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Lesson 5-1 Monomials

Lesson 5-2 Polynomials

Lesson 5-3 Dividing Polynomials

Lesson 5-4 Factoring Polynomials

Lesson 5-5 Roots of Real Numbers

Lesson 5-6 Radical Expressions

Lesson 5-7 Rational Exponents

Lesson 5-8 Radical Equations and Inequalities

Lesson 5-9 Complex Numbers

Example 1 Simplify Expressions with Multiplication

Example 2 Simplify Expressions with Division

Example 3 Simplify Expressions with Powers

Example 4Simplify Expressions Using Several Properties

Example 5 Express Numbers in Scientific Notation

Example 6 Multiply Numbers in Scientific Notation

Example 7 Divide Numbers in Scientific Notation

Multiplying Monomials:

• When multiplying monomials you must ADD exponents.

• Example: 2x3 3x5 2x x x 3x x x x x 6x8

Commutative Property

Answer: Definition of exponents

Definition of exponents

Answer:

Try TheseMultiply the following monomials.

1. a2 • a6

3. (-3b3c)(7b2c2)

2. 3x2 • 7x4

4. 2x2(6y3)(2x2y)

Try TheseMultiply the following monomials.

1. a2 • a6

3. (-3b3c)(4b2c2)

2. 3x2 • 7x4

4. 2x2(6y3)(2x2y)

a8 21x6

-12b5c3 24x4y4

Dividing Monomials:

• When dividing monomials you must SUBTRACT exponents.

• Example:

•

• Cancel x’s 3xx 3x2

5

7

2

6

x

xxxxxx

xxxxxxx

2

6

Subtract exponents.

Remember that a simplified expression cannot contain negative exponents.

Answer: Simplify.

1 1

1 1

Answer:

Try TheseDivide the following monomials.

1.

3.

2.

4.

5

62

an

na52

75

zy

zy

cba

cba73

335

9

324

523

30

)3(2

dc

dcdc

Try TheseDivide the following monomials.

1.

3.

2.

4.

5

62

an

na52

75

zy

zy

cba

cba73

335

9

324

523

30

)3(2

dc

dcdc

an -y3z2

Power to a Power:

• When raising a power to a power you must MULTIPLY exponents.

• Example: (x3)5 This means 5 groups of (x3). (x3) (x3) (x3) (x3) (x3)

• (xxx)(xxx)(xxx)(xxx)(xxx)

• x15

Product to a Power:

• When raising a product to a power you raise every number/variable to that power.

• Example: (2x2y3) 6 (2x2y3) (2x2y3) (2x2y3) (2x2y3) (2x2y3) (2x2y3)

which can be written as:

(2xxyyy) (2xxyyy) (2xxyyy) (2xxyyy) (2xxyyy) (2xxyyy)

64x12y18

Quotient to a Power

• When raising a quotient to a power you raise the numerator & denominator to that power.

12555

3

3

33xxx

Power of a power

Answer:

Power of a powerAnswer:

Power of a product

Power of a quotient

Answer:

Negative exponent

Power of a quotient

Answer:

Simplify each expression.

a.

b.

c.

d.

Answer:

Answer:

Answer:

Answer:

Try TheseSimplify each monomial.

1. (n4)4

3. (-2r2s) 3 (3rs2)

2. (2x)4

4. 3

34

42

)3

6(

yx

yx

Try TheseSimplify each monomial.

1. (n4)4

3. (-2r2s) 3 (3rs2)

2. (2x)4

4. 3

34

42

)3

6(

yx

yx

n1616x4

-24r7n5

Negative Exponents

• To make a negative exponent positive, move the number/variable that is being raised to that exponent from the numerator to the denominator or vice versa.

• Example: x-3 3

1

x

Try TheseSimplify each monomial.

1.

3. (a3b3)(ab)-2

2.

4. 22)(

y

xy

62

42

5

15

yx

yx

62

42

5

15

yx

yx

Try TheseSimplify each monomial.

1.

3. (a3b3)(ab)-2

2.

4. 22)(

y

xy

2

4

6

28x

x

62

42

5

15

yx

yx

Method 1 Raise the numerator and the denominator to the fifth power before simplifying.

Answer:

Method 2 Simplify the fraction before raising to the fifth power.

Answer:

Answer:

Express 4,560,000 in scientific notation.

4,560,000

Answer: Write 1,000,000 as a power of 10.

Express 0.000092 in scientific notation.

Use a negative exponent.Answer:

Express each number in scientific notation.

a. 52,000

b. 0.00012

Answer:

Answer:

Express the result in scientific notation.

Associative and Commutative Properties

Answer:

Associative and Commutative Properties

Express the result in scientific notation.

Answer:

Evaluate. Express the result in scientific notation.

a.

b.

Answer:

Answer:

Divide the number of red blood cells in the sample by the number of red blood cells in 1 milliliter of blood.

Answer: There are about 1.66 milliliters of blood in the sample.

Biology There are about red blood cells in one milliliter of blood. A certain blood sample contains red blood cells. About how many milliliters of blood are in the sample?

number of red blood cells in sample

number of red blood cells in 1 milliliter

Answer:

Biology A petri dish startedwith germs in it. A half hour later, there are

How many times asgreat is the amount a half hour later?

Assignment:

Page 226 #26, 32, 36, 40

Example 1 Degree of a Polynomial

Example 2 Subtract and Simplify

Example 3 Multiply and Simplify

Example 4 Multiply Two Binomials

Example 5 Multiply Polynomials

Polynomials

• Polynomial: The sum of terms such as 5x, 3x2, 4xy, 5

• Polynomial Terms have variable and whole number exponents. There are no square roots of exponents, no fractional powers, and no variables in the denominators.

Polynomials

6x-2 Not a polynomial term

Has a negative exponent

1/x2 Not a polynomial term

Has variable in the denominator.

Not a polynomial term

Has variable in radical.

4x2 Is a polynomial term

Determine whether is a polynomial. If it is a polynomial, state the degree of the polynomial.

Answer: This expression is not a polynomial

because is not a monomial.

Answer: This expression is a polynomial because each term is a monomial. The degree of the first term is 5 and the degree of the second term is 2 + 7 or 9. The degree of the polynomial is 9.

Determine whether is a polynomial.

If it is a polynomial, state the degree of the polynomial.

Determine whether each expression is a polynomial. If it is a polynomial, state the degree of the polynomial.

a.

b.

Answer: yes, 5

Answer: no

Adding Polynomials

• Add:  (2x2 - 4) + (x2 + 3x - 3)

1. Remove parentheses.  

2. Identify like terms. 

3. Add the like terms.

• (2x2 - 4) + (x2 + 3x - 3)

• = 2x2 - 4 + x2 + 3x - 3

• = 3x2 + 3x - 7

Subtracting Polynomials

• Subtract:  (2x2 - 4) - (x2 + 3x - 3) 1. Remove parentheses.   2. Change the signs of ALL of the terms being

subtracted. 3. Change the subtraction sign to addition.  4. Follow the rules for adding signed numbers.

• (2x2 - 4) - (x2 + 3x - 3)• (Change the signs of terms being subtracted)• = (2x2 - 4) + (-x2 - 3x + 3)  • = 2x2 - 4 + -x2 - 3x + 3• = x2 - 3x - 1

Simplify

Distribute the –1.

Group like terms.

Combine like terms.Answer:

Simplify

Answer:

Try TheseAdd or subtract as indicated.

1. (3x2 – x – 2) + (x2 + 4x – 9)

2. (5y + 3y2) + (– 8y – 6y2)

3. (9r2 + 6r + 16) – (8r2 + 7r + 10)

4. (10x2 – 3xy + 4y2) – (3x2 + 5xy)

Try TheseAdd or subtract as indicated.

1. (3x2 – x – 2) + (x2 + 4x – 9)

2. (5y + 3y2) + (– 8y – 6y2)

3. (9r2 + 6r + 16) – (8r2 + 7r + 10)

4. (10x2 – 3xy + 4y2) – (3x2 + 5xy)

4x2 + 3x - 11

-3y – 3y2

r2 – r + 6

7x2 -8xy + 4y2

Multiplying Polynomials

• Simply multiply each term from the first polynomial by each term of the second polynomial. 

• Example:

• (x + 3)(x² + 2x + 4)

•  = x³ + 2x² + 4x + 3x² + 6x + 12

•  = x³ + 2x² + 3x² + 4x + 6x + 12 

• = x³ + 5x² + 10x + 12

Distributive Property

Answer:Multiply the monomials.

Answer:

Outer terms Inner terms Last termsFirst terms

Answer: Multiply monomials and add like terms.

+ + +

Answer:

Distributive Property

Distributive Property

Multiply monomials.

Answer: Combine like terms.

Answer:

Try TheseMultiply the polynomials.

1. 4b(cb – zd)

2. 2xy(3xy3 – 4xy + 2y4)

3. (3x + 8)(2x + 6)

4. (x – 3y)2

5. (x2 + xy + y2)(x – y)

Try TheseMultiply the polynomials.

1. 4b(cb – zd)

2. 2xy(3xy3 – 4xy + 2y4)

3. (3x + 8)(2x + 6)

4. (x – 3y)2

5. (x2 + xy + y2)(x – y)

4b2c – 4bdz

6x2y4 – 8x2y2 + 4xy5

6x2 + 34x + 48

x2 – 6xy + 9y2

x3 – y3

Assignment:

Page 231-232 #25, 26, 30, 42

Example 1 GCF

Example 2 Grouping

Example 3 Two or Three Terms

Example 4 Quotient of Two Trinomials

Factoring Lesson #1

• Greatest Common Factor

• Polynomials in the form x2 + bx + c

Greatest Common Factor

• The first thing you should always do when factoring is to take out a common factor. This is the simplest technique of factoring, but it is important even when you learn fancier techniques, because you will make your later work much easier if you always look for common factors first. Taking out common factors is using the distributive property backwards. The distributive property says:

a(b+c)=ab+ac

• The idea behind taking out a common factor is to look for something that all terms have “in common.” Look at thr right side of the above equation. There is a common factor, a.

Greatest Common Factor

• A good trick for finding the greatest common factor to factor polynomials is to find the greatest common factor of the numbers and the smaller power of the variable, so here the greatest common factor of the numbers is 4 and the smallest power of x is 3, so we can take out 4x3 as a common factor.

Example:The polynomial:

4x5+12x4-8x3

Can be factored into:4x3(x2+3x-8)

Example 1:

Factor the polynomial:

2x2 + 6x4

by taking out a common factor.

Solution: Choose the common factor. 2x2.

2x2 (1 + ___ )

2x2 (1 + 3x2)

Now Check your work:

2x2 (1 + 3x2)

Multiply back together:

2x2 + 6x4

Example 2:

Factor the polynomial:

15x2y – 10xy2

by taking out a common factor.

Solution: Choose the common factor: 5xy.

5xy (3x – ___ )

5xy (3x – 2y)

Now Check your work:

5xy (3x – 2y)

Multiply back together:

15x2y – 10xy2

Example 3:

Factor the polynomial:16a3b 5 – 24a2b4 – 8a4b7c

by taking out a common factor.

Solution: Choose the common factor: 8a2b4.

8a2b4 (2ab – ___ – ___ ) 8a2b4 (2ab – 3 – ___ )

8a2b4 (2ab – 3 – a2b3c )

Now Check your work: 8a2b4 (2ab – 3 – a2b3c )

Multiply back together:

16a3b 5 – 24a2b4 – 8a4b7c

Now Try These:

Factor the following polynomials and check your work.

a. 6x2y3 + 8x2y5 Solution: 2x2y3 (3 + 4y2)

b. 12a4b2c3 – 18ab2c4 + 24a5b3c4

Solution: 6ab2c3 (2a3 – 3c + 4a4bc)

Factoring Polynomials in the form x2 + bx + c (General Quadratics)

Examples of these “General Quadratics” are:

a. x2 + 7x + 10 b. x2 + 13x - 30

c. x2 - 8x + 15 d. x2 - 8x - 20

Rules for Factoring General Quadratics

If the constant term is positive:

- Choose factors of the constant term whose SUM is the middle term.

- Use the same signs – the sign of the middle term.

• Example:

x2 + 10x + 16

( x )( x ) Choose factors of 16 whose sum is 10

(8 and 2)

( x 8 )( x 2 )

Use the same signs – sign of middle term (+)

( x + 8 )( x + 2 )

Rules for Factoring General Quadratics

If the constant term is negative:

- Choose factors of the constant term whose DIFFERENCE is the middle term.

- Use different signs – the larger factor gets the sign of the middle term.

• Example:

x2 - 2x - 24

( x )( x ) Choose factors of 24 whose difference

is 2 (6 and 4)

( x 6 )( x 4 )

Use different signs – the six gets the sign of middle term (-)

( x - 6 )( x + 4 )

Now Try These:

Factor the following polynomials and check your work.

a. x2 + 7x + 10 b. x2 + 13x - 30 Answer: (x + 5)(x + 2) Answer: (x + 15)(x – 2)

c. x2 - 8x + 15 d. x2 - 8x – 20

Answer: (x - 5)(x – 3) Answer: (x – 10)(x + 2)

Part I

Warm Up – Section 5-4 #1

Factor and check.

1. 10x2y + 15x3y2

2. 16a2b4c5 + 48a3bc2 – 12ab4c3

3. y2 + 11y + 24

4. y2 - 15y + 36

5. y2 + 7y - 30

6. y2 - 4y - 45

Factoring Lesson #2

• Polynomials in the form ax2 + bx + c

Factoring Polynomials in the form ax2 + bx + c (Trial and Error)

Examples of these “Trial and Error” Quadratics are:

a. 4x2 - 8x - 45 b. 12x2 + 13x - 14

c. 15x2 - 26x + 7 d. 25x2 +15x + 2

Rules for Factoring General Quadratics in the form ax2 + bx + c

- List all of the possible factors of the first term and the last term.

- Choose the combination that will allow you to get the correct middle term.

- Check your work!!!

• Example: 4x2 - 24x + 35

4 x 1 5 x 72 x 2 35 x 1

Choose factors whose combination will give you the middle term (-24x). You may have to try different combinations before finding the one that works.

( 2x )( 2x )

Use the same signs – sign of middle term (-)

( 2x - 7 )( 2x - 5 )

Now Try These:

Factor the following polynomials and check your work.

a. 2x2 + 7x + 6 b. 3x2 + 10x + 3 Answer: (2x + 3)(x + 2) Answer: (3x + 1)(x + 3)

c. 15x2 - 38x + 7 d. 10x2 - 3x – 27

Answer: (5x - 1)(3x – 7) Answer: (5x – 9)(2x + 3)

Part II

Warm Up – Section 5-4 #2

Factor using trial and error or the junk method and check your work.

1. 2x2 + 11x + 142. 14y2 – 19y – 3 3. 3a2 – 22a + 24

Factor.4. 25r2s4t + 100rs2t3

5. x2 – 11x + 24 6. x2 + 2x – 35

Factoring Lesson #3

• Difference of two perfect squares x2 - y2

• Factoring by grouping

• Factoring Completely

Factoring the difference of two perfect squares

Examples of these polynomials are:

a. 4x2 – 9 b. 16x2 – 36

c. x2 – 4 d. 25x2 – 16y2

Rules for Factoring the difference of two perfect squares

- The square root of the first term becomes the first term of each binomial.

- The square root of the second term becomes the second term of each binomial.

- Use different signs.

Example: x2 - 64

Since the square root of x2 is x, x is the first term of each binomial.

( x )( x )

Since the square root of 64 is 8, 8 is the second term of each binomial.

( x 8 )( x 8 )

Use different signs.

( x + 8 )( x - 8 )

Now Try These:

Factor the following polynomials and check your work.

a. 4x2 – 9 b. 16x2 – 36 Answer: (2x + 3)(2x - 3) Answer: (4x + 6)(4x - 6)

c. x2 – 4: d. 25x2 – 16y2

Answer: (x - 2)(x + 2) Answer: (5x – 4y)(5x + 4y)

Factoring by grouping

Examples of polynomials that are factored by grouping are:

a. 6x2 + 3xy + 2xz + yz

b. 6x2 + 2xy – 3xz – yz

Note: You will see 4 terms when using the grouping method.

Rules for Factoring by grouping:

- Group terms so that there is a common factor in each group.

- Take out the common factor in both groups.

- Combine like groups.

Example: 10a2 + 2ab + 5ad + bd

I will group the first two terms and the last two terms since both of those groups contain a common factor. Note: I am adding these groups.

(10a2 + 2ab) + (5ad + bd)

Take out a common factor.

2a(5a + b) + d(5a + b)

Combine like groups:

( 2a + d )( 5a + b )

Examples of factoring by grouping:

Factor the polynomial:a3 – 4a2 + 3a – 12

Group:(a3 – 4a2)+ (3a – 12)Factor:a2 (a – 4) + 3(a – 4)Combine:(a2 + 3)(a – 4)

Factor the polynomial:7ac2 + 2bc2 – 7ad2 – 2bd2

Group:(7ac2 + 2bc2) + (– 7ad2 – 2bd2)Factor:c2(7a + 2b) + d2 (– 7a – 2b)Factor a negative out of second

group:c2(7a + 2b) - d2 ( 7a + 2b)Now groups match – so,

Combine:(c2 - d2) ( 7a + 2b)

Now Try These:

Factor the following polynomials and check your work.

a. 6x2 + 3xy + 2xz + yz Answer: (3x + z)(2x + y)

b. 6x2 + 2xy – 3xz – yz

Answer: (2x - z)(3x + y)

Factoring Completely

Some polynomials can be factored more than once. This may not be apparent from the beginning.

Just as integers can be factored into primes, polynomials can too, and it may take more than one step.

Rules for Factoring Completely

- Factor a polynomial using the appropriate method.

- Check each factor to see if you can factor it again.

- If so, do it until all polynomials are prime.

Example: 3x2 – 21x + 30

Here, I notice that I have a common factor of 3, so take it out.

3(x2 – 7x + 10)

Now x2 – 7x + 10 can be factored.

3(x – 5)(x – 2)

Now all terms are prime.

Now Try These:

Factor the following polynomials and check your work.

a. 2x2 + 12x + 18 b. 3x2 – 21x – 54 Answer: 2(x + 3)(x + 3) Answer: 3(x + 2)(x - 9)

c. 5x2 – 20: d. 25x2 – 100y2

Answer: 5(x - 2)(x + 2) Answer: 25(x – 2y)(x + 2y)

Part III

Warm Up Section 5-4 #3

Factor.

1. 7c3 – 28c2d + 35cd3

2. x2 – 5x – 14

3. x2 – 15x + 54

4. 3x2 – 22x + 35

5. 64x2 – 81

6. 3r + 3s + 5r3s + 5r2s2

The GCF is 5ab.

Answer: Distributive Property

Factor

Answer:

Factor

Group to find the GCF.

Factor the GCF of each binomial.

Factor

Answer:Distributive Property

Answer:

Factor

Rewrite the expression using –5y and 3y in place of –2y and factor by grouping.

To find the coefficient of the y terms, you must find two numbers whose product is 3(–5) or –15 and whose sum is –2. The two coefficients must be 3 and –5 since

and .

Substitute –5y + 3y for –2y.

Factor

Factor out the GCF of each group.

Answer:Distributive Property

Associative Property

Answer: p2 – 9 is the difference of two squares.

Factor out the GCF.

Factor

This is the sum of two cubes.

Answer:Simplify.

Sum of two cubes formula with and

Factor

This polynomial could be considered the difference of two squares or the difference of two cubes. The difference of two squares should always be done before the difference of two cubes.

Difference of two squares

Answer:Sum and differenceof two cubes

Factor

Factor each polynomial.a.

b.

c.

d.Answer:

Answer:

Answer:

Answer:

Factor the numerator and the denominator.

Divide. Assume a –5, –2.

Answer: Therefore,

Simplify

Simplify

Answer:

Example 1 Divide a Polynomial by a Monomial

Example 2 Division Algorithm

Example 3 Quotient with Remainder

Example 4 Synthetic Division

Example 5 Divisor with First Coefficient Other than 1

Steps for Dividing a Polynomial by a Monomial

• 1. Divide each term of the polynomial by the monomial.

a)  Divide numbersb)  Subtract exponents

• 2.  Remember to write the appropriate sign in between the terms.

Example: 

Answer:

Sum of quotients

Divide.

Answer:

Answer:

Try TheseDivide the following polynomials.

1.

2.

Try TheseDivide the following polynomials.

1.

2.

332 32 yxxyx

332 532 abaab

Use factoring to find

Answer:

Answer: x + 2

Use factoring to find

Try TheseDivide the following polynomials by factoring.

1. 2.

3. 4.

Try TheseDivide the following polynomials by factoring.

1. 2.

3. 4.

2x 2x

3x 5x

Warm Up Section 5-5Covering lessons 5.1-5.4

1. Simplify: 5x2y(4x4y3)

2. Simplify: (4a5b4c2) 3

3. Simplify:

4. Multiply: (3x + 7)(x – 4)

5. Factor: x2 – 11x + 18

6. Factor: 3x2 + 7x – 20

84

36

12

18

yx

yx

Warm Up Section 5-5Covering lessons 5.1-5.4

1. Simplify: 5x2y(4x4y3) 20x6y4

2. Simplify: (4a5b4c2) 3 64a15b12c6

3. Simplify:

4. Multiply: (3x + 7)(x – 4) 3x2 – 5x – 28

5. Factor: x2 – 11x + 18 (x – 9)(x – 2)

6. Factor: 3x2 + 7x – 20 (3x – 5)(x + 4)

84

36

12

18

yx

yx5

2

2

3

y

x

Example 1 Find Roots

Example 2 Simplify Using Absolute Value

Example 3 Approximate a Square Root

Simplifying Radicals

• When working with the simplification of radicals you must remember some basic information about perfect square numbers.

Perfect Squares

1 = 1 x 1

4 = 2 x 2

9 = 3 x 3

16 = 4 x 4

25 = 5 x 5

36 = 6 x 6

49 = 7 x 7

64 = 8 x 8

81 = 9 x 9

100 = 10 x 10

Perfect Squares Containing Variables

a2 = a x aa4 = a2 x a2

a6 = a3 x a3

a8 = a4 x a4

a10 = a5 x a5

So, a variable is a “perfect square” if it has an even exponent.

To take the square root, just divide the exponent by 2.

Simplifying Radical Expressions

To simplify means to find another expression with the same value.  It does not mean to find a decimal approximation.

Example: and, although it is equivalent to 5.65, we do not use the decimal value since the radical value is exact and the decimal is an estimate.

To simplify (or reduce) a radical:• 1.  Find the largest perfect

square which will divide evenly into the number under your radical sign.  This means that when you divide, you get no remainders, no decimals, no fractions.

• 2.  Write the number appearing under your radical as the product (multiplication) of the perfect square and your answer from dividing.

• 3.  Give each number in the product its own radical sign.

• 4.  Reduce the "perfect" radical which you have now created.

Example:

• Reduce : the largest perfect square that divides evenly into 48 is 16.

• Find the largest perfect square which will divide evenly into 48.

• Give each number in the product its own radical sign.

Example Continued

• Reduce the "perfect" radical which you have now created.

Answer: The square roots of 16x6 are 4x3.

Simplify

Simplify

Answer: The fifth root is 3a2b3.

Simplify

Answer: You cannot take the square root of a negative number.

Thus, is not a real number.

Simplify

Answer: 3x4

Answer: 2xy2

Answer: not a real number

Answer:

Simplify.

a.

b.

c.

d.

Try These

225 2)7( 3 27

16

1 25.0 4 8z

4)5( g 48169 yx 2)4( yx

Try These

15 Not real # -3

1/4 0.5 z2

25g2 13x4y2 4x - y

225 2)7( 3 27

16

1 25.0 4 8z

4)5( g 48169 yx 2)4( yx

Note that t is a sixth root of t6. The index is even, so the principal root is nonnegative. Since t could be negative, you must take the absolute value of t to identify the principal root.

Answer:

Simplify

Since the index is odd, you do not need absolute value.

Answer:

Simplify

Answer:

Simplify.

a.

b. Answer:

Try These

169 2)4( 3 125

169

25 81.0 4 12z

8)2( x 10636 yx 2)63( x

Try These

-13 4 5

5/13 0.9 z3

16x4 6x3y5 3x+6

169 2)4( 3 125

169

25 81.0 4 12z

8)2( x 10636 yx 2)63( x

Assignment:

• P248 #40, 42, 46, 50

Example 1 Square Root of a Product

Example 2 Simplify Quotients

Example 3 Multiply Radicals

Example 4 Add and Subtract Radicals

Example 5 Multiply Radicals

Example 6Use a Conjugate to Rationalize a Denominator

Factor into squares where possible.

Product Property of Radicals

Answer: Simplify.

Simplify

Answer:

Simplify

Simplify

Quotient Property

Factor into squares.

Product Property

Rationalize the denominator.

Answer:

Quotient Property

Rationalize the denominator.

Product Property

Simplify

Multiply.

Answer:

Answer:

Simplify each expression.

a.

b. Answer:

Try These

72 3 54 4 96

3 316y 3 54242 nm

5 76

32

1zw

9

84

t

r10

520

y

x

5432 yx

Try These

72 3 54 4 96

3 316y 3 54242 nm

5 76

32

1zw

9

84

t

r10

520

y

x

5432 yx

26 3 23 4 62

yyx 24 22 3 22y 3 234 mnmn

5 2

2

1wzwz 5

42

t

tr5

2 52

y

xx

Warm Up 5-6Simplify.1.

2.

3.

4.

5.

6.

36

6281 yx

7681 yx

11440 yx

3 5324 ba

4 7432 ts

Warm Up 5-6Simplify.1.

2.

3.

4.

5.

6.

36

6281 yx

7681 yx

11440 yx

3 5324 ba

4 7432 ts

639xy

yyx 339

yyx 102 52

3 232 bab

4 322 tst

Factor into cubes.

Product Property of Radicals

Answer: Multiply.

Simplify

Product Property of Radicals

Answer: 24a

Simplify

Try These

1.

2.

)212)(123(

)205)(243(

Try These

1.

2.

)212)(123(

)205)(243(

736

3060

Product Property

Multiply.

Combine like radicals.

Simplify

Factor using squares.

Answer:

Answer:

Simplify

Try These

1.

2.

274812

54718024205

Try These

1.

2.

274812

54718024205

33

62354

F O I L

Product Property

Answer:

Simplify

FOIL

Multiply.

Answer: Subtract.

Simplify

Simplify each expression.

a.

b.

Answer: 41

Answer:

Multiply by since

is the conjugate

of

FOIL

Difference ofsquares

Simplify

Multiply.

Answer: Combine like terms.

Answer:

Simplify

Assignment

• P 254 #16-46 even

Example 1 Radical Form

Example 2 Exponential Form

Example 3Evaluate Expressions with Rational Exponents

Example 4Rational Exponent with Numerator Other Than 1

Example 5Simplify Expressions with Rational Exponents

Example 6 Simplify Radical Expressions

Write in radical form.

Answer: Definition of

Write in radical form.

Answer: Definition of

Write each expression in radical form.

a.

b.

Answer:

Answer:

Write using rational exponents.

Definition of Answer:

Write using rational exponents.

Definition of Answer:

Write each radical using rational exponents.

a.

b.

Answer:

Answer:

Evaluate

Method 1

Answer: Simplify.

Multiply exponents.

Method 2

Answer:

Power of a Power

Answer: The root is 4.

Evaluate .

Method 1 Factor.

Power of a Power

Expand the square.

Find the fifth root.

Answer: The root is 4.

Power of a Power

Multiply exponents.

Method 2

Evaluate each expression.

a.

b.

Answer: 8

Answer:

According to the formula, what is the maximum that U.S. Weightlifter Oscar Chaplin III can lift if he weighs 77 kilograms?

Answer: The formula predicts that he can lift at most 372 kg.

Weight Lifting The formula can be used to estimate the maximum total mass that a weight lifter of mass B kilograms can lift in two lifts, the snatch and the clean and jerk, combined.

Original formula

Use a calculator.

Oscar Chaplin’s total in the 2000 Olympics was 355 kg. Compare this to the value predicted by the formula.

Answer: The formula prediction is somewhat higher than his actual total.

Weight Lifting The formula can be used to estimate the maximum total mass that a weight lifter of mass B kilograms can lift in two lifts, the snatch and the clean and jerk, combined.

Answer: 380 kg

Answer: The formula prediction is slightly higher than hisactual total.

Weight Lifting Use the formula where M is the maximum total mass that a weight lifter of mass B kilograms can lift.

a. According to the formula, what is the maximum that a weight lifter can lift if he weighs 80 kilograms?

b. If he actually lifted 379 kg, compare this to the valuepredicted by the formula.

Simplify .

Multiply powers.

Answer: Add exponents.

Simplify .

Multiply by

Answer:

Simplify each expression.

a.

b.

Answer:

Answer:

Simplify .

Rational exponents

Power of a Power

Quotient of Powers

Answer: Simplify.

Simplify .

Rational exponents

Power of a Power

Answer: Simplify.

Multiply.

Answer: Multiply.

Simplify .

is the conjugate

of

Answer: 1

Simplify each expression.

a.

b.

c. Answer:

Answer:

Example 1 Solve a Radical Equation

Example 2 Extraneous Solution

Example 3 Cube Root Equation

Example 4 Radical Inequality

Solving Radical Equations

1. Isolate the radical2. Raise each side to the

appropriate power to eliminate the radical.

3. Solve for the variable.

• Example:Solvefor x.

1. Isolate radical by adding 2 to both sides.

2.

Square both sides.

3.

So, x = 95

Solve

Original equation

Add 1 to each side to isolate the radical.

Square each side to eliminate the radical.

Find the squares.

Add 2 to each side.

Check

Answer: The solution checks. The solution is 38.

Replace y with 38.

Original equation

Simplify.

Answer: 67

Solve

Try TheseSolve each equation.

Try TheseSolve each equation.

25 144

1 -11

Solve

Original equation

Square each side.

Find the squares.

Isolate the radical.

Divide each side by –4.

Answer: The solution does not check, so there is no real solution.

Check

Square each side.

Evaluate the squares.

Original equation

Evaluate the square roots.

Replace x with 16.

Simplify.

Solve .

Answer: no real solution

Solve

In order to remove the power, or cube root, you must

first isolate it and then raise each side of the equation to

the third power.

Original equation

Subtract 5 from each side.

Cube each side.

Evaluate the cubes.

Answer: The solution is –42.

Divide each side by 3. Check

Original equation

Add.

Replace y with –42.

Simplify.

The cube root of –125 is –5.

Subtract 1 from each side.

Answer: 13

Solve

Try TheseSolve each equation.

Try TheseSolve each equation.

49 5

9 -20

Assignment

P 266 #16, 19, 22, 24

Example 1 Square Roots of Negative Numbers

Example 2 Multiply Pure Imaginary Numbers

Example 3 Simplify a Power of i

Example 4 Equation with Imaginary Solutions

Example 5 Equate Complex Numbers

Example 6 Add and Subtract Complex Numbers

Example 7 Multiply Complex Numbers

Example 8 Divide Complex Numbers

Keep in Mind:

• The square root of a negative number does not exist.

• Example: is not 5 or -5 since

5 x 5 = 25 and -5 x -5 = 25.

• So up until now, we could not simplify .

i

i is defined to have the property that:

i2 = -1

therefore, we could say that square root of -1 is i.

This allows us to simplify the square roots of negative numbers such as .

Examples

1. Simplify:

Since is 5 and is i, our answer is 5i.

2. Simplify

6ix2

Simplify .

Answer:

Simplify .

Answer:

Simplify.

a.

b.

Answer:

Answer:

Answer: = 6

Simplify .

Answer:

Simplify .

Answer: –15

Answer:

Simplify.

a.

b.

Simplify

Multiplying powers

Power of a Power

Answer:

Answer: i

Simplify .

Solve

Answer:

Original equation

Subtract 20 from each side.

Divide each side by 5.

Take the square root of each side.

Solve

Answer:

Find the values of x and y that make the equationtrue.

Set the real parts equal to each other and the imaginary parts equal to each other.

Real parts

Divide each side by 2.

Imaginary parts

Answer:

Find the values of x and y that make the equationtrue.

Answer:

Simplify .

Answer:

Commutative and AssociativeProperties

Simplify .

Commutative and Associative Properties

Answer:

Simplify.

a.

b.

Answer:

Answer:

Answer: The voltage is volts.

Electricity In an AC circuit, the voltage E, current I, and impedance Z are related by the formulaFind the voltage in a circuit with current 1 + 4 j ampsand impedance 3 – 6 j ohms.

Electricity formula

FOIL

Multiply.

Add.

Electricity In an AC circuit, the voltage E, current I, and impedance Z are related by the formula E = I • Z. Find the voltage in a circuit with current 1 – 3 j ampsand impedance 3 + 2 j ohms.

Answer: 9 – 7 j

andare conjugates.

Multiply.

Answer: Standard form

Simplify .

Simplify .

Multiply.

Answer: Standard form

Multiply by

Simplify.

a.

b.

Answer:

Answer:

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