1. Warm Up Divide using long division, if there is a remainder put it as a fraction. 1. 161 ÷ 7 2. 104 ÷ 3 3. 4. 2x + 5y 23 7a – b Divide. 6x – 15y 3 7a 2 – ab a

2. 3.3 Dividing Polynomials Part 1 – Long Division

3. Polynomial long division is a method for dividing a polynomial by another polynomial of a lower degree. It is very similar to dividing numbers.

4. 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

5. 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. y – 3 2y 3 – y 2 + 0 y + 25 Step 4 Write the final answer. –y 2 + 2y 3 + 25 y – 3 = 2y 2 + 5y + 15 + 70 y – 3

6. 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. Step 3 Divide. 5x5x –(15x 2 + 5x) 3x – 12 + 1 –(3x + 1) –13 3x + 1 15x 2 + 8x – 12 Step 4 Write the final answer. 15x 2 + 8x – 12 3x + 1 = 5x + 1 – 13 3x + 1

7. Divide using long division. (x 2 + 5x – 28) ÷ (x – 3) = x + 8 – 4 x – 3

8. Use Long Division

9. Homework 3.1 #15-18 3.2 #31-34 3.3 #5-7, 13-18

10. Warm Up

11. 3.3 Dividing Polynomial Part 2 – Synthetic Division

12. Synthetic Division Shorthand method of dividing a polynomial by a linear binomial Uses only the coefficients Must haves: 1.Standard form 2.Include all terms (use 0 coefficient, if necessary) 3.Divisor in the form: (x-a)

14. Divide using synthetic division. (3x 4 – x 3 + 5x – 1) ÷ (x + 2) 3 –1 0 5 –1–2 –6 345 3x 3 – 7x 2 + 14x – 23 + 45 x + 2 46–2814 –2314–7 (6x 2 – 5x – 6) ÷ (x + 3) 6 –5 –6–3 –18 663 6x – 23 + 63 x + 3 –23 69

15. Divide using synthetic division. (x 2 – 3x – 18) ÷ (x – 6) 1 –3 –18 6 6 10 x + 3 18 3

16. Divide using synthetic division. (3x 2 + 9x – 2) ÷ (x – ) Step 1 Find a. Then write the coefficients and a in the synthetic division format. 1 3 Step 2 Bring down the first coefficient. Then multiply and add for each column. 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

17. 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 –

18. 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.

19. 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

20. 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 P( ) = 7 3 2–39 –69–27

21. Use synthetic substitution to evaluate the polynomial for the given value and check your answer. 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

22. 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. 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

23. 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

24. 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

25. HA2 Homework 3.3: pp 170-171: #21-28, 39-45 (odd)