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6-2. Multiplying Polynomials. Warm Up. Lesson Presentation. Lesson Quiz. Holt Algebra 2. Warm Up Multiply. 1. x ( x 3 ). x 4. 2. 3 x 2 ( x 5 ). 3 x 7. 3. 2(5 x 3 ). 10 x 3. 4. x (6 x 2 ). 6 x 3. 5. xy (7 x 2 ). 7 x 3 y. 6. 3 y 2 (–3 y ). –9 y 3. Objectives.

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6-2

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  1. 6-2 Multiplying Polynomials Warm Up Lesson Presentation Lesson Quiz Holt Algebra 2

  2. Warm Up Multiply. 1. x(x3) x4 2. 3x2(x5) 3x7 3. 2(5x3) 10x3 4. x(6x2) 6x3 5. xy(7x2) 7x3y 6. 3y2(–3y) –9y3

  3. Objectives Multiply polynomials. Use binomial expansion to expand binomial expressions that are raised to positive integer powers.

  4. Example 1: Multiplying a Monomial and a Polynomial Find each product. A. 4y2(y2+ 3) 4y2(y2+ 3) 4y2  y2+ 4y2 3 Distribute. 4y4 + 12y2 Multiply. B. fg(f4 + 2f3g – 3f2g2 + fg3) fg(f4 + 2f3g – 3f2g2 + fg3) Distribute. fgf4 + fg 2f3g – fg  3f2g2 + fgfg3 f5g + 2f4g2 – 3f3g3 + f2g4 Multiply.

  5. Check It Out! Example 1 Find each product. a. 3cd2(4c2d– 6cd + 14cd2) 3cd2(4c2d– 6cd + 14cd2) 3cd2  4c2d– 3cd2 6cd + 3cd2 14cd2 Distribute. 12c3d3 – 18c2d3 + 42c2d4 Multiply. b. x2y(6y3 + y2 – 28y + 30) x2y(6y3 + y2 – 28y + 30) Distribute. x2y 6y3 + x2yy2 – x2y  28y + x2y 30 6x2y4 + x2y3 – 28x2y2 + 30x2y Multiply.

  6. To multiply any two polynomials, use the Distributive Property and multiply each term in the second polynomial by each term in the first. Keep in mind that if one polynomial has m terms and the other has n terms, then the product has mn terms before it is simplified.

  7. Example 2A: Multiplying Polynomials Find the product. (a – 3)(2 – 5a + a2) a3 – 8a2+ 17a – 6 Combine like terms.

  8. y2 –y–3 y2 –7y 5 Example 2B: Multiplying Polynomials Find the product. (y2 – 7y + 5)(y2 – y – 3) Multiply each term of one polynomial by each term of the other. Use a table to organize the products. The top left corner is the first term in the product. Combine terms along diagonals to get the middle terms. The bottom right corner is the last term in the product. y4+ (–7y3 – y3 ) + (5y2 + 7y2 – 3y2) + (–5y + 21y) – 15 y4 – 8y3 + 9y2 + 16y – 15

  9. Check It Out! Example 2a Find the product. (3b – 2c)(3b2 – bc – 2c2) 9b3 – 9b2c – 4bc2 + 4c3 Combine like terms.

  10. x2 –4x1 x2 5x –2 Check It Out! Example 2b Find the product. (x2 – 4x + 1)(x2 + 5x – 2) Multiply each term of one polynomial by each term of the other. Use a table to organize the products. The top left corner is the first term in the product. Combine terms along diagonals to get the middle terms. The bottom right corner is the last term in the product. x4+ (–4x3 + 5x3) + (–2x2 – 20x2 + x2) + (8x + 5x) – 2 x4 + x3 – 21x2 + 13x – 2

  11. Example 3: Business Application A standard Burly Box is p ft by 3p ft by 4p ft. A large Burly Box has 1.5 ft added to each dimension. Write a polynomial V(p) in standard form that can be used to find the volume of a large Burly Box. The volume of a large Burly Box is the product of the area of the base and height. V(p) = A(p)  h(p) The area of the base of the large Burly Box is the product of the length and width of the box. A(p) = l(p)  w(p) The length, width, and height of the large Burly Box are greater than that of the standard Burly Box. l(p) = p + 1.5, w(p) = 3p + 1.5, h(p) = 4p + 1.5

  12. Example 3: Business Application Solve A(p) = l(p)  w(p). Solve V(p) = A(p)  h(p). p + 1.5 3p2+ 6p + 2.25 3p+ 1.5 4p+ 1.5 1.5p+ 2.25 4.5p2 + 9p+ 3.375 3p2+ 4.5p 12p3+ 24p2 + 9p 3p2+ 6p + 2.25 12p3+ 28.5p2 + 18p + 3.375 The volume of a large Burly Box can be modeled byV(p) = 12p3+ 28.5p2 + 18p + 3.375

  13. Notice the coefficients of the variables in the final product of (a+ b)3. these coefficients are the numbers from the third row of Pascal's triangle. Each row of Pascal’s triangle gives the coefficients of the corresponding binomial expansion. The pattern in the table can be extended to apply to the expansion of any binomial of the form (a + b)n, where n is a whole number.

  14. This information is formalized by the Binomial Theorem, which you will study further in Chapter 11.

  15. Lesson Quiz Find each product. 1. 5jk(k – 2j) 2. (2a3– a + 3)(a2+ 3a – 5) 5jk2– 10j2k 2a5 + 6a4 – 11a3+ 14a – 15 3. The number of items is modeled by 0.3x2 + 0.1x + 2, and the cost per item is modeled by g(x) = –0.1x2 – 0.3x + 5. Write a polynomial c(x) that can be used to model the total cost. –0.03x4 – 0.1x3 + 1.27x2 – 0.1x + 10 4. Find the product. (y – 5)4 y4 – 20y3 + 150y2 – 500y + 625 5. Expand the expression. (3a – b)3 27a3 – 27a2b + 9ab2 – b3

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