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2-3 The Remainder and Factor Theorems

2-3 The Remainder and Factor Theorems Factor each polynomial completely using the given Factor and long division. 1. x3 + 2x2 23x 60; x + 4. SOLUTION: 3 2 2. So, x + 2x 23x 60 = (x + 4)(x 2x 15). 3 2. Factoring the quadratic expression yields x + 2x 23x 60 = (x + 4)(x 5)(x + 3). 2. x3 + 2x2 21x + 18; x 3. SOLUTION: 3 2 2. So, x + 2x 21x + 18 = (x 3)(x + 5x 6). 3 2. Factoring the quadratic expression yields x + 2x 21x + 18 = (x 3)(x + 6)(x 1). 3. x3 + 3x2 18x 40; x 4. SOLUTION: 3 2 2. So, x + 3x 18x 40 = (x 4)(x + 7x + 10). 3 2. Factoring the quadratic expression yields x + 3x 18x 40 = (x 4)(x + 2)(x + 5). eSolutions Manual - Powered by Cognero Page 1.

Use the Remainder Theorem to find the distance traveled after 45 seconds. 62/87,21 To find the distance traveled after 45 seconds, use synthetic substitution to evaluate d(t) for t = 45. The remainder is 540, so d(45) = 540. Therefore, 540 meters were traveled in 45 seconds.

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Transcription of 2-3 The Remainder and Factor Theorems

1 2-3 The Remainder and Factor Theorems Factor each polynomial completely using the given Factor and long division. 1. x3 + 2x2 23x 60; x + 4. SOLUTION: 3 2 2. So, x + 2x 23x 60 = (x + 4)(x 2x 15). 3 2. Factoring the quadratic expression yields x + 2x 23x 60 = (x + 4)(x 5)(x + 3). 2. x3 + 2x2 21x + 18; x 3. SOLUTION: 3 2 2. So, x + 2x 21x + 18 = (x 3)(x + 5x 6). 3 2. Factoring the quadratic expression yields x + 2x 21x + 18 = (x 3)(x + 6)(x 1). 3. x3 + 3x2 18x 40; x 4. SOLUTION: 3 2 2. So, x + 3x 18x 40 = (x 4)(x + 7x + 10). 3 2. Factoring the quadratic expression yields x + 3x 18x 40 = (x 4)(x + 2)(x + 5). eSolutions Manual - Powered by Cognero Page 1.

2 4. 4x3 + 20x2 8x 96; x + 3. SOLUTION: 3 2 2. So, x + 2x 21x + 18 = (x 3)(x + 5x 6). 2-3 The Remainder and Factor Theorems 3 2. Factoring the quadratic expression yields x + 2x 21x + 18 = (x 3)(x + 6)(x 1). 3. x3 + 3x2 18x 40; x 4. SOLUTION: 3 2 2. So, x + 3x 18x 40 = (x 4)(x + 7x + 10). 3 2. Factoring the quadratic expression yields x + 3x 18x 40 = (x 4)(x + 2)(x + 5). 4. 4x3 + 20x2 8x 96; x + 3. SOLUTION: 3 2 2. So, 4x + 20x 8x 96 = (x + 3)(4x + 8x 32). 3 2. Factoring the quadratic expression yields 4x + 20x 8x 96 = 4(x + 3)(x + 4)(x 2). 5. 3x3 + 15x2 + 108x 540; x 6. SOLUTION: 3 2 2. So, 3x + 15x + 108x 540 = (x 6)( 3x 3x + 90). 3 2. Factoring the quadratic expression yields 3x + 15x + 108x 540 = 3(x 6)(x + 6)(x 5).

3 6. 6x3 7x2 29x 12; 3x + 4. eSolutions Manual - Powered by Cognero Page 2. SOLUTION: 3 2 2. So, 3x + 15x + 108x 540 = (x 6)( 3x 3x + 90). 2-3 The Remainder and Factor Theorems 3 2. Factoring the quadratic expression yields 3x + 15x + 108x 540 = 3(x 6)(x + 6)(x 5). 6. 6x3 7x2 29x 12; 3x + 4. SOLUTION: 3 2 2. So, 6x 7x 29x 12 = (3x + 4)(2x 5x 3). 3 2. Factoring the quadratic expression yields 6x 7x 29x 12 = (3x + 4)(2x + 1)(x 3). 7. x4 + 12x3 + 38x2 + 12x 63; x2 + 6x + 9. SOLUTION: 4 3 2 2 2. So, x + 12x + 38x + 12x 63 = (x + 6x + 9)(x + 6x 7). 4 3 2 2. Factoring both quadratic expressions yield x + 12x + 38x + 12x 63 = (x + 3) (x + 7)(x 1).

4 8. x4 3x3 36x2 + 68x + 240; x2 4x 12. SOLUTION: 4 3 2 2 2. So, x 3x 36x + 68x + 240 = (x 4x 12)(x + x 20). 4 3 2. Factoring both quadratic expressions yield x + 12x + 38x + 12x 63 = (x 6)(x + 2)(x + 5)(x 4). Divide using long division. eSolutions4 Manual3- Powered 2 by Cognero Page 3. 9. (5x 3x + 6x x + 12) (x 4). SOLUTION: 4 3 2 2 2. So, x 3x 36x + 68x + 240 = (x 4x 12)(x + x 20). 2-3 The Remainder and Factor Theorems 4 3 2. Factoring both quadratic expressions yield x + 12x + 38x + 12x 63 = (x 6)(x + 2)(x + 5)(x 4). Divide using long division. 9. (5x4 3x3 + 6x2 x + 12) (x 4). SOLUTION: 3 2. So, = 5x + 17x + 74x + 295 + . 10. (x6 2x5 + x4 x3 + 3x2 x + 24) (x + 2).

5 SOLUTION: 5 4 3 2. So, = x 4x + 9x 19x + 41x 83 + . 11. (4x4 8x3 + 12x2 6x + 12) (2x + 4). SOLUTION: eSolutions Manual - Powered by Cognero Page 4. 5 4 3 2. 2-3 So, The Remainder and Factor = x Theorems 4x + 9x 19x + 41x 83 + . 11. (4x4 8x3 + 12x2 6x + 12) (2x + 4). SOLUTION: 3 2. The Remainder can be written as . So, = 2x 8x + 22x 47 + . 12. (2x4 7x3 38x2 + 103x + 60) (x 3). SOLUTION: 3 2. So, = 2x x 41x 20. 13. (6x6 3x5 + 6x4 15x3 + 2x2 + 10x 6) (2x 1). SOLUTION: eSolutions Manual - Powered by Cognero Page 5. 3 2. 2-3 So, The Remainder and Factor = 2x Theorems x 41x 20. 13. (6x6 3x5 + 6x4 15x3 + 2x2 + 10x 6) (2x 1). SOLUTION: So, = 14.

6 (108x5 36x4 + 75x2 + 36x + 24) (3x + 2). SOLUTION: 4 3 2. So, = 36x 36x + 24x + 9x + 6 + . 15. (x4 + x3 + 6x2 + 18x 216) (x3 3x2 + 18x 54). SOLUTION: eSolutions Manual - Powered by Cognero Page 6. 4 3 2. 2-3 So, The Remainder and Factor = 36x Theorems 36x + 24x + 9x + 6 + . 15. (x4 + x3 + 6x2 + 18x 216) (x3 3x2 + 18x 54). SOLUTION: So, = x + 4. 16. (4x4 14x3 14x2 + 110x 84) (2x2 + x 12). SOLUTION: 2. So, = 2x 8x + 9 + . 17. SOLUTION: 2. So, = 2x 4x + 2 + . 18. eSolutions Manual - Powered by Cognero SOLUTION: Page 7. 2. 2-3 So, The Remainder and Factor Theorems = 2x 4x + 2 + . 18. SOLUTION: The Remainder can be written as . So, = Divide using synthetic division.

7 19. (x4 x3 + 3x2 6x 6) (x 2). SOLUTION: Because x 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure. 3 2. The quotient is x + x + 5x + 4 + . 20. (2x4 + 4x3 2x2 + 8x 4) (x + 3). SOLUTION: Because x + 3, c = 3. Set up the synthetic division as follows. Then follow the synthetic division procedure. 3 2. The quotient is 2x 2x + 4x 4 + . 21. (3x4 9x3 24x 48) (x 4). SOLUTION: 2. Because x 4, c = 4. Set up the synthetic division as follows, using a zero placeholder for the missing x -term in the dividend. Then follow the synthetic division procedure. eSolutions Manual - Powered by Cognero Page 8. 3 2.

8 The quotient is 3x + 3x + 12x + 24 + . 3 2. 2-3 The Thequotient Remainder is 2x 2xand+ 4xFactor 4 + Theorems . 21. (3x4 9x3 24x 48) (x 4). SOLUTION: 2. Because x 4, c = 4. Set up the synthetic division as follows, using a zero placeholder for the missing x -term in the dividend. Then follow the synthetic division procedure. 3 2. The quotient is 3x + 3x + 12x + 24 + . 22. (x5 3x3 + 6x2 + 9x + 6) (x + 2). SOLUTION: 4. Because x + 2, c = 2. Set up the synthetic division as follows, using a zero placeholder for the missing x -term in the dividend. Then follow the synthetic division procedure. 4 3 2. The quotient is x 2x + x + 4x + 1 + . 23. (12x5 + 10x4 18x3 12x2 8) (2x 3).

9 SOLUTION: Rewrite the division expression so that the divisor is of the form x c. Because c = . Set up the synthetic division as follows, using a zero placeholder for the missing x-term in the dividend. Then follow the synthetic division procedure. 4 3 2. The Remainder can be written as . So, the quotient is 6x + 14x + 12x + 12x + 18 + . 24. (36x4 6x3 + 12x2 30x 12) (3x + 1). SOLUTION: Rewrite the division expression so that the divisor is of the form x c. eSolutions Manual - Powered by Cognero Page 9. 4 3 2. The Remainder can be written as . So, the quotient is 6x + 14x + 12x + 12x + 18 + . 2-3 The Remainder and Factor Theorems 24. (36x4 6x3 + 12x2 30x 12) (3x + 1).

10 SOLUTION: Rewrite the division expression so that the divisor is of the form x c. Because Set up the synthetic division as follows. Then follow the synthetic division procedure. 3 2. The quotient is 12x 6x + 6x 12. 25. (45x5 + 6x4 + 3x3 + 8x + 12) (3x 2). SOLUTION: Rewrite the division expression so that the divisor is of the form x c. 2. Because c = . Set up the synthetic division as follows, using a zero placeholder for the missing x -term in the dividend. Then follow the synthetic division procedure. 4 3 2. The Remainder can be written as . So, the quotient is 15x + 12x + 9x + 6x + + . 26. (48x5 + 28x4 + 68x3 + 11x + 6) (4x + 1). SOLUTION: Rewrite the division expression so that the divisor is of the form x c.

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