c.) 9x x 2 – 6

3-22 © 2011 Pearson Prentice Hall. All rights reserved Section 6.6 Factoring Trinomials Name one way that a factorization can be checked. a.) By adding the factors to see that the product is the original polynomial b.) By subtracting the factors to see that the product is the original polynomial c.) By multiplying the factors to see that the product is the original polynomial

3-23 © 2011 Pearson Prentice Hall. All rights reserved Section 6.6 Factoring Trinomials Name one way that a factorization can be checked. a.) By adding the factors to see that the product is the original polynomial b.) By subtracting the factors to see that the product is the original polynomial c.) By multiplying the factors to see that the product is the original polynomial

3-24 © 2011 Pearson Prentice Hall. All rights reserved Section 6.7 Factoring by Special Products 8x 3 – 125y 3 factors as (2x – 5y)( ? )? a.) 4x xy + 25y 2 b.) 4x 2 – 10xy + 25y 2 c.) 2x 2 – 10xy + 25y 2

3-25 © 2011 Pearson Prentice Hall. All rights reserved Section 6.7 Factoring by Special Products 8x 3 – 125y 3 factors as (2x – 5y)( ? )? a.) 4x xy + 25y 2 b.) 4x 2 – 10xy + 25y 2 c.) 2x 2 – 10xy + 25y 2

3-26 © 2011 Pearson Prentice Hall. All rights reserved Section 6.7 Factoring by Special Products Is (x – 4)(y 2 – 9) completely factored? a.) yes b.) no; x – 4 can be factored c.) no; y 2 – 9 can be factored

3-27 © 2011 Pearson Prentice Hall. All rights reserved Section 6.7 Factoring by Special Products Is (x – 4)(y 2 – 9) completely factored? a.) yes b.) no; x – 4 can be factored c.) no; y 2 – 9 can be factored

3-28 © 2011 Pearson Prentice Hall. All rights reserved Section 6.8 Solving Equations by Factoring and Problem Solving Which solutions strategies are incorrect? a.) Solve (y – 2)(y + 2) = 4 by setting each factor equal to 4. b.) Solve (x + 1)(x + 3) = 0 by setting each factor equal to 0. c.) Solve z 2 + 5z + 6 = 0 by factoring z 2 + 5z + 6 and setting each factor equal to 0.

3-29 © 2011 Pearson Prentice Hall. All rights reserved Section 6.8 Solving Equations by Factoring and Problem Solving Which solutions strategies are incorrect? a.) Solve (y – 2)(y + 2) = 4 by setting each factor equal to 4. b.) Solve (x + 1)(x + 3) = 0 by setting each factor equal to 0. c.) Solve z 2 + 5z + 6 = 0 by factoring z 2 + 5z + 6 and setting each factor equal to 0.