Welcome to Interactive Chalkboard Algebra 2 Interactive Chalkboard Copyright © by The McGraw-Hill Companies, Inc. Send all inquiries to: GLENCOE DIVISION Glencoe/McGraw-Hill 8787 Orion Place Columbus, Ohio 43240 Welcome to Interactive Chalkboard

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Lesson 5-3 Dividing Polynomials Lesson 5-4 Factoring Polynomials 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 Contents

Example 1 Simplify Expressions with Multiplication Example 2 Simplify Expressions with Division Example 3 Simplify Expressions with Powers Example 4 Simplify 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 Lesson 1 Contents

Definition of exponents Commutative Property Answer: Definition of exponents Example 1-1a

Answer: Example 1-1b

Subtract exponents. Remember that a simplified expression cannot contain negative exponents. 1 Answer: Simplify. Example 1-2a

Answer: Example 1-2b

Power of a power Answer: Example 1-3a

Power of a power Answer: Example 1-3b

Power of a quotient Power of a product Answer: Example 1-3c

Negative exponent Power of a quotient Answer: Example 1-3d

Simplify each expression. a. b. d. Answer: Answer: Answer: Answer: Example 1-3e

Method 1 Raise the numerator and the denominator to the fifth power before simplifying. Example 1-4a

Answer: Example 1-4a

Simplify the fraction before raising to the fifth power. Method 2 Simplify the fraction before raising to the fifth power. Answer: Example 1-4b

Answer: Example 1-4c

4,560,000 Express 4,560,000 in scientific notation. Answer: Write 1,000,000 as a power of 10. Example 1-5a

Express 0.000092 in scientific notation. Use a negative exponent. Answer: Example 1-5b

Express each number in scientific notation. a. 52,000 Answer: Answer: Example 1-5c

Express the result in scientific notation. Associative and Commutative Properties Answer: Example 1-6a

Express the result in scientific notation. Associative and Commutative Properties Answer: Example 1-6b

Evaluate. Express the result in scientific notation. a. b. Answer: Answer: Example 1-6c

 number of red blood cells in 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? Divide the number of red blood cells in the sample by the number of red blood cells in 1 milliliter of blood.  number of red blood cells in sample  number of red blood cells in 1 milliliter Answer: There are about 1.66 milliliters of blood in the sample. Example 1-7a

Biology A petri dish started with. germs in it Biology A petri dish started with germs in it. A half hour later, there are How many times as great is the amount a half hour later? Answer: Example 1-7b

End of Lesson 1

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 Lesson 2 Contents

Answer: This expression is not a polynomial because is not a monomial. 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. Example 2-1a

Determine whether is a polynomial. If it is a polynomial, state the degree of the polynomial. 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. Example 2-1b

Determine whether each expression is a 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 Example 2-1c

Simplify Distribute the –1. Group like terms. Combine like terms. Answer: Example 2-2a

Simplify Answer: Example 2-2b

Distributive Property Answer: Multiply the monomials. Example 2-3a

Answer: Example 2-3b

Multiply monomials and add like terms. + Outer terms Inner terms Last terms First terms Answer: Multiply monomials and add like terms. Example 2-4a

Answer: Example 2-4b

Distributive Property Multiply monomials. Answer: Combine like terms. Example 2-5a

Answer: Example 2-5b

End of Lesson 2

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 Lesson 3 Contents

Sum of quotients Divide. Answer: Example 3-1a

Answer: Example 3-1b

Use long division to find Answer: Example 3-2a

Use long division to find Answer: x + 2 Example 3-2b

Multiple-Choice Test Item Which expression is equal to A B C D Example 3-3a

Read the Test Item Since the second factor has an exponent of –1, this is a division problem. Solve the Test Item Example 3-3b

The quotient is –a + 3 and the remainder is –3. Therefore, Answer: D Example 3-3c

Multiple-Choice Test Item Which expression is equal to A B C D Answer: B Example 3-3d

Use synthetic division to find Step 1 Write the terms of the dividend so that the degrees of the terms are in descending order. Then write just the coefficients as shown. x3 – 4x2 + 6x – 4     1 – 4 6 – 4 Step 2 Write the constant r of the divisor x – r to the left. In this case, r = 2. Bring the first coefficient, 1, down as shown. 1 – 4 6 – 4 1 Example 3-4a

Step 3. Multiply the first coefficient by. Write the product Step 3 Multiply the first coefficient by . Write the product under the second coefficient. Then add the product and the second coefficient. 1 1 –4 6 – 4 2 –2 Step 4 Multiply the sum, –2, by Write the product under the next coefficient and add: 2 1 –4 6 – 4 –2 1 –4 2 Example 3-4b

Step 5 Multiply the sum, 2, by. Write the product Step 5 Multiply the sum, 2, by Write the product under the next coefficient and add: The remainder is 0. 2 1 –4 6 – 4 –2 1 –4 4 2 The numbers along the bottom are the coefficients of the quotient. Start with the power of x that is one less than the degree of the dividend. Answer: The quotient is Example 3-4c

Use synthetic division to find Answer: x + 7 Example 3-4d

Use synthetic division to find Use division to rewrite the divisor so it has a first coefficient of 1. Divide numerator and denominator by 2. Simplify the numerator and denominator. Example 3-5a

Since the numerator does not have a y3 term, use a coefficient of 0 for y3. 2 1 –2 0 2 The result is Now simplify the fraction. Example 3-5b

Rewrite as a division expression. Multiply by the reciprocal. Multiply. Answer: The solution is Example 3-5c

Use synthetic division to find Answer: Example 3-5d

End of Lesson 3

Example 3 Two or Three Terms Example 4 Quotient of Two Trinomials Example 1 GCF Example 2 Grouping Example 3 Two or Three Terms Example 4 Quotient of Two Trinomials Lesson 4 Contents

Distributive Property Factor The GCF is 5ab. Answer: Distributive Property Example 4-1a

Factor Answer: Example 4-1b

Factor the GCF of each binomial. Group to find the GCF. Factor the GCF of each binomial. Answer: Distributive Property Example 4-2a

Factoring with boxes Factor Answer: Remove common factor from top row. Fill in the circles by finding the missing factors. Answer: Example 4-2a

Factor Answer: Example 4-2b

Factor 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 . Rewrite the expression using –5y and 3y in place of –2y and factor by grouping. Substitute –5y + 3y for –2y. Example 4-3a

Factor out the GCF of each group. Associative Property Factor out the GCF of each group. Answer: Distributive Property Example 4-3b

p2 – 9 is the difference of two squares. Factor Factor out the GCF. Answer: p2 – 9 is the difference of two squares. Example 4-3c

This is the sum of two cubes. Factor This is the sum of two cubes. Sum of two cubes formula with and Answer: Simplify. Example 4-3d

Difference of two squares 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 difference of two cubes Example 4-3e

Factor each polynomial. a. b. c. d. Answer: Answer: Answer: Answer: Example 4-3f

Factor the numerator and the denominator. Simplify Factor the numerator and the denominator. Divide. Assume a  –5, –2. Answer: Therefore, Example 4-4a

Simplify Answer: Example 4-4b

End of Lesson 4

Example 2 Simplify Using Absolute Value Example 1 Find Roots Example 2 Simplify Using Absolute Value Example 3 Approximate a Square Root Lesson 5 Contents

Answer: The square roots of 16x6 are  4x3. Simplify Answer: The square roots of 16x6 are  4x3. Example 5-1a

Answer: The opposite of the principal square root of Simplify Answer: The opposite of the principal square root of Example 5-1b

Answer: The principal fifth root is 3a2b3. Simplify Answer: The principal fifth root is 3a2b3. Example 5-1c

Answer: n is even and b is negative. Thus, is not a real number. Simplify Answer: n is even and b is negative. Thus, is not a real number. Example 5-1d

Answer: not a real number Simplify. a. b. c. d. Answer:  3x4 Answer: Answer: 2xy2 Answer: not a real number Example 5-1e

Simplify 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: Example 5-2a

Since the index is odd, you do not need absolute value. Simplify Since the index is odd, you do not need absolute value. Answer: Example 5-2b

Simplify. a. b. Answer: Answer: Example 5-2c

given by the formula where L is the length Physics The time T in seconds that it takes a pendulum to make a complete swing back and forth is given by the formula where L is the length of the pendulum in feet and g is the acceleration due to gravity, 32 feet per second squared. Find the value of T for a 1.5-foot-long pendulum. Explore You are given the values of L and g and must find the value of T. Since the units on g are feet per second squared, the units on the time T should be seconds. Plan Substitute the values for L and g into the formula. Use a calculator to evaluate. Example 5-3a

Original formula Solve Use a calculator. Answer: It takes the pendulum about 1.36 seconds to make a complete swing. Example 5-3b

Examine The closest square to and  is approximately 3, so the answer should be close to The answer is reasonable. Example 5-3c

given by the formula where L is the length Physics The time T in seconds that it takes a pendulum to make a complete swing back and forth is given by the formula where L is the length of the pendulum in feet and g is the acceleration due to gravity, 32 feet per second squared. Find the value of T for a 2-foot-long pendulum. Answer: about 1.57 seconds Example 5-3d

End of Lesson 5

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 6 Use a Conjugate to Rationalize a Denominator Lesson 6 Contents

Factor into squares where possible. Simplify Factor into squares where possible. Product Property of Radicals Answer: Simplify. Example 6-1a

Simplify Answer: Example 6-1b

Simplify Quotient Property Factor into squares. Product Property Example 6-2a

Rationalize the denominator. Answer: Example 6-2b

Rationalize the denominator. Simplify Quotient Property Rationalize the denominator. Product Property Example 6-2c

Multiply. Answer: Example 6-2d

Simplify each expression. b. Answer: Answer: Example 6-2e

Product Property of Radicals Simplify Product Property of Radicals Factor into cubes. Product Property of Radicals Answer: Multiply. Example 6-3a

Simplify Answer: 24a Example 6-3b

Simplify Answer: Factor using squares. Product Property Multiply. Combine like radicals. Answer: Example 6-4a

Simplify Answer: Example 6-4b

Simplify F O I L Product Property Answer: Example 6-5a

Simplify FOIL Multiply. Answer: Subtract. Example 6-5b

Simplify each expression. a. b. Answer: Answer: 41 Example 6-5c

Simplify Multiply by since is the conjugate of FOIL Difference of squares Example 6-6a

Multiply. Answer: Combine like terms. Example 6-6b

Simplify Answer: Example 6-6c

End of Lesson 6

Example 2 Exponential Form Example 1 Radical Form Example 2 Exponential Form Example 3 Evaluate Expressions with Rational Exponents Example 4 Rational Exponent with Numerator Other Than 1 Example 5 Simplify Expressions with Rational Exponents Example 6 Simplify Radical Expressions Lesson 7 Contents

Write in radical form. Answer: Definition of Example 7-1a

Write in radical form. Answer: Definition of Example 7-1b

Write each expression in radical form. a. b. Answer: Answer: Example 7-1c

Write using rational exponents. Definition of Answer: Example 7-2a

Write using rational exponents. Definition of Answer: Example 7-2b

Write each radical using rational exponents. a. b. Answer: Answer: Example 7-2c

Evaluate Method 1 Answer: Simplify. Example 7-3a

Method 2 Power of a Power Multiply exponents. Answer: Example 7-3b

Evaluate . Method 1 Factor. Power of a Power Expand the square. Find the fifth root. Answer: The root is 4. Example 7-3c

Method 2 Power of a Power Multiply exponents. Answer: The root is 4. Example 7-3d

Evaluate each expression. a. b. Answer: Answer: 8 Example 7-3e

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. According to the formula, what is the maximum that U.S. Weightlifter Oscar Chaplin III can lift if he weighs 77 kilograms? Original formula Use a calculator. Answer: The formula predicts that he can lift at most 372 kg. Example 7-4a

Weight Lifting The formula 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. 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. Example 7-4b

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 value predicted by the formula. Answer: 380 kg Answer: The formula prediction is slightly higher than his actual total. Example 7-4c

Simplify . Multiply powers. Answer: Add exponents. Example 7-5a

Simplify . Multiply by Example 7-5b

Answer: Example 7-5c

Simplify each expression. a. b. Answer: Answer: Example 7-5d

Simplify . Rational exponents Power of a Power Example 7-6a

Quotient of Powers Answer: Simplify. Example 7-6b

Simplify . Rational exponents Power of a Power Multiply. Answer: Example 7-6c

Simplify . is the conjugate of Answer: Multiply. Example 7-6d

Simplify each expression. a. b. c. Answer: 1 Answer: Answer: Example 7-6e

End of Lesson 7

Example 1 Solve a Radical Equation Example 2 Extraneous Solution Example 3 Cube Root Equation Example 4 Radical Inequality Lesson 8 Contents

Add 1 to each side to isolate the radical. 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. Example 8-1a

Answer: The solution checks. The solution is 38. Original equation Replace y with 38. Simplify. Answer: The solution checks. The solution is 38. Example 8-1b

Solve Answer: 67 Example 8-1c

Solve Original equation Square each side. Find the squares. Isolate the radical. Divide each side by –4. Example 8-2a

Evaluate the square roots. Square each side. Evaluate the squares. Check Original equation Replace x with 16. Simplify. Evaluate the square roots. Answer: The solution does not check, so there is no real solution. Example 8-2b

Answer: no real solution Solve . Answer: no real solution Example 8-2c

In order to remove the power, or cube root, you must 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. Example 8-3a

Subtract 1 from each side. Divide each side by 3. Check Original equation Replace y with –42. Simplify. The cube root of –125 is –5. Add. Answer: The solution is –42. Example 8-3b

Solve Answer: 13 Example 8-3c

Solve Since the radicand of a square root must be greater than or equal to zero, first solve to identify the values of x for which the left side of the inequality is defined. Example 8-4a

Answer: The solution is Now solve . Original inequality Isolate the radical. Eliminate the radical. Add 6 to each side. Divide each side by 3. Answer: The solution is Example 8-4b

Only the values in the interval satisfy the inequality. Check Test some x values to confirm the solution. Let Use three test values: one less than 2, one between 2 and 5, and one greater than 5. Since is not a real number, the inequality is not satisfied. Since the inequality is satisfied. Since the inequality is not satisfied. Only the values in the interval satisfy the inequality. Example 8-4c

Solve Answer: Example 8-4d

End of Lesson 8

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 Lesson 9 Contents

Simplify . Answer: Example 9-1a

Simplify . Answer: Example 9-1b

Simplify. a. b. Answer: Answer: Example 9-1c

Simplify . Answer: = 6 Example 9-2a

Simplify . Answer: Example 9-2b

Simplify. a. b. Answer: –15 Answer: Example 9-2c

Simplify Multiplying powers Power of a Power Answer: Example 9-3a

Simplify . Answer: i Example 9-3b

Subtract 20 from each side. Solve Original equation Subtract 20 from each side. Divide each side by 5. Take the square root of each side. Answer: Example 9-4a

Solve Answer: Example 9-4b

Find the values of x and y that make the equation true. 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: Example 9-5a

Find the values of x and y that make the equation true. Answer: Example 9-5b

Commutative and Associative Properties Simplify . Commutative and Associative Properties Answer: Example 9-6a

Commutative and Associative Properties Simplify . Commutative and Associative Properties Answer: Example 9-6b

Simplify. a. b. Answer: Answer: Example 9-6c

Answer: The voltage is volts. Electricity In an AC circuit, the voltage E, current I, and impedance Z are related by the formula Find the voltage in a circuit with current 1 + 4 j amps and impedance 3 – 6 j ohms. Electricity formula FOIL Multiply. Add. Answer: The voltage is volts. Example 9-7a

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 amps and impedance 3 + 2 j ohms. Answer: 9 – 7 j Example 9-7b

Simplify . and are conjugates. Multiply. Answer: Standard form Example 9-8a

Simplify . Multiply by Multiply. Answer: Standard form Example 9-8b

Simplify. a. b. Answer: Answer: Example 9-8c

End of Lesson 9

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