7.3.1 Products and Factors of Polynomials

1. 7.3.1 Products and Factors of Polynomials Objectives: Multiply and factor polynomials Use the Factor Theorem to solve problems

2. Real-World Application • Objective: Multiply and factor polynomials

3. Real-World Application • Objective: Multiply and factor polynomials

4. Collins Type 1 If I wanted to maximize the volume of this open-top box, what do you hypothesize I would need to do? In other words, what important information do I need to find? • Objective: Multiply and factor polynomials

5. Example 1 Write the function f(x) = (x – 1)(x + 4)(x – 3) as a polynomial function in standard form. (x – 1)(x + 4)(x – 3) = (x – 1) [(x + 4)(x – 3)] = (x – 1) (x2 + x – 12) = x(x2+ x – 12) – 1(x2 + x – 12) = x3 + x2 – 12x – x2 – x + 12 = x3 – 13x + 12 f(x) = x3 – 13x + 12 • Objective: Multiply and factor polynomials

6. Example 2 Factor each polynomial. a) x3 – 16x2 + 64x x x x = x(x2 – 16x + 64) = x(x – 8)(x – 8) b) x3 + 6x2 – 5x - 30 = (x3 + 6x2) + (-5x – 30) = x2(x + 6) – 5(x + 6) (x + 6) (x + 6) = (x + 6)(x2 – 5) • Objective: Multiply and factor polynomials

7. Factoring the Sum and Difference of Two Cubes (a + b)(a2 – ab + b2) a3 + b3 = (a - b)(a2 + ab + b2) a3 - b3 = • Objective: Multiply and factor polynomials

8. Example 3 Factor each polynomial. a) x3 + 125 = x3 + 53 = (x + 5)(x2 – 5x + 25) b) x3 - 27 = x3 - 33 = (x - 3)(x2 + 3x + 9) • Objective: Multiply and factor polynomials

9. Factor Theorem x – r is a factor of the polynomial expression that defines the function P iff r is a solution of P(x) = 0, that is, iff P(r) = 0. • Objective: Use the Factor Theorem to solve problems

10. Example 4 Use substitution to determine whether x – 1 is a factor of x3 – x2 – 5x – 3. Let x3 – x2 – 5x – 3 = 0 f(1) = (1)3 – (1)2 – 5(1) - 3 f(1) = 1 – 1 – 5 - 3 f(1) = -8 Since f(1) does not equal zero, x – 1 is not a factor. • Objective: Use the Factor Theorem to solve problems

11. Practice 1) Factor each polynomial. x3 + 1000 x3 - 125 2) Use substitution to determine whether x + 3 is a factor of x3 – 3x2 – 6x + 8. • Objective: Use the Factor Theorem to solve problems

12. Collins Type 2 If p(-2) = 0, what does that tell you about the graph of p(x)? • Objective: Use the Factor Theorem to solve problems

13. Homework Lesson 7.3 Exercises 51-69 odd

14. 7.3.2 Products and Factors of Polynomials Objectives: Divide one polynomial by another synthetic division Divide one polynomial by another using long division

15. Example 1 Use synthetic division to find the quotient: (6 – 3x2 + x + x3) ÷ (x – 3) Are the conditions for synthetic division met? Step 1: Write the opposite of the constant of the divisor on the shelf, and the coefficients of the dividend (in order) on the right. -3 6 1 1 3 • Objective: Divide one polynomial by another using synthetic division

16. Example 1 Use synthetic division to find the quotient: (x3 – 3x2 + x + 6) ÷ (x – 3) Step 2: Bring down the first coefficient under the line. -3 6 1 1 3 1 • Objective: Divide one polynomial by another using synthetic division

17. Example 1 Use synthetic division to find the quotient: (x3 – 3x2 + x + 6) ÷ (x – 3) Step 3: Multiply the number on the shelf, 3, by the number below the line and write the product below the next coefficient. -3 6 1 1 3 3 1 • Objective: Divide one polynomial by another using synthetic division

18. Example 1 Use synthetic division to find the quotient: (x3 – 3x2 + x + 6) ÷ (x – 3) Step 4: Write the sum of -3 and 3 below the line. -3 6 1 1 3 3 1 0 • Objective: Divide one polynomial by another using synthetic division

19. Example 1 Use synthetic division to find the quotient: (x3 – 3x2 + x + 6) ÷ (x – 3) Repeat steps 3 and 4. -3 6 1 1 3 3 0 1 0 1 • Objective: Divide one polynomial by another using synthetic division

20. Example 1 Use synthetic division to find the quotient: (x3 – 3x2 + x + 6) ÷ (x – 3) Repeat steps 3 and 4. -3 6 1 1 3 3 0 3 1 0 1 9 • Objective: Divide one polynomial by another using synthetic division

21. 9 x2 + 1 + x – 3 Example 1 Use synthetic division to find the quotient: (x3 – 3x2 + x + 6) ÷ (x – 3) The remainder is 9 and the resulting numbers are the coefficients of the quotient. -3 6 1 1 3 3 0 3 1 0 1 9 Remainder Answer: • Objective: Divide one polynomial by another using synthetic division

22. Practice Group 1 & 5: Divide: (x3 + 3x2 – 13x - 15) ÷ (x – 3) Group 2 & 6: Divide: (x3 - 2x2 – 22x + 40) ÷ (x – 4) Group 3 & 7: Divide: (x3 - 27) ÷ (x – 3) Group 4 & 8: Divide: (x5 + 6x3 - 5x4 + 5x - 15) ÷ (x – 3) • Objective: Divide one polynomial by another using synthetic division

23. Do you remember long division? Using long division: 745 ÷ 3 2 4 8 745 3 1 - 6 Answer: 248 3 4 1 1 2 - 2 5 - 24 1 • Objective: Divide one polynomial by another using long division

24. - 16 x – 4 x2 + 2x – 14 – 16 x – 4 Example 2 Using long division: (x3 – 2x2 – 22x + 40) ÷ (x – 4) x2 + 2x – 14 x – 4 x3 – 2x2 – 22x + 40 (x3 – 4x2) - 2x2 – 22x (2x2 – 8x) - + 40 –14x (–14x + 56) – – 16 Answer: • Objective: Divide one polynomial by another using long division

25. Example 3 Use long division to determine if x2 + 3x + 2 is a factor of x3 + 6x2 + 11x + 6. x + 3 x2 + 3x + 2 x3 + 6x2 + 11x + 6 + 6 (x3 + 3x2 + 2x ) - 3x2 + 9x (3x2 + 9x + 6) - x2 + 3x + 2 is a factor because the remainder is 0 0 • Objective: Divide one polynomial by another using long division

26. Practice Group 1 & 5: Divide: (x3 - 8) ÷ (x2 – 2x + 4) Group 2 & 6: Divide: (10x - 5x2 + x3 - 24) ÷ (x2 – x + 6) Group 3 & 7: Divide: (x3 + 6x2 – x - 30) ÷ (x2 + 8x + 15) Group 4 & 8: Divide: (x3 + 3x2 – 13x - 15) ÷ (x2 – 2x – 3) • Objective: Divide one polynomial by another using long division

27. Collins Type 1 When dividing x3 + 11x2 + 39x + 45 by x + 5, would you use synthetic division or long division? Explain why. • Objective: Divide one polynomial by another

28. Homework Lesson 7.3 Read Textbook Pages 442-444 Exercises 71-89 odd

29. 2 Example 3 Given that 2 is a zero of P(x) = x3 – 3x2 + 4, use division to factor x3 – 3x2 + 4. Since 2 is a zero, x = 2 , so x – 2 = 0 , which means x – 2 is a factor of x3 – 3x2 + 4. (x3 – 3x2 + 4) ÷ (x – 2) Method 2 Method 1 x2 - x – 2 1 -3 0 4 x – 2 x3 – 3x2 + 0x + 4 2 -2 -4 - (x3 – 2x2) 1 -1 -2 0 -x2 + 0x - (-x2 + 2x) x3 – 3x2 + 4 = (x – 2)(x2 – x – 2) –2x + 4 - (–2x + 4) 0 • Objective: Divide one polynomial by another

30. Practice Given that -3 is a zero of P(x) = x3 – 13x - 12, use division to factor x3 – 13x – 12. Groups 1-4 use Method 1 (Long Division) Groups 5-8 use Method 2 (Synthetic Division) • Objective: Divide one polynomial by another

31. Remainder Theorem If the polynomial expression that defines the function of P is divided by x – a, then the remainder is the number P(a). • Objective: Use the Remainder Theorem to solve problems

32. 6 Example 6 Given P(x) = 3x3 – 4x2 + 9x + 5 is divided by x – 6, find the remainder. Method 2 Method 1 3 -4 9 5 P(6) = 3(6)3 – 4(6)2 + 9(6) + 5 = 3(216) – 4(36) + 54 + 5 18 84 558 3 14 93 563 = 648 – 144 + 54 + 5 = 563 • Objective: Use the Remainder Theorem to solve problems

33. Practice Given P(x) = 3x3 + 2x2 + 3x + 1 is divided by x + 2, find the remainder. • Objective: Use the Remainder Theorem to solve problems

34. A company manufactures cardboard boxes in the following way: they begin with 12"-by-18" pieces of cardboard, cut an x"-by-x" square from each of the four corners, then fold up the four flaps to make an open-top box. a. Sketch a picture or pictures of the manufacturing process described above. Label all segments in your diagram with their lengths (these will be formulas in terms of x). b. What are the length, width, and height of the box, in terms of x? c. Write a function V(x) expressing the volume of the box. d. Only some values of x would be meaningful in this problem. What is the interval of appropriate x-values? e. Using the interval you just named, make the graph V(x) on your calculator, then sketch it on paper. f. What value of x would produce a box with maximum volume? g. What are the dimensions and the volume for the box of maximum volume?