1. Division of Polynomials Digital Lesson

2. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 2 Dividing Polynomials Long division of polynomials is similar to long division of whole numbers. dividend = (quotient divisor) + remainder The result is written in the form: quotient + When you divide two polynomials you can check the answer using the following:

3. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 3 + 2 Example: Divide & Check Example: Divide x 2 + 3x – 2 by x – 1 and check the answer. x x 2 + x 2x2x– 2 2x + 2 – 4– 4 remainder Check: 1. 2. 3. 4. 5. 6. correct (x + 2) quotient (x + 1) divisor + (– 4) remainder = x 2 + 3x – 2 dividend Answer: x + 2 + – 4– 4

4. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 4 Example: Divide & Check Example: Divide 4x + 2x 3 – 1 by 2x – 2 and check the answer. Write the terms of the dividend in descending order. 1. x2x2 2. 2x 3 – 2x 2 3. 2x22x2 + 4x 4. + x 5. 2x 2 – 2x 6. 6x6x – 1 7. + 3 8. 6x – 6 9. 5 Check: (x 2 + x + 3)(2x – 2) + 5 = 4x + 2x 3 – 1 Answer: x 2 + x + 3 5 Since there is no x 2 term in the dividend, add 0x 2 as a placeholder.

5. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 5 Example: Division With Zero Remainder x x 2 – 2x – 3x + 6 – 3 – 3x + 6 0 Answer: x – 3 with no remainder. Check: (x – 2)(x – 3) = x 2 – 5x + 6 Example: Divide x 2 – 5x + 6 by x – 2.

6. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 6 Example: Division With Nonzero Remainder Example: Divide x 3 + 3x 2 – 2x + 2 by x + 3 and check the answer. x2x2 x 3 + 3x 2 0x20x2 – 2x – 2 – 2x – 6 8 Check: (x + 3)(x 2 – 2) + 8 = x 3 + 3x 2 – 2x + 2 Answer: x 2 – 2 + 8 + 2 Note: the first subtraction eliminated two terms from the dividend. Therefore, the quotient skips a term. + 0x

7. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 7 16 Synthetic Division Synthetic division is a shorter method of dividing polynomials. This method can be used only when the divisor is of the form x – a. It uses the coefficients of each term in the dividend. Example: Divide 3x 2 + 2x – 1 by x – 2 using synthetic division. 3 2 – 1 2 Since the divisor is x – 2, a = 2. 3 1. Bring down 3 2. (2 3) = 6 6 815 3. (2 + 6) = 8 4. (2 8) = 16 5. (–1 + 16) = 15 coefficients of quotient remainder value of a coefficients of the dividend 3x + 8Answer: 15

8. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 8 Example: Synthetic Division Example: Divide x 3 – 3x + 4 by x + 3 using synthetic division. Notice that the degree of the first term of the quotient is one less than the degree of the first term of the dividend. remainder a coefficients of quotient – 3– 3 Since, x – a = x + 3, a = – 3. 1 0 – 3 4 1 – 3– 3 – 3– 3 9– 18 6– 14 coefficients of dividend = x 2 – 3x + 6 – 14 Insert zero coefficient as placeholder for the missing x 2 term.

9. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 9 Remainder Theorem Remainder Theorem: The remainder of the division of a polynomial f (x) by x – a is f (a). Example: Using the remainder theorem, evaluate f(x) = x 4 – 4x – 1 when x = 3. 9 1 0 0 – 4 – 1 3 1 3 39 6927 2368 The remainder is 68 at x = 3, so f (3) = 68. You can check this using substitution:f(3) = (3) 4 – 4(3) – 1 = 68. value of x

10. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 10 Example: Using Remainder Theorem Example: Using synthetic division and the remainder theorem, evaluate f (x) = x 2 – x at x = – 2. 6 1 – 1 0 – 2– 2 1 – 2– 2 – 3– 36 Then f (– 2) = 6 and (– 2, 6) is a point on the graph of f(x) = x 2 – x. f(x) = x 2 – x x y 2 4 (– 2, 6) remainder