1. Holt Algebra 2 6-3 Dividing Polynomials 6-3 Dividing Polynomials Holt Algebra 2 Warm Up Warm Up Lesson Presentation Lesson Presentation Lesson Quiz Lesson Quiz

2. Holt Algebra 2 6-3 Dividing Polynomials Warm Up Divide using long division. 1. 161 ÷ 7 2. 12.18 ÷ 2.1 3. 4. 2x + 5y 23 5.8 7a – b Divide. 6x – 15y 3 7a 2 – ab a

3. Holt Algebra 2 6-3 Dividing Polynomials Use long division and synthetic division to divide polynomials. Objective

4. Holt Algebra 2 6-3 Dividing Polynomials Polynomial long division is a method for dividing a polynomial by another polynomials of a lower degree. It is very similar to dividing numbers.

5. Holt Algebra 2 6-3 Dividing Polynomials Divide using long division. Example 1: Using Long Division to Divide a Polynomial (–y 2 + 2y 3 + 25) ÷ (y – 3) 2y 3 – y 2 + 0y + 25 Step 1 Write the dividend in standard form, including terms with a coefficient of 0. Step 2 Write division in the same way you would when dividing numbers. y – 3 2y 3 – y 2 + 0 y + 25

6. Holt Algebra 2 6-3 Dividing Polynomials Notice that y times 2y 2 is 2y 3. Write 2y 2 above 2y 3. Step 3 Divide. 2y22y2 –(2y 3 – 6y 2 ) Multiply y – 3 by 2y 2. Then subtract. Bring down the next term. Divide 5y 2 by y. 5y 2 + 0y + 5y –(5y 2 – 15y) Multiply y – 3 by 5y. Then subtract. Bring down the next term. Divide 15y by y. 15y + 25 –(15y – 45) 70 Find the remainder. + 15 Multiply y – 3 by 15. Then subtract. Example 1 Continued y – 3 2y 3 – y 2 + 0 y + 25

7. Holt Algebra 2 6-3 Dividing Polynomials Step 4 Write the final answer. Example 1 Continued –y 2 + 2y 3 + 25 y – 3 = 2y 2 + 5y + 15 + 70 y – 3

8. Holt Algebra 2 6-3 Dividing Polynomials Check It Out! Example 1a Divide using long division. (15x 2 + 8x – 12) ÷ (3x + 1) 15x 2 + 8x – 12 Step 1 Write the dividend in standard form, including terms with a coefficient of 0. Step 2 Write division in the same way you would when dividing numbers. 3x + 1 15x 2 + 8x – 12

9. Holt Algebra 2 6-3 Dividing Polynomials Check It Out! Example 1a Continued Step 4 Write the final answer. 15x 2 + 8x – 12 3x + 1 = 5x + 1 – 13 3x + 1

10. Holt Algebra 2 6-3 Dividing Polynomials Check It Out! Example 1a Continued Step 3 Divide. 5x5x –(15x 2 + 5x) 3x – 12 + 1 –(3x + 1) –13 3x + 1 15x 2 + 8x – 12

11. Holt Algebra 2 6-3 Dividing Polynomials Check It Out! Example 1b Divide using long division. (x 2 + 5x – 28) ÷ (x – 3) x 2 + 5x – 28 Step 1 Write the dividend in standard form, including terms with a coefficient of 0. Step 2 Write division in the same way you would when dividing numbers. x – 3 x 2 + 5x – 28

12. Holt Algebra 2 6-3 Dividing Polynomials Step 3 Divide. x –(x 2 – 3x) 8x – 28 + 8 –(8x – 24) –4 Check It Out! Example 1b Continued x – 3 x 2 + 5x – 28

13. Holt Algebra 2 6-3 Dividing Polynomials Check It Out! Example 1b Continued Step 4 Write the final answer. x 2 + 5x – 28 x – 3 = x + 8 – 4 x – 3

14. Holt Algebra 2 6-3 Dividing Polynomials

15. Holt Algebra 2 6-3 Dividing Polynomials Synthetic division is a shorthand method of dividing a polynomial by a linear binomial by using only the coefficients. For synthetic division to work, the polynomial must be written in standard form, using 0 and a coefficient for any missing terms, and the divisor must be in the form (x – a).

16. Holt Algebra 2 6-3 Dividing Polynomials

17. Holt Algebra 2 6-3 Dividing Polynomials Divide using synthetic division. Example 2A: Using Synthetic Division to Divide by a Linear Binomial (3x 2 + 9x – 2) ÷ (x – ) Step 1 Find a. Then write the coefficients and a in the synthetic division format. Write the coefficients of 3x 2 + 9x – 2. 1 3 For (x – ), a =. 1 3 1 3 1 3 a = 1 3 3 9 –2

18. Holt Algebra 2 6-3 Dividing Polynomials Example 2A Continued Step 2 Bring down the first coefficient. Then multiply and add for each column. Draw a box around the remainder, 1. 1 3 1 3 3 9 –2 1 3 Step 3 Write the quotient. 3x + 10 + 1 1 3 1 3 x – 10 1 3 1 1 3 3

19. Holt Algebra 2 6-3 Dividing Polynomials Example 2A Continued 3x + 10 + 1 1 3 1 3 x – Check Multiply (x – ) 1 3 = 3x 2 + 9x – 2 (x – ) 1 3 1 3 1 3 3x + 10 + 1 1 3 1 3 x –

20. Holt Algebra 2 6-3 Dividing Polynomials Divide using synthetic division. (3x 4 – x 3 + 5x – 1) ÷ (x + 2) Step 1 Find a. Use 0 for the coefficient of x 2. For (x + 2), a = –2. a = –2 Example 2B: Using Synthetic Division to Divide by a Linear Binomial 3 – 1 0 5 –1–2 Step 2 Write the coefficients and a in the synthetic division format.

21. Holt Algebra 2 6-3 Dividing Polynomials Example 2B Continued Draw a box around the remainder, 45. 3 –1 0 5 –1–2 Step 3 Bring down the first coefficient. Then multiply and add for each column. –6 345 Step 4 Write the quotient. 3x 3 – 7x 2 + 14x – 23 + 45 x + 2 Write the remainder over the divisor. 46–2814 –2314–7

22. Holt Algebra 2 6-3 Dividing Polynomials Check It Out! Example 2a Divide using synthetic division. (6x 2 – 5x – 6) ÷ (x + 3) Step 1 Find a. a = –3 –3 6 –5 –6 Step 2 Write the coefficients and a in the synthetic division format.

23. Holt Algebra 2 6-3 Dividing Polynomials Check It Out! Example 2a Continued 6 –5 –6–3 Step 3 Bring down the first coefficient. Then multiply and add for each column. –18 663 Step 4 Write the quotient. 6x – 23 + 63 x + 3 –23 69

24. Holt Algebra 2 6-3 Dividing Polynomials Check It Out! Example 2b Divide using synthetic division. (x 2 – 3x – 18) ÷ (x – 6) Step 1 Find a. Write the coefficients of x 2 – 3x – 18. For (x – 6), a = 6. a = 6 6 1 –3 –18 Step 2 Write the coefficients and a in the synthetic division format.

25. Holt Algebra 2 6-3 Dividing Polynomials Check It Out! Example 2b Continued 1 –3 –18 6 Step 3 Bring down the first coefficient. Then multiply and add for each column. 6 10 Step 4 Write the quotient. x + 3 18 3

26. Holt Algebra 2 6-3 Dividing Polynomials You can use synthetic division to evaluate polynomials. This process is called synthetic substitution. The process of synthetic substitution is exactly the same as the process of synthetic division, but the final answer is interpreted differently, as described by the Remainder Theorem.

27. Holt Algebra 2 6-3 Dividing Polynomials Example 3A: Using Synthetic Substitution Use synthetic substitution to evaluate the polynomial for the given value. P(x) = 2x 3 + 5x 2 – x + 7 for x = 2. Write the coefficients of the dividend. Use a = 2. 2 5 –1 7 2 4 241 P(2) = 41 Check Substitute 2 for x in P(x) = 2x 3 + 5x 2 – x + 7. P(2) = 2(2) 3 + 5(2) 2 – (2) + 7 P(2) = 41 3418 179

28. Holt Algebra 2 6-3 Dividing Polynomials Example 3B: Using Synthetic Substitution Use synthetic substitution to evaluate the polynomial for the given value. P(x) = 6x 4 – 25x 3 – 3x + 5 for x = –. 6 –25 0 –3 5 –2 67 1 3 – 1 3 Write the coefficients of the dividend. Use 0 for the coefficient of x 2. Use a =. 1 3 P( ) = 7 1 3 2–39 –69–27

29. Holt Algebra 2 6-3 Dividing Polynomials Check It Out! Example 3a Use synthetic substitution to evaluate the polynomial for the given value. P(x) = x 3 + 3x 2 + 4 for x = –3. 1 3 0 4 –3 14 P(–3) = 4 Check Substitute –3 for x in P(x) = x 3 + 3x 2 + 4. P(–3) = (–3) 3 + 3(–3) 2 + 4 P(–3) = 4 00 00

30. Holt Algebra 2 6-3 Dividing Polynomials Check It Out! Example 3b Use synthetic substitution to evaluate the polynomial for the given value. P(x) = 5x 2 + 9x + 3 for x =. 5 9 3 1 55 1 5 1 5 P( ) = 5 1 5 2 10

31. Holt Algebra 2 6-3 Dividing Polynomials Example 4: Geometry Application Write an expression that represents the area of the top face of a rectangular prism when the height is x + 2 and the volume of the prism is x 3 – x 2 – 6x. Substitute. Use synthetic division. The volume V is related to the area A and the height h by the equation V = A  h. Rearranging for A gives A =. V h x 3 – x 2 – 6x x + 2 A(x) = 1 –1 –6 0 –2 10 The area of the face of the rectangular prism can be represented by A(x)= x 2 – 3x. 0 6 0 –3

32. Holt Algebra 2 6-3 Dividing Polynomials Check It Out! Example 4 Write an expression for the length of a rectangle with width y – 9 and area y 2 – 14y + 45. The area A is related to the width w and the length l by the equation A = l  w. y 2 – 14y + 45 y – 9 l(x) = 1 –14 45 9 9 10 The length of the rectangle can be represented by l(x)= y – 5. –45 –5

33. Holt Algebra 2 6-3 Dividing Polynomials 4. Find an expression for the height of a parallelogram whose area is represented by 2x 3 – x 2 – 20x + 3 and whose base is represented by (x + 3). Lesson Quiz 2. Divide by using synthetic division. (x 3 – 3x + 5) ÷ (x + 2) 1. Divide by using long division. (8x 3 + 6x 2 + 7) ÷ (x + 2) 2x 2 – 7x + 1 194; –4 3. Use synthetic substitution to evaluate P(x) = x 3 + 3x 2 – 6 for x = 5 and x = –1. 8x 2 – 10x + 20 – 33 x + 2 x 2 – 2x + 1 + 3 x + 2