2-3 The Remainder and Factor Theorems
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1 Factor each polynomial completely using the given factor and long division. 1. x 3 + 2x 2 23x 60; x + 4 So, x 3 + 2x 2 23x 60 = (x + 4)(x 2 2x 15). Factoring the quadratic expression yields x 3 + 2x 2 23x 60 = (x + 4)(x 5)(x + 3). 3. x 3 + 3x 2 18x 40; x 4 So, x 3 + 3x 2 18x 40 = (x 4)(x 2 + 7x + 10). Factoring the quadratic expression yields x 3 + 3x 2 18x 40 = (x 4)(x + 2)(x + 5). esolutions Manual - Powered by Cognero Page 1
2 5. 3x x x 540; x 6 So, 3x x x 540 = (x 6)( 3x 2 3x + 90). Factoring the quadratic expression yields 3x x x 540 = 3(x 6)(x + 6)(x 5). 7. x x x x 63; x 2 + 6x + 9 So, x x x x 63 = (x 2 + 6x + 9)(x 2 + 6x 7). Factoring both quadratic expressions yield x x x x 63 = (x + 3) 2 (x + 7)(x 1). esolutions Manual - Powered by Cognero Page 2
3 Divide using long division. 9. (5x 4 3x 3 + 6x 2 x + 12) (x 4) 5x x x (4x 4 8x x 2 6x + 12) (2x + 4) The remainder can be written as. 2x 3 8x x 47 + esolutions Manual - Powered by Cognero Page 3
4 13. (6x 6 3x 5 + 6x 4 15x 3 + 2x x 6) (2x 1) 15. (x 4 + x 3 + 6x x 216) (x 3 3x x 54) x + 4 esolutions Manual - Powered by Cognero Page 4
5 17. 2x 2 4x Divide using synthetic division. 19. (x 4 x 3 + 3x 2 6x 6) (x 2) Because x 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure. The quotient is x 3 + x 2 + 5x (3x 4 9x 3 24x 48) (x 4) Because x 4, c = 4. Set up the synthetic division as follows, using a zero placeholder for the missing x 2 -term in the dividend. Then follow the synthetic division procedure. The quotient is 3x 3 + 3x x esolutions Manual - Powered by Cognero Page 5
6 23. (12x x 4 18x 3 12x 2 8) (2x 3) 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. The remainder can be written as. So, the quotient is 6x x x x esolutions Manual - Powered by Cognero Page 6
7 25. (45x 5 + 6x 4 + 3x 3 + 8x + 12) (3x 2) 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 2 -term in the dividend. Then follow the synthetic division procedure. The remainder can be written as. So, the quotient is 15x x 3 + 9x 2 + 6x + +. Find each f (c) using synthetic substitution. 31. f (x) = 4x 5 3x 4 + x 3 6x 2 + 8x 15; c = 3 The remainder is 711. Therefore, f (3) = f (x) = 2x 6 + 5x 5 3x 4 + 6x 3 9x 2 + 3x 4; c = 5 The remainder is 45,536. Therefore, f (5) = 45,536. esolutions Manual - Powered by Cognero Page 7
8 35. f (x) = 10x 5 + 6x 4 8x 3 + 7x 2 3x + 8; c = 6 The remainder is 67,978. Therefore, f ( 6) = 67, f (x) = 2x 8 + 6x 5 4x x 3 6x + 24; c = 4 The remainder is 125,184. Therefore, f (4) = 125,184. Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x). 39. f (x) = x 4 + 2x 3 5x 2 + 8x + 12; (x 1), (x + 3) Use synthetic division to test each factor, (x 1) and (x + 3). Because the remainder when f (x) is divided by (x 1) is 18, (x 1) is not a factor. Because the remainder when f (x) is divided by (x + 3) is 30, (x + 3) is not a factor. esolutions Manual - Powered by Cognero Page 8
9 41. f (x) = 3x 4 22x x x 40; (3x 1), (x 5) Use synthetic division to test each factor, (3x 1) and (x 5). For (3x 1), 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. Because the remainder when f (x) is divided by (3x 1) is 0, (3x 1) is a factor. Test the second factor, (x 5), with the depressed polynomial x 3 7x 2 + 2x Because the remainder when the depressed polynomial is divided by (x 5) is 0, (x 5) is a factor of f (x). Because (3x 1) and (x 5) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x) = (3x 1)(x 5)(x 2 2x 8). Factoring the quadratic expression yields f (x) = (3x 1)(x 5)(x 4)(x + 2). esolutions Manual - Powered by Cognero Page 9
10 43. f (x) = 3x 4 35x x x + 64; (3x 2), (x + 2) Use synthetic division to test each factor, (3x 2) and (x + 2). For (3x 2), 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. Because the remainder when f (x) is divided by (3x 2) is, (3x 2) is not a factor. Test (x + 2). Because the remainder when f (x) is divided by (x + 2) is 432, (x + 2) is not a factor. esolutions Manual - Powered by Cognero Page 10
11 45. f (x) = 4x 5 9x x x x + 63; (4x + 3), (x 1) Use synthetic division to test each factor, (4x + 3) and (x 1). For (4x + 3), 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. Because the remainder when f (x) is divided by (4x + 3) is 0, (4x + 3) is a factor. Test the second factor, (x 1), with the depressed polynomial x 4 3x x 2 3x Because the remainder when the depressed polynomial is divided by (x 1) is 28, (x 1) is not a factor of f (x). Because (4x + 3) is a factor of f (x), we can use the quotient of f (x) (4x + 3) to write a factored form of f (x) as f (x) = (4x + 3) (x 4 3x x 2 3x + 21). esolutions Manual - Powered by Cognero Page 11
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