1. Splash Screen

2. Five-Minute Check (over Lesson 5–1) CCSS Then/Now New Vocabulary
Example 1: Divide a Polynomial by a Monomial
Example 2: Division Algorithm
Example 3: Standardized Test Example: Divide Polynomials
Key Concept: Synthetic Division
Example 4: Synthetic Division
Example 5: Divisor with First Coeffecient Other than 1
Lesson Menu

3. Simplify b2 ● b5 ● b3.
A. b5
B. b8
C. b10
D. b30
5-Minute Check 1

4. Simplify b2 ● b5 ● b3.
A. b5
B. b8
C. b10
D. b30
5-Minute Check 1

5. A. B. C. D.
5-Minute Check 2

6. A. B. C. D.
5-Minute Check 2

7. Simplify (10a2 – 6ab + b2) – (5a2 – 2b2).
A. 15a2 + 8ab + 3b2
B. 10a2 – 6ab – b2
C. 5a2 + 6ab – 3b2
D. 5a2 – 6ab + 3b2
5-Minute Check 3

8. Simplify (10a2 – 6ab + b2) – (5a2 – 2b2).
A. 15a2 + 8ab + 3b2
B. 10a2 – 6ab – b2
C. 5a2 + 6ab – 3b2
D. 5a2 – 6ab + 3b2
5-Minute Check 3

9. Simplify 7w(2w2 + 8w – 5). A. 14w3 + 56w2 – 35w B. 14w2 + 15w – 35
C. 9w2 + 15w – 12
D. 2w2 + 15w – 5
5-Minute Check 4

10. Simplify 7w(2w2 + 8w – 5). A. 14w3 + 56w2 – 35w B. 14w2 + 15w – 35
C. 9w2 + 15w – 12
D. 2w2 + 15w – 5
5-Minute Check 4

11. State the degree of 6xy2 – 12x3y2 + y4 – 26.
B. 7
C. 5
D. 4
5-Minute Check 5

12. State the degree of 6xy2 – 12x3y2 + y4 – 26.
B. 7
C. 5
D. 4
5-Minute Check 5

13. Find the product of 3y(2y2 – 1)(y + 4).
A. 18y5 + 72y4 – 9y3 – 36y2
B. 6y4 + 24y3 – 3y2 – 12y
C. –18y3 – 3y2 + 12y
D. 6y3 – 2y + 4
5-Minute Check 6

14. Find the product of 3y(2y2 – 1)(y + 4).
A. 18y5 + 72y4 – 9y3 – 36y2
B. 6y4 + 24y3 – 3y2 – 12y
C. –18y3 – 3y2 + 12y
D. 6y3 – 2y + 4
5-Minute Check 6

15. Mathematical Practices 6 Attend to precision.
Content Standards
A.APR.6 Rewrite simple rational expressions in different forms; write a(x)/b(x) in the form q(x) + r(x)/b(x), where a(x), b(x), q(x), and r(x) are polynomials with the degree of r(x) less than the degree of b(x), using inspection, long division, or, for the more complicated examples, a computer algebra system.
Mathematical Practices
6 Attend to precision.
CCSS

16. Divide polynomials using long division.
You divided monomials.
Divide polynomials using long division.
Divide polynomials using synthetic division.
Then/Now

17. synthetic division
Vocabulary

18. = a – 3b2 + 2a2b3 a1 – 1 = a0 or 1 and b1 – 1 = b0 or 1
Divide a Polynomial by a Monomial
Sum of quotients
Divide.
= a – 3b2 + 2a2b3 a1 – 1 = a0 or 1 and b1 – 1 = b0 or 1
Answer:
Example 1

19. = a – 3b2 + 2a2b3 a1 – 1 = a0 or 1 and b1 – 1 = b0 or 1
Divide a Polynomial by a Monomial
Sum of quotients
Divide.
= a – 3b2 + 2a2b3 a1 – 1 = a0 or 1 and b1 – 1 = b0 or 1
Answer: a – 3b2 + 2a2b3
Example 1

20. A. 2x3y – 3x5y2 B. 1 + 2x3y – 3x5y2 C. 6x4y2 + 9x7y3 – 6x9y4
D x7y3 – 3x9y4
Example 1

21. A. 2x3y – 3x5y2 B. 1 + 2x3y – 3x5y2 C. 6x4y2 + 9x7y3 – 6x9y4
D x7y3 – 3x9y4
Example 1

22. Use long division to find (x2 – 2x – 15) ÷ (x – 5).
Division Algorithm
Use long division to find (x2 – 2x – 15) ÷ (x – 5).
3(x – 5) = 3x – 15
x(x – 5) = x2 – 5x
–2x – (–5x) = 3x
Answer:
Example 2

23. Use long division to find (x2 – 2x – 15) ÷ (x – 5).
Division Algorithm
Use long division to find (x2 – 2x – 15) ÷ (x – 5).
3(x – 5) = 3x – 15
x(x – 5) = x2 – 5x
–2x – (–5x) = 3x
Answer: The quotient is x + 3. The remainder is 0.
Example 2

24. Use long division to find (x2 + 5x + 6) ÷ (x + 3).
A. x + 2
B. x + 3
C. x + 2x
D. x + 8
Example 2

25. Use long division to find (x2 + 5x + 6) ÷ (x + 3).
A. x + 2
B. x + 3
C. x + 2x
D. x + 8
Example 2

26. Which expression is equal to (a2 – 5a + 3)(2 – a)–1?
Divide Polynomials
Which expression is equal to (a2 – 5a + 3)(2 – a)–1?
A a + 3 B C D
Example 3

27. Divide Polynomials
Read the Test Item
Since the second factor has an exponent of –1, this is a division problem.
Solve the Test Item
3(–a + 2) = –3a + 6
–a(–a + 2) = a2 – 2a
Rewrite 2 – a as –a + 2.
–5a – (–2a) = –3a
Subtract. 3 – 6 = –3
Example 3

28. The quotient is –a + 3 and the remainder is –3.
Divide Polynomials
The quotient is –a + 3 and the remainder is –3.
Therefore, .
Answer:
Example 3

29. The quotient is –a + 3 and the remainder is –3.
Divide Polynomials
The quotient is –a + 3 and the remainder is –3.
Therefore, .
Answer: The answer is D.
Example 3

30. Which expression is equal to (x2 – x – 7)(x – 3)–1?B. C. D.
Example 3

31. Which expression is equal to (x2 – x – 7)(x – 3)–1?B. C. D.
Example 3

32. Concept

33. Use synthetic division to find (x3 – 4x2 + 6x – 4) ÷ (x – 2).
Step 1 Write the terms of the dividend so that the degrees of the terms are in descending order. Then write just the coefficients as shown.
x3 – 4x2 + 6x – 4
   
1 –4 6 –4
Step 2 Write the constant r of the divisor x – r to the left. In this case, r = 2. Bring the first coefficient, 1, down as shown.
1 –4 6 –4 1
Example 4

34. Synthetic Division
Step 3 Multiply the first coefficient by r : 1 ● 2 = 2. Write the product under the second coefficient. Then add the product and the second coefficient: –4 + 2 = –2. 1
1 –4 6 –4 2 –2 2 –2 1
1 –4 6 –4
Step 4 Multiply the sum, –2, by r : –2 ● 2 = –4. Write the product under the next coefficient and add: (–4) = 2. –4 2
Example 4

35. Synthetic Division 2 –2 1 –4
1 –4 6 –4
Step 5 Multiply the sum, 2, by r : 2 ● 2 = 4. Write the product under the next coefficient and add: –4 + 4 = 0. The remainder is 0. 4
The numbers along the bottom are the coefficients of the quotient. Start with the power of x that is one less than the degree of the dividend.
Answer:
Example 4

36. Answer: The quotient is x2 – 2x + 2.
Synthetic Division 2 –2 1 –4
1 –4 6 –4
Step 5 Multiply the sum, 2, by r : 2 ● 2 = 4. Write the product under the next coefficient and add: –4 + 4 = 0. The remainder is 0. 4
The numbers along the bottom are the coefficients of the quotient. Start with the power of x that is one less than the degree of the dividend.
Answer: The quotient is x2 – 2x + 2.
Example 4

37. Use synthetic division to find (x2 + 8x + 7) ÷ (x + 1).
A. x + 9
B. x + 7
C. x + 8
D. x + 7
Example 4

38. Use synthetic division to find (x2 + 8x + 7) ÷ (x + 1).
A. x + 9
B. x + 7
C. x + 8
D. x + 7
Example 4

39. Use synthetic division to find (4y3 – 6y2 + 4y – 1) ÷ (2y – 1).
Divisor with First Coefficient Other than 1
Use synthetic division to find (4y3 – 6y2 + 4y – 1) ÷ (2y – 1).
Rewrite the divisor so it has a leading coefficient of 1.
Divide numerator and denominator by 2.
Simplify the numerator and denominator.
Example 5

40. Divisor with First Coefficient Other than 1
The result is
Example 5

41. Divisor with First Coefficient Other than 1
Answer:
Example 5

42. Answer: The solution is .
Divisor with First Coefficient Other than 1
Answer: The solution is
Check: Divide using long division.
2y2
–2y
+ 1
4y3 – 2y2
–4y2 + 4y
–4y2 + 2y
2y – 1
2y – 1
The result is 
Example 5

43. Use synthetic division to find (8y3 – 12y2 + 4y + 10) ÷ (2y + 1).A. B. C. D.
Example 5

44. Use synthetic division to find (8y3 – 12y2 + 4y + 10) ÷ (2y + 1).A. B. C. D.
Example 5

45. End of the Lesson