Presentation is loading. Please wait.

Presentation is loading. Please wait.

1 Topic 6.2.1 Rules of Exponents. 2 Lesson 1.1.1 California Standards: 2.0 Students understand and use such operations as taking the opposite, finding.

Similar presentations


Presentation on theme: "1 Topic 6.2.1 Rules of Exponents. 2 Lesson 1.1.1 California Standards: 2.0 Students understand and use such operations as taking the opposite, finding."— Presentation transcript:

1 1 Topic 6.2.1 Rules of Exponents

2 2 Lesson 1.1.1 California Standards: 2.0 Students understand and use such operations as taking the opposite, finding the reciprocal, taking a root, and raising to a fractional power. They understand and use the rules of exponents. 10.0 Students add, subtract, multiply, and divide monomials and polynomials. Students solve multistep problems, including word problems, by using these techniques. What it means for you: You’ll multiply and divide algebraic expressions using the rules of exponents. Rules of Exponents Topic 6.2.1 Key words: exponent

3 3 Lesson 1.1.1 You learned about the rules of exponents in Topic 1.3.1. Rules of Exponents Topic 6.2.1 In this Topic, you’ll apply those same rules to monomials and polynomials. We’ll start with a quick recap of the rules of exponents to make sure you remember them all.

4 4 1) x a · x b = x a + b 2) x a ÷ x b = x a – b (if x  0) 3) ( x a ) b = x ab 4) ( cx ) b = c b x b 5) x 0 = 1 6) x – a = (if x  0) 7) xaxa 1 Lesson 1.1.1 Use the Rules of Exponents to Simplify Expressions Rules of Exponents Topic 6.2.1 These are the same rules you learned in Chapter 1, but this time you’ll use them to simplify algebraic expressions: Rules of Exponents

5 5 (–2 x 2 m )(–3 x 3 m 3 ) = (–2)(–3)( x 2 )( x 3 )( m )( m 3 ) = 6 x 2+3 · m 1+3 = 6 x 5 m 4 Rules of Exponents Example 1 Topic 6.2.1 Simplify the expression (–2 x 2 m )(–3 x 3 m 3 ). Solution Solution follows… Put all like variables together Use Rule 1 and add the powers Rule 1) x a · x b = x a + b

6 6 (3 a 2 xb 3 ) 2 = 3 2 · a 2·2 · x 2 · b 3·2 = 9 a 4 x 2 b 6 Rules of Exponents Example 2 Topic 6.2.1 Simplify the expression (3 a 2 xb 3 ) 2. Solution Solution follows… Use Rules 3 and 4 Rule 3) ( x a ) b = x ab Rule 4) ( cx ) b = c b x b

7 7 = 2 xm 2 Rules of Exponents Example 3 Topic 6.2.1 Solution Solution follows… From Rule 5, anything to the power 0 is 1 Simplify the expression. Separate the expression into parts that have only one variable = =Use Rule 2 and subtract the powers Rule 2) x a ÷ x b = x a – b (if x  0) Rule 5) x 0 = 1

8 8 1. –3 at (4 a 2 t 3 ) 2. (–5 x 3 yt 2 )(–2 x 2 y 3 t ) 3. (–2 x 2 y 3 ) 3 4. –2 mx (3 m 2 x – 4 m 2 x + m 3 x 3 ) 5. (–3 x 2 t ) 3 (–2 x 3 t 2 ) 2 6. –2 mc (–3 m 2 c 3 + 5 mc ) Simplify each expression. Lesson 1.1.1 Guided Practice Rules of Exponents Topic 6.2.1 Solution follows… (–3 4)( a a 2 )( t t 3 ) = –12 a (1 + 2) t (1 + 3) (Rule 1) = –12 a 3 t 4 (–5 –2)( x 3 x 2 )( y y 3 )( t 2 t ) = 10 x (3 + 2) y (1 + 3) t (2 + 1) (Rule 1) = 10 x 5 y 4 t 3 (–2) (1 3) x (2 3) y (3 3) (Rule 3) = (–2) 3 x 6 y 9 = –8 x 6 y 9 –2 mx (– m 2 x + m 3 x 3 ) = 2 m (1 + 2) x (1 + 1) – 2 m (1 + 3) x (1 + 3) (Rule 1) = 2 m 3 x 2 – 2 m 4 x 4 ((–3) 3 x (2 3) t 3 )((–2) 2 x (3 2) t (2 2) ) (Rule 3) = (–27 x 6 t 3 )(4 x 6 t 4 ) = –108 x (6 + 6) t (3 + 4) (Rule 1) = –108 x 12 t 7 6 m (1 + 2) c (1 + 3) – 10 m (1 + 1) c (1 + 1) (Rule 1) = 6 m 3 c 4 – 10 m 2 c 2

9 9 Simplify each expression. Lesson 1.1.1 Guided Practice Rules of Exponents Topic 6.2.1 Solution follows… 7. 8. 9. 10. = 5 m (3 – 2) n (8 – 3) z (6 – 1) (Rule 2) = 5 mn 5 z 5 = (14 ÷ 4) a (2 – 7) b (4 – 4) c (8 – 0) (Rule 2) = a –5 c 8 (Rule 5) = (Rule 6) = (12 ÷ 8) j (8 – 2) k (–8 – –10) m (–1 – 4) (Rule 2) = j 6 k 2 m –5 = (Rule 6) = (16 ÷ 32) b (9 – 5 2) a (4 – 3 2) c (–1 2) j 4 (Rule 2) = b –1 a –2 c –2 j 4 = (Rule 6)

10 10 1. 2. 3. 4 a 2 ( a 2 – b 2 )4. 4 m 2 x 2 ( x 2 + x + 1) 5. a ( a + 4) + 4( a + 4)6. 2 a ( a – 4) – 3( a – 4) 7. m 2 n 3 ( mx 2 + 3 nx + 2) – 4 m 2 n 3 8. 4 m 2 n 2 ( m 3 n 8 + 4) – 3 m 3 n 10 ( m 2 + 2 n 3 ) Simplify. Rules of Exponents Independent Practice Solution follows… Topic 6.2.1 4 a 4 – 4 a 2 b 2 a 2 + 8 a + 16 m 3 n 3 x 2 + 3 m 2 n 4 x – 2 m 2 n 3 2 a 2 – 11 a + 12 m 5 n 10 + 16 m 2 n 2 – 6 m 3 n 13 4 m 2 x 4 + 4 m 2 x 3 + 4 m 2 x 2 1

11 11 9. 10. 11. 12. Simplify. Rules of Exponents Independent Practice Solution follows… Topic 6.2.1

12 12 13. m ? ( m 4 + 2 m 3 ) = m 6 + 2 m 5 14. m 4 a 6 (3 m ? a 8 + 4 m 2 a ? ) = 3 m 7 a 14 + 4 m 6 a 9 Find the value of ? that makes these statements true. Rules of Exponents Independent Practice Solution follows… Topic 6.2.1 ? = 2 ? = 3 ? = 4 ? = 7 15. 16.

13 13 Topic 6.2.1 Round Up Rules of Exponents You can apply the rules of exponents to any algebraic values. In this Topic you just dealt with monomials, but the rules work with expressions with more than one term too.


Download ppt "1 Topic 6.2.1 Rules of Exponents. 2 Lesson 1.1.1 California Standards: 2.0 Students understand and use such operations as taking the opposite, finding."

Similar presentations


Ads by Google