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

To multiply a polynomial by a monomial, use the Distributive Property and the Properties of Exponents.

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) fg  f4 + fg  2f3g – fg  3f2g2 + fg  fg3 Distribute. f5g + 2f4g2 – 3f3g3 + f2g4 Multiply.

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) x2y  6y3 + x2y  y2 – x2y  28y + x2y  30 Distribute. 6x2y4 + x2y3 – 28x2y2 + 30x2y Multiply.

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.

Example 2A: Multiplying Polynomials Find the product. (a – 3)(2 – 5a + a2) Method 1 Multiply horizontally. (a – 3)(a2 – 5a + 2) Write polynomials in standard form. Distribute a and then –3. a(a2) + a(–5a) + a(2) – 3(a2) – 3(–5a) –3(2) a3 – 5a2 + 2a – 3a2 + 15a – 6 Multiply. Add exponents. a3 – 8a2 + 17a – 6 Combine like terms.

Example 2A: Multiplying Polynomials Find the product. (a – 3)(2 – 5a + a2) Method 2 Multiply vertically. a2 – 5a + 2 a – 3 Write each polynomial in standard form. – 3a2 + 15a – 6 Multiply (a2 – 5a + 2) by –3. a3 – 5a2 + 2a Multiply (a2 – 5a + 2) by a, and align like terms. a3 – 8a2 + 17a – 6 Combine like terms.

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. y2 –y –3 y2 –7y 5 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 –y3 –3y2 –7y3 7y2 21y 5y2 –5y –15 y4 + (–7y3 – y3 ) + (5y2 + 7y2 – 3y2) + (–5y + 21y) – 15 y4 – 8y3 + 9y2 + 16y – 15

Multiply horizontally. Write polynomials in standard form. Check It Out! Example 2a Find the product. (3b – 2c)(3b2 – bc – 2c2) Multiply horizontally. Write polynomials in standard form. (3b – 2c)(3b2 – 2c2 – bc) Distribute 3b and then –2c. 3b(3b2) + 3b(–2c2) + 3b(–bc) – 2c(3b2) – 2c(–2c2) – 2c(–bc) Multiply. Add exponents. 9b3 – 6bc2 – 3b2c – 6b2c + 4c3 + 2bc2 9b3 – 9b2c – 4bc2 + 4c3 Combine like terms.

x4 –4x3 x2 5x3 5x –2x2 8x –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. x2 –4x 1 x2 5x –2 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 x2 5x3 –20x2 5x –2x2 8x –2 x4 + (–4x3 + 5x3) + (–2x2 – 20x2 + x2) + (8x + 5x) – 2 x4 + x3 – 21x2 + 13x – 2

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

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 by V(p) = 12p3 + 28.5p2 + 18p + 3.375

Check It Out! Example 3 Mr. Silva manages a manufacturing plant. From 1990 through 2005 the number of units produced (in thousands) can be modeled by N(x) = 0.02x2 + 0.2x + 3. The average cost per unit (in dollars) can be modeled by C(x) = –0.004x2 – 0.1x + 3. Write a polynomial T(x) that can be used to model the total costs. Total cost is the product of the number of units and the cost per unit. T(x) = N(x)  C(x)

Check It Out! Example 3 Multiply the two polynomials. 0.02x2 + 0.2x + 3  –0.004x2 – 0.1x + 3 0.06x2 + 0.6x + 9 –0.002x3 – 0.02x2 – 0.3x –0.00008x4 – 0.0008x3 – 0.012x2 –0.00008x4 – 0.0028x3 + 0.028x2 + 0.3x + 9 Mr. Silva’s total manufacturing costs, in thousands of dollars, can be modeled by T(x) = –0.00008x4 – 0.0028x3 + 0.028x2 + 0.3x + 9

Example 4: Expanding a Power of a Binomial Find the product. (a + 2b)3 (a + 2b)(a + 2b)(a + 2b) Write in expanded form. (a + 2b)(a2 + 4ab + 4b2) Multiply the last two binomial factors. Distribute a and then 2b. a(a2) + a(4ab) + a(4b2) + 2b(a2) + 2b(4ab) + 2b(4b2) a3 + 4a2b + 4ab2 + 2a2b + 8ab2 + 8b3 Multiply. a3 + 6a2b + 12ab2 + 8b3 Combine like terms.

Multiply the last two binomial factors. Check It Out! Example 4a Find the product. (x + 4)4 (x + 4)(x + 4)(x + 4)(x + 4) Write in expanded form. (x + 4)(x + 4)(x2 + 8x + 16) Multiply the last two binomial factors. (x2 + 8x + 16)(x2 + 8x + 16) Multiply the first two binomial factors. Distribute x2 and then 8x and then 16. x2(x2) + x2(8x) + x2(16) + 8x(x2) + 8x(8x) + 8x(16) + 16(x2) + 16(8x) + 16(16) Multiply. x4 + 8x3 + 16x2 + 8x3 + 64x2 + 128x + 16x2 + 128x + 256 x4 + 16x3 + 96x2 + 256x + 256 Combine like terms.

Check It Out! Example 4b Find the product. (2x – 1)3 (2x – 1)(2x – 1)(2x – 1) Write in expanded form. (2x – 1)(4x2 – 4x + 1) Multiply the last two binomial factors. Distribute 2x and then –1. 2x(4x2) + 2x(–4x) + 2x(1) – 1(4x2) – 1(–4x) – 1(1) 8x3 – 8x2 + 2x – 4x2 + 4x – 1 Multiply. 8x3 – 12x2 + 6x – 1 Combine like terms.

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.

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

Example 5: Using Pascal’s Triangle to Expand Binomial Expressions Expand each expression. A. (k – 5)3 1 3 3 1 Identify the coefficients for n = 3, or row 3. [1(k)3(–5)0] + [3(k)2(–5)1] + [3(k)1(–5)2] + [1(k)0(–5)3] k3 – 15k2 + 75k – 125 B. (6m – 8)3 1 3 3 1 Identify the coefficients for n = 3, or row 3. [1(6m)3(–8)0] + [3(6m)2(–8)1] + [3(6m)1(–8)2] + [1(6m)0(–8)3] 216m3 – 864m2 + 1152m – 512

Identify the coefficients for n = 3, or row 3. Check It Out! Example 5 Expand each expression. a. (x + 2)3 1 3 3 1 Identify the coefficients for n = 3, or row 3. [1(x)3(2)0] + [3(x)2(2)1] + [3(x)1(2)2] + [1(x)0(2)3] x3 + 6x2 + 12x + 8 b. (x – 4)5 1 5 10 10 5 1 Identify the coefficients for n = 5, or row 5. [1(x)5(–4)0] + [5(x)4(–4)1] + [10(x)3(–4)2] + [10(x)2(–4)3] + [5(x)1(–4)4] + [1(x)0(–4)5] x5 – 20x4 + 160x3 – 640x2 + 1280x – 1024

Identify the coefficients for n = 4, or row 4. Check It Out! Example 5 Expand the expression. c. (3x + 1)4 1 4 6 4 1 Identify the coefficients for n = 4, or row 4. [1(3x)4(1)0] + [4(3x)3(1)1] + [6(3x)2(1)2] + [4(3x)1(1)3] + [1(3x)0(1)4] 81x4 + 108x3 + 54x2 + 12x + 1

3. The number of items is modeled by Lesson Quiz Find each product. 1. 5jk(k – 2j) 5jk2 – 10j2k 2. (2a3 – a + 3)(a2 + 3a – 5) 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