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Chapter 6
Exponents, Polynomials,
and Polynomial Functions
Active Learning
Questions
© 2011 Pearson Prentice Hall.
All rights reserved
3-1
Section 6.1
Exponents and Scientific Notation
xt
Simplify: t 1
x
a.)
1
x
b.)
x
c.)
x
t
t 1
© 2011 Pearson Prentice Hall. All rights reserved
3-2
Section 6.1
Exponents and Scientific Notation
xt
Simplify: t 1
x
a.)
1
x
b.)
x
c.)
x
t
t 1
© 2011 Pearson Prentice Hall. All rights reserved
3-3
Section 6.1
Exponents and Scientific Notation
Find and correct the error in the following:
6
y
1
6 2
8
y  8
2  y
y
y
y 6 6(2) 4 1
y  4
a.) y 2  y
y
6
y
2(6)
4

y

y
b.) 2
y
6
y
2(6)
8
y
c.) 2  y
y
© 2011 Pearson Prentice Hall. All rights reserved
3-4
Section 6.1
Exponents and Scientific Notation
Find and correct the error in the following:
6
y
1
6 2
8
y  8
2  y
y
y
y 6 6(2) 4 1
y  4
a.) y 2  y
y
6
y
2(6)
4

y

y
b.) 2
y
6
y
2(6)
8
y
c.) 2  y
y
© 2011 Pearson Prentice Hall. All rights reserved
3-5
Section 6.1
Exponents and Scientific Notation
Which of the following numbers have values that
are less than 1?
a.) 3.5  105
b.) 3.5  10 – 5
c.) – 3.5  105
© 2011 Pearson Prentice Hall. All rights reserved
3-6
Section 6.1
Exponents and Scientific Notation
Which of the following numbers have values that
are less than 1?
a.) 3.5  105
b.) 3.5  10 – 5
c.) – 3.5  105
© 2011 Pearson Prentice Hall. All rights reserved
3-7
Section 6.2
More Work with Exponents and
Scientific Notation
1 simplifies to:
2
3
a.) – 6
b.) – 9
c.) 9
© 2011 Pearson Prentice Hall. All rights reserved
3-8
Section 6.2
More Work with Exponents and
Scientific Notation
1 simplifies to:
2
3
a.) – 6
b.) – 9
c.) 9
© 2011 Pearson Prentice Hall. All rights reserved
3-9
Section 6.3
Polynomials and Polynomial Functions
Simplify: (2x2 + x + xy3) – (x2 + 5xy3)
a.) x2 + x – 4xy3
b.) x2 + x + 6xy3
c.) 2 + x – 4xy3
© 2011 Pearson Prentice Hall. All rights reserved
3-10
Section 6.3
Polynomials and Polynomial Functions
Simplify: (2x2 + x + xy3) – (x2 + 5xy3)
a.) x2 + x – 4xy3
b.) x2 + x + 6xy3
c.) 2 + x – 4xy3
© 2011 Pearson Prentice Hall. All rights reserved
3-11
Section 6.3
Polynomials and Polynomial Functions
Which polynomial is the opposite of 16x3 – 5x + 7?
a.) – 16x3 – 5x + 7
b.) 16x3 + 5x + 7
c.) – 16x3 + 5x – 7
© 2011 Pearson Prentice Hall. All rights reserved
3-12
Section 6.3
Polynomials and Polynomial Functions
Which polynomial is the opposite of 16x3 – 5x + 7?
a.) – 16x3 – 5x + 7
b.) 16x3 + 5x + 7
c.) – 16x3 + 5x – 7
© 2011 Pearson Prentice Hall. All rights reserved
3-13
Section 6.4
Multiplying Polynomials
Find the product: (x + 1)(x – 1)(x2 – 1)
a.) x4 – 1
b.) x4 – 2x2 + 1
c.) x2 – 1
© 2011 Pearson Prentice Hall. All rights reserved
3-14
Section 6.4
Multiplying Polynomials
Find the product: (x + 1)(x – 1)(x2 – 1)
a.) x4 – 1
b.) x4 – 2x2 + 1
c.) x2 – 1
© 2011 Pearson Prentice Hall. All rights reserved
3-15
Section 6.4
Multiplying Polynomials
Find the error. What should the correct answer be?
4x(x – 5) + 2x = 4x(x) + 4x(– 5) + 4x(2x)
a.) 4x(–x) + 4x(5) + 2x
b.) 4x(x) + 4x(– 5) + 2x
c.) 4x(x) + (– 5) + 2x
© 2011 Pearson Prentice Hall. All rights reserved
3-16
Section 6.4
Multiplying Polynomials
Find the error. What should the correct answer be?
4x(x – 5) + 2x = 4x(x) + 4x(– 5) + 4x(2x)
a.) 4x(–x) + 4x(5) + 2x
b.) 4x(x) + 4x(– 5) + 2x
c.) 4x(x) + (– 5) + 2x
© 2011 Pearson Prentice Hall. All rights reserved
3-17
Section 6.5
The Greatest Common Factor and
Factoring by Grouping
Which factorization of 12x2 + 9x – 3 is correct?
a.) 3(4x2 + 3x + 1)
b.) 3(4x2 + 3x – 1)
c.) 3(4x2 + 3x)
© 2011 Pearson Prentice Hall. All rights reserved
3-18
Section 6.5
The Greatest Common Factor and
Factoring by Grouping
Which factorization of 12x2 + 9x – 3 is correct?
a.) 3(4x2 + 3x + 1)
b.) 3(4x2 + 3x – 1)
c.) 3(4x2 + 3x)
© 2011 Pearson Prentice Hall. All rights reserved
3-19
Section 6.6
Factoring Trinomials
Substitute x for y2 in 9y4 + 53y2 – 6.
a.) 9x2 + 53x – 6
b.) 3x2 + 53x – 6
c.) 9x4 + 53x2 – 6
© 2011 Pearson Prentice Hall. All rights reserved
3-20
Section 6.6
Factoring Trinomials
Substitute x for y2 in 9y4 + 53y2 – 6.
a.) 9x2 + 53x – 6
b.) 3x2 + 53x – 6
c.) 9x4 + 53x2 – 6
© 2011 Pearson Prentice Hall. All rights reserved
3-21
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
© 2011 Pearson Prentice Hall. All rights reserved
3-22
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
© 2011 Pearson Prentice Hall. All rights reserved
3-23
Section 6.7
Factoring by Special Products
8x3 – 125y3 factors as (2x – 5y)(
?
)?
a.) 4x2 + 10xy + 25y2
b.) 4x2 – 10xy + 25y2
c.) 2x2 – 10xy + 25y2
© 2011 Pearson Prentice Hall. All rights reserved
3-24
Section 6.7
Factoring by Special Products
8x3 – 125y3 factors as (2x – 5y)(
?
)?
a.) 4x2 + 10xy + 25y2
b.) 4x2 – 10xy + 25y2
c.) 2x2 – 10xy + 25y2
© 2011 Pearson Prentice Hall. All rights reserved
3-25
Section 6.7
Factoring by Special Products
Is (x – 4)(y2 – 9) completely factored?
a.) yes
b.) no; x – 4 can be factored
c.) no; y2 – 9 can be factored
© 2011 Pearson Prentice Hall. All rights reserved
3-26
Section 6.7
Factoring by Special Products
Is (x – 4)(y2 – 9) completely factored?
a.) yes
b.) no; x – 4 can be factored
c.) no; y2 – 9 can be factored
© 2011 Pearson Prentice Hall. All rights reserved
3-27
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 z2 + 5z + 6 = 0 by factoring z2 + 5z + 6
and setting each factor equal to 0.
© 2011 Pearson Prentice Hall. All rights reserved
3-28
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 z2 + 5z + 6 = 0 by factoring z2 + 5z + 6
and setting each factor equal to 0.
© 2011 Pearson Prentice Hall. All rights reserved
3-29