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Polynomial Division Methods Lesson Presentation

Learn to divide polynomials using long division and synthetic division. Understand the vocabulary and steps involved. Practice with examples to master the concept. Check your understanding with a quiz.

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Polynomial Division Methods Lesson Presentation

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  1. 6-3 Dividing Polynomials Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 2 Holt Algebra 2

  2. 6x– 15y 7a2 – ab 3 a Warm Up Divide using long division. 1. 161 ÷ 7 23 2. 12.18 ÷ 2.1 5.8 Divide. 3. 2x + 5y 4. 7a – b

  3. Objective Use long division and synthetic division to divide polynomials.

  4. Vocabulary synthetic division

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

  6. y – 3 2y3 – y2 + 0y + 25 Example 1: 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.

  7. y – 3 2y3 – y2 + 0y + 25 Example 1 Continued Step 3 Divide. 2y2 + 5y + 15 Notice that y times 2y2 is 2y3. Write 2y2 above 2y3. Multiply y – 3 by 2y2. Then subtract. Bring down the next term. Divide 5y2 by y. –(2y3 – 6y2) 5y2 + 0y –(5y2 – 15y) Multiply y – 3 by 5y. Then subtract. Bring down the next term. Divide 15y by y. 15y + 25 Multiply y – 3 by 15. Then subtract. –(15y – 45) 70 Find the remainder.

  8. 70 –y2 + 2y3 + 25 = 2y2 + 5y + 15 + y – 3 y – 3 Example 1 Continued Step 4 Write the final answer.

  9. 3x + 1 15x2 + 8x – 12 Check It Out! Example 1a Divide using long division. (15x2 + 8x– 12) ÷ (3x + 1) Step 1 Write the dividend in standard form, including terms with a coefficient of 0. 15x2 + 8x– 12 Step 2 Write division in the same way you would when dividing numbers.

  10. 3x + 1 15x2 + 8x – 12 Check It Out! Example 1a Continued Step 3 Divide. 5x + 1 Notice that 3x times 5x is 15x2. Write 5x above 15x2. Multiply 3x + 1 by 5x. Then subtract. Bring down the next term. Divide 3x by 3x. –(15x2 + 5x) 3x – 12 –(3x + 1) Multiply 3x + 1 by 1. Then subtract. –13 Find the remainder.

  11. 13 15x2 + 8x– 12 = 5x + 1 – 3x + 1 3x + 1 Check It Out! Example 1a Continued Step 4 Write the final answer.

  12. x – 3 x2 + 5x – 28 Check It Out! Example 1b 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.

  13. x – 3 x2 + 5x – 28 Check It Out! Example 1b Continued Step 3 Divide. x + 8 Notice that x times x is x2. Write x above x2. Multiply x – 3 by x. Then subtract. Bring down the next term. Divide 8x by x. –(x2 – 3x) 8x – 28 –(8x – 24) Multiply x – 3 by 8. Then subtract. –4 Find the remainder.

  14. 4 x2 + 5x– 28 = x + 8 – x – 3 x – 3 Check It Out! Example 1b Continued Step 4 Write the final answer.

  15. Synthetic division is 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 and a coefficient for any missing terms, and the divisor must be in the form (x – a).

  16. 1 1 1 3 3 3 a = For (x – ), a = . 3 9 –2 1 1 3 3 Example 2A: Using Synthetic Division to Divide by a Linear Binomial Divide using synthetic division. (3x2 + 9x – 2) ÷ (x – ) Step 1 Find a. Then write the coefficients and a in the synthetic division format. Write the coefficients of 3x2 + 9x – 2.

  17. 1 1 1 1 1 3 3 3 3 3 Draw a box around the remainder, 1 . 3 1 3 9 –2 1 1 3 3x + 10 + x – Example 2A Continued Step 2 Bring down the first coefficient. Then multiply and add for each column. 1 3 10 Step 3 Write the quotient.

  18. Check Multiply (x – ) 1 1 1 1 1 1 1 1 3 3 3 3 3 3 3 3 (x – ) (x – ) (x – ) 3x + 10 + 1 1 3x + 10 + x – x – Example 2A Continued = 3x2 + 9x – 2

  19. –2 3 – 1 0 5 –1 Example 2B: Using Synthetic Division to Divide by a Linear Binomial Divide using synthetic division. (3x4 – x3+ 5x – 1) ÷ (x + 2) Step 1 Find a. a = –2 For (x + 2), a = –2. Step 2 Write the coefficients and a in the synthetic division format. Use 0 for the coefficient of x2.

  20. 45 3x3 – 7x2 + 14x – 23 + x + 2 Example 2B Continued Step 3 Bring 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 3 –7 14 –23 45 Step 4 Write the quotient. Write the remainder over the divisor.

  21. 6 –5 –6 –3 Check It Out! Example 2a Divide using synthetic division. (6x2 – 5x – 6) ÷ (x + 3) Step 1 Find a. a = –3 For (x + 3), a = –3. Step 2 Write the coefficients and a in the synthetic division format. Write the coefficients of 6x2 – 5x – 6.

  22. 63 6x – 23 + x + 3 Check It Out! Example 2a Continued Step 3 Bring 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 4 Write the quotient. Write the remainder over the divisor.

  23. 1 –3 –18 6 Check It Out! Example 2b Divide using synthetic division. (x2 – 3x – 18) ÷ (x – 6) Step 1 Find a. a = 6 For (x – 6), a = 6. Step 2 Write the coefficients and a in the synthetic division format. Write the coefficients of x2 – 3x – 18.

  24. Check It Out! Example 2b Continued Step 3 Bring 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 4 Write the quotient. x + 3

  25. 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.

  26. Example 3A: Using Synthetic Substitution 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

  27. 1 1 1 1 3 3 3 3 Write the coefficients of the dividend. Use 0 for the coefficient of x2. Use a = . – P( ) = 7 Example 3B: Using Synthetic Substitution Use synthetic substitution to evaluate the polynomial for the given value. P(x) = 6x4 – 25x3 – 3x + 5 for x = – . 6 –25 0 –3 5 –2 9 –3 2 6 –27 9 –6 7

  28. Check It Out! Example 3a 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

  29. 1 1 1 1 5 5 5 5 Write the coefficients of the dividend. Use a = . P( ) = 5 Check It Out! Example 3b Use synthetic substitution to evaluate the polynomial for the given value. P(x) = 5x2 + 9x + 3 for x = . 5 9 3 1 2 5 10 5

  30. V h The volume V is related to the area A and the height h by the equation V = A h. Rearrangingfor A gives A = . x3 – x2 – 6x A(x) = x + 2 Example 4: 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 x3 – x2 – 6x. Substitute. –2 1 –1 –6 0 Use synthetic division. The area of the face of the rectangular prism can be represented by A(x)= x2 – 3x. –2 6 0 1 0 –3 0

  31. y2 – 14y + 45 l(x) = y – 9 Check It Out! Example 4 Write an expression for the length of a rectangle with width y – 9 and area y2 – 14y + 45. The area A is related to the width w and the length l by the equation A = l w. Substitute. 9 1 –14 45 Use synthetic division. 9 –45 1 0 –5 The length of the rectangle can be represented by l(x)= y – 5.

  32. 8x2– 10x + 20 – x2– 2x + 1 + 33 3 x + 2 x + 2 Lesson Quiz 1. Divide by using long division.(8x3 + 6x2 + 7) ÷ (x + 2) 2. Divide by using synthetic division. (x3– 3x + 5) ÷ (x+ 2) 3. Use synthetic substitution to evaluate P(x) = x3 + 3x2 – 6 for x = 5 and x = –1. 194; –4 4. Find an expression for the height of a parallelogram whose area is represented by 2x3 – x2– 20x + 3 and whose base is represented by (x + 3). 2x2 – 7x + 1

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