Dividing Polynomials

1. Use long division to divide polynomials • Use synthetic division to divide polynomials • Evaluate a polynomial using the Remainder Theorem • Use the Factor Theorem to solve a polynomial equation Dividing Polynomials Objectives

2. 6x– 15y 7a2 – ab 3 a • Divide using long division. 161 ÷ 7 23 12.18 ÷ 2.1 5.8 2x + 5y 7a – b

3. Long Division Vocabulary Reminders

4. Examples with variables

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

6. Arrange the terms of both the dividend and the divisor in descending order of degree (or power). • Divide the first term in the dividend by the first term in the divisor. The result is the first term of the quotient. • Multiply every term in the divisor by the first term in the quotient. Write the resulting product beneath the dividend with like terms lined up. • Subtract the product from the dividend. • Bring down the next term in the original dividend and write it next to the remainder to form a new dividend. • Usethis new expression as the dividend and repeat this process until the remainder can no longer be divided. • Check your answer by multiplying the quotient by the divisor to get the dividend. Long Division of Polynomials

7. y – 3 2y3 – y2 + 0y + 25 Using Long Division to Divide a Polynomial Divide using long division. (–y2 + 2y3+ 25) ÷ (y – 3) Step 1:Write the dividend in standard form, including terms with a coefficient of 0. 2y3 – y2 + 0y + 25 Step 2:Write division in the same way you would when dividing numbers.

8. y – 3 2y3 – y2 + 0y + 25 Step 3 Divide. Notice that y times 2y2 is 2y3. Write 2y2 above 2y3. 2y2 + 5y + 15 Multiply 2y2 by y – 3. Then subtract. Bring down the next term. Divide 5y2 by y. –(2y3 – 6y2) 5y2 + 0y Multiply 5y by y – 3. Then subtract. Bring down the next term. Divide 15y by y. –(5y2 – 15y) 15y + 25 Multiply 15 by y – 3. Then subtract. –(15y – 45) 70 Find the remainder.

9. 70 –y2 + 2y3 + 25 = 2y2 + 5y + 15 + y – 3 y – 3 Step 4 Write the final answer.

10. 3x + 1 15x2 + 8x – 12 Divide using long division. (15x2 + 8x– 12) ÷ (3x + 1) Step 1:Make sure the dividend and divisor are in standard form. Include terms with a coefficient of 0. 15x2 + 8x– 12 Step 2:Write division in the same way you would when dividing numbers.

11. 3x + 1 15x2 + 8x – 12 Step 3 Divide. Divide by3x. Notice that 3x times 5x is 15x2. Write 5x above 15x2. 5x + 1 Multiply 5x by 3x + 1. Then subtract. –(15x2 + 5x) 3x Bring down the next term. Divide 3x by 3x. –(3x + 1) –13 Multiply 1 by 3x + 1. Then subtract. Find the remainder.

12. 13 15x2 + 8x– 12 = 5x + 1 – 3x + 1 3x + 1 Step 4Write the final answer.

13. x – 3 x2 + 5x – 28 Divide using long division. (x2 + 5x– 28) ÷ (x – 3) Step 1:Write the dividend in standard form, including terms with a coefficient of 0. x2 + 5x– 28 Step 2:Write division in the same way you would when dividing numbers.

14. x – 3 x2 + 5x – 28 Step 3 Divide. Notice that x times x is x2. Write x above x2. x + 8 Multiply x by x – 3. Then subtract. Bring down the next term. Divide 8x by x. –(x2 – 3x) 8x – 28 –(8x – 24) Multiply 8 by x –3. Then subtract. –4 Find the remainder.

15. 4 x2 + 5x– 28 = x + 8 – x – 3 x – 3 Step 4 Write the final answer.

16. Synthetic divisionis 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 as a coefficient for any missing terms as well as a missing constant, and the divisor must be in the form (x – a). Dividing Polynomials Using Synthetic Division

17. The degree of the first term of the quotient is one less than the degree of the first term of the dividend.

18. –2 3 – 1 0 5 –1 Divide using synthetic division. (3x4 – x3+ 5x – 1) ÷ (x + 2) Step 1 Find a. a = –2 For (x + 2), a = –2. Step 2Write the coefficients and a in the synthetic division format. Use 0 for the coefficient of x2.

19. 45 3x3 – 7x2 + 14x – 23 + x + 2 Step 3Bring down the first coefficient. Then multiply and add for each column. –2 3 –1 0 5 –1 Draw a box around the remainder, 45. –6 14 –28 46 –23 3 14 45 –7 Step 4Write the quotient. Write the remainder over the divisor.

20. 6 –5 –6 –3 Divide using synthetic division. (6x2 – 5x – 6) ÷ (x + 3) Step 1Find a. a = –3 For (x + 3), a = –3. Step 2Write the coefficients and a in the synthetic division format. Write the coefficients of 6x2 – 5x – 6.

21. 63 6x – 23 + x + 3 Step 3Bring down the first coefficient. Then multiply and add for each column. –3 6 –5 –6 Draw a box around the remainder, 63. –18 69 6 –23 63 Step 4Write the quotient. Write the remainder over the divisor.

22. 1 –3 –18 6 Divide using synthetic division. (x2 – 3x – 18) ÷ (x – 6) Step 1Find a. a = 6 For (x – 6), a = 6. Step 2Write the coefficients and a in the synthetic division format. Write the coefficients of x2 – 3x – 18.

23. Step 3Bring down the first coefficient. Then multiply and add for each column. 6 1 –3 –18 There is no remainder. 6 18 1 3 0 Step 4Write the quotient. x + 3

24. 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.

25. Use synthetic substitution to evaluate the polynomial for the given value. P(x) = 2x3 + 5x2 – x + 7 for x = 2. 2 2 5 –1 7 Write the coefficients of the dividend. Use a = 2. 4 18 34 2 9 17 41 P(2) = 41 Check Substitute 2 for x in P(x) = 2x3 + 5x2 – x + 7. P(2) = 2(2)3 + 5(2)2 – (2) + 7  P(2) = 41

26. Use synthetic substitution to evaluate the polynomial for the given value. P(x) = x3 + 3x2 + 4 for x = –3. –3 1 3 0 4 Write the coefficients of the dividend. Use 0 for the coefficient of x2 Use a = –3. –3 0 0 1 0 0 4 P(–3) = 4 Check Substitute –3 for x in P(x) = x3 + 3x2 + 4. P(–3) = (–3)3 + 3(–3)2 + 4  P(–3) = 4

27. The Factor Theorem Let be a polynomial. If , then is a factor of . If is a factor of ,then . Example: Solve the equation given that is a zero of this function. Since 3 is a zero of the function, then . The Factor Theorem tells us that is a factor of this polynomial function.

28. Use synthetic division to divide by . The result is . The remainder, 0, verifies that is a factor of . Is this factored completely? No. We have to factor further or use the quadratic formula to solve.

29. We need to try factoring or use the quadratic formula to solve for . Now we can solve by setting up each expression = 0. The solution set is