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Section 6.2

Section 6.2. Polynomials and Linear Factors. Polynomials and Linear Factors. ALGEBRA 2 LESSON 6-2. (For help, go to Lessons 5-1 and 5-4.). Factor each quadratic expression. 1. x 2 + 7 x + 12 2. x 2 + 8 x – 20 3. x 2 – 14 x + 24 Find each product.

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Section 6.2

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  1. Section 6.2 Polynomials and Linear Factors

  2. Polynomials and Linear Factors ALGEBRA 2 LESSON 6-2 (For help, go to Lessons 5-1 and 5-4.) Factor each quadratic expression. 1.x2 + 7x + 12 2.x2 + 8x – 20 3.x2 – 14x + 24 Find each product. 4.x(x + 4) 5. (x + 1)26. (x – 3)2(x + 2) Check Skills You’ll Need 6-2

  3. Polynomials and Linear Factors ALGEBRA 2 LESSON 6-2 Solutions 1. Factors of 12 with a sum of 7: 4 and 3 x2 + 7x + 12 = (x + 4)(x + 3) 2. Factors of –20 with a sum of 8: 10 and –2 x2 + 8x – 20 = (x + 10)(x – 2) 3. Factors of 24 with a sum of –14: –12 and –2 x2 – 14x + 24 = (x – 12)(x – 2) 4.x(x + 4) = x(x) + x(4) = x2 + 4x 5. (x + 1)2 = x2 + 2(1)x + 12 = x2 + 2x + 1 6. (x – 3)2(x + 2) = (x2 – 2(3)x + 32)(x + 2) = (x2 – 6x + 9)(x + 2) = (x2 – 6x + 9)(x) + (x2 – 6x + 9)(2) = (x3 – 6x2 + 9x) + (2x2 – 12x + 18) = x3 + (–6 + 2)x2 + (9 – 12)x + 18 = x3 – 4x2 – 3x + 18 6-2

  4. Polynomials and Linear Factors ALGEBRA 2 LESSON 6-2 Write (x – 1)(x + 3)(x + 4) as a polynomial in standard form. (x – 1)(x + 3)(x + 4) = (x – 1)(x2 + 4x + 3x + 12) Multiply (x + 3) and (x + 4). = (x – 1)(x2 + 7x + 12) Simplify. = x(x2 + 7x + 12) – 1(x2 + 7x + 12) Distributive Property = x3 + 7x2 + 12x – x2 – 7x – 12 Multiply. = x3 + 6x2 + 5x – 12 Simplify. The expression (x – 1)(x + 3)(x + 4) is the factored form of x3 + 6x2 + 5x – 12. Quick Check 6-2

  5. Polynomials and Linear Factors ALGEBRA 2 LESSON 6-2 Write 3x3 – 18x2 + 24x in factored form. 3x3 – 18x2 + 24x = 3x(x2 – 6x + 8) Factor out the GCF, 3x. = 3x(x – 4)(x – 2) Factor x2 – 6x + 8. Check: 3x(x – 4)(x – 2) = 3x(x2 – 6x + 8) Multiply (x – 4)(x – 2). = 3x3 – 18x2 + 24xDistributive Property Quick Check 6-2

  6. Relate: Volume = depth • length • width Define: Let x = depth. Then x + 10 = length, and 50 – (depth + length) = width. Write: V(x) = x ( x + 10 )( 50 – (x + x + 10) ) = x (x + 10)(40 – 2x) The x-intercepts of the function are x = 0, x = –10, x = 20. These values of x produce a volume of zero. Polynomials and Linear Factors ALGEBRA 2 LESSON 6-2 Another airline has different carry-on luggage regulations. The sum of the length, width, and depth may not exceed 50 in. a. Assume that the sum of the length, width, and depth is 50 in. and the length is 10 in. greater than the depth. Graph the function relating the volume v to depth x. Find the x-intercepts. What do they represent? Graph the function for volume. 6-2

  7. Polynomials and Linear Factors ALGEBRA 2 LESSON 6-2 (continued) b. Describe a realistic domain for V(x). The function has values over the set of all real numbers x. Since x represents the depth of the luggage, x > 0. Since the volume must be positive, x < 20. A realistic domain is 0 < x < 20. c. What is the maximum possible volume of the box? What are the corresponding dimensions of the box? Look for the greatest value of y that occurs within the domain 0 < x < 20. Use the Maximum feature of a graphing calculator to find the maximum volume. A volume of approximately 4225 in.3 occurs for a depth of about 12.2 in. Then the length is about 22.2 in. and the width is about 15.6 in. Quick Check 6-2

  8. Polynomials and Linear Factors ALGEBRA 2 LESSON 6-2 Find the zeros of y = (x + 1)(x – 1)(x + 3). Then graph the function using a graphing calculator. Using the Zero Product Property, find a zero for each linear factor. x + 1 = 0    or     x – 1 = 0    or    x + 3 = 0 x = –1 x = 1 x = –3 The zeros of the function are –1, 1, –3. Now sketch and label the function. Quick Check 6-2

  9. 2 –3 0Zeros ƒ(x) = (x – 2)(x + 3)(x) Write a linear factor for each zero. Polynomials and Linear Factors ALGEBRA 2 LESSON 6-2 Write a polynomial in standard form with zeros at 2, –3, and 0. = (x – 2)(x2 + 3x) Multiply (x + 3)(x). = x(x2 + 3x) – 2(x2 + 3x) Distributive Property = x3 + 3x2 – 2x2 – 6xMultiply. = x3 + x2 – 6xSimplify. The function ƒ(x) = x3 + x2 – 6x has zeros at 2, –3, and 0. Quick Check 6-2

  10. Polynomials and Linear Factors ALGEBRA 2 LESSON 6-2 Find any multiple zeros of ƒ(x) = x5 – 6x4 + 9x3 and state the multiplicity. ƒ(x) = x5 – 6x4 + 9x3 ƒ(x) = x3(x2 – 6x + 9) Factor out the GCF, x3. ƒ(x) = x3(x – 3)(x – 3) Factor x2 – 6x + 9. Since you can rewrite x3 as (x – 0)(x – 0)(x – 0), or (x – 0)3, the number 0 is a multiple zero of the function, with multiplicity 3. Since you can rewrite (x – 3)(x – 3) as (x – 3)2, the number 3 is a multiple zero of the function with multiplicity 2. Quick Check 6-2

  11. 400 27 x(x + 3)(x – 5); , –36 Polynomials and Linear Factors ALGEBRA 2 LESSON 6-2 1. Find the zeros of y = (x + 7)(x – 3)(x – 2). Then write the polynomial in standard form. 2. Factor x3 – 2x2 – 15x. Find the relative maximum and relative minimum. 3. Write a polynomial function in standard form with zeros at –5, –4, and 3. 4. Find any multiple zeros of ƒ(x) = x4 – 25x2 and state the multiplicity. –7, 2, 3; x3 + 2x2 – 29x + 42 Answers may vary. Sample: ƒ(x) = x3 + 6x2 – 7x – 60 The number 0 is a zero with multiplicity 2. 6-2

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