1. 1 Topic 6.3.1 Division by Monomials

2. 2 Lesson 1.1.1 California Standard: 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 learn how to use the rules of exponents to divide a polynomial by a monomial. Division by Monomials Topic 6.3.1 Key words: polynomial monomial exponent distributive property

3. 3 Lesson 1.1.1 The rules of exponents that you saw in Topic 6.2.1 really are useful. Division by Monomials Topic 6.3.1 In this Topic you’ll use them to divide polynomials by monomials. 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 Rules of Exponents

4. 4 Lesson 1.1.1 Dividing a Polynomial by a Monomial Division by Monomials Topic 6.3.1 To divide a polynomial by a monomial, you need to use the rules of exponents. The particular rule that’s useful here is: = x a – b provided x  0

5. 5 Division by Monomials Example 1 Topic 6.3.1 Divide 2 x 2 by x. Solution Solution follows… 2 x 2 x 2 x 2 ÷ x = = x a – b provided x  0 Use the rule to simplify the expression = 2 x = 2 x 1 = 2 x 2–1

6. 6 = (2 x 3–1  y 1–1 ) + ( x 1–1  y 2– 1 ) Division by Monomials Example 2 Topic 6.3.1 Divide 2 x 3 y + xy 2 by xy. Solution Solution follows… Divide each term in the expression by xy, using the distributive property = 2 x 2 + y = (2 x 2  1) + (1  y 1 ) Simplify using the rule = x a – b provided x  0

7. 7 Division by Monomials Example 3 Topic 6.3.1 Solution = (–1  m 3–1 ) – (–2  m 2–1  c 3–2 ) + (–5  c 4–2  v 3–1 ) Solution follows… Simplify. = – m 2 + 2 mc – 5 c 2 v 2

8. 8 Lesson 1.1.1 Guided Practice Division by Monomials Topic 6.3.1 Solution follows… Simplify each of these quotients. 1. 9 m 3 c 2 v 4 ÷ (–3 m 2 cv 3 ) 2. 3. –3 m 3 – 2 c 2 – 1 v 4 – 3 = –3 mcv 3 x 5 – 3 y 6 – 5 z 4 – 2 = 3 x 2 yz 2 2 m 3 – 3 x 2 – 2 – 3 m 4 – 3 x 3 – 2 = 2 – 3 mx

9. 9 Lesson 1.1.1 Guided Practice Division by Monomials Topic 6.3.1 Solution follows… Simplify each of these quotients. 4. 5. 6. –2 x 4 – 3 y 5 – 3 t 3 – 2 + 4 x 3 – 3 y 4 – 3 t 2 – 2 + x 5 – 3 y 3 – 3 t 3 – 2 = –2 xy 2 t + 4 y + x 2 t –2 x 5 – 4 y 8 – 3 a 4 – 0 z 12 – 9 = –2 xy 5 a 4 z 3 –2 a 9 – 5 d 1 – 1 f 9 – 0 k 3 – 2 + 3 a 8 – 5 d 6 – 1 f 5 – 0 k 3 – 2 – 7 c 1 – 0 a 8 – 5 d 8 – 1 k 4 – 2 = –2 a 4 f 9 k + 3 a 3 d 5 f 5 k – 7 ca 3 d 7 k 2

10. 10 Division by Monomials Independent Practice Solution follows… Topic 6.3.1 Simplify each of these quotients. 1. 2. 3. 4. –2 x 2 + x – 3 2 x 3 – x 2 + 3 x – 5 3 mv 2 – cv + 4 4 yz + 8 xy 2 z 2

11. 11 Division by Monomials Independent Practice Solution follows… Topic 6.3.1 5. Divide 15 x 5 – 10 x 3 + 25 x 2 by –5 x 2. –3 x 3 + 2 x – 5 6. Divide 20 a 6 b 4 – 14 a 7 b 5 + 10 a 3 b 7 by 2 a 3 b 4. 7. Divide 4 m 5 x 7 v 6 – 12 m 4 c 2 x 8 v 4 + 16 a 3 m 6 c 2 x 9 v 7 by –4 m 4 x 7 v 4. 10 a 3 – 7 a 4 b + 5 b 3 – mv 2 + 3 c 2 x – 4 a 3 m 2 c 2 x 2 v 3

12. 12 Division by Monomials Independent Practice Solution follows… Topic 6.3.1 Find the missing exponent in the quotients. 8. 9. ? = 3 ? = 4

13. 13 Topic 6.3.1 Round Up Division by Monomials This leads on to the next few Topics, where you’ll divide one polynomial by another polynomial. First, you’ll learn how to find the multiplicative inverse of a polynomial in Topic 6.3.2.