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2-4 Zeros of Polynomial Functions List all possible rational zeros of each function. Then determine which, if any, are zeros. 1. g(x) = x4 – 6x3 – 31x2 + 216x − 180 SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −180. Therefore, the possible rational zeros of g are . By using synthetic division, it can be determined that x = 1 is a rational zero. By using synthetic division on the depressed polynomial, it can be determined that x = 5 is a rational zero. Because (x − 1) and (x − 5) are factors of g(x), we can use the final quotient to write a factored form of g(x) as g 2 (x) = (x − 1)(x − 5)(x − 36). Factoring the quadratic expression yields f (x) = (x – 1)(x – 5)(x – 6)(x + 6). Thus, the rational zeros of g are 1, 5, 6, and −6. 3. g(x) = x4 – x3 – 31x2 + x + 30 SOLUTION: Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 30. Therefore, the possible rational zeros of g are ±1, ±2, ±3, ±5, ±6, ±10, ±15, and ±30. By using synthetic division, it can be determined that 1 is a rational zero. By using synthetic division on the depressed polynomial, it can be determined that 6 is a rational zero. Because (x − 1) and (x − 6) are factors of g(x), we can use the final quotient to write a factored form of g(x) as g 2 (x) = (x − 1)(x − 6)(x + 6x + 5). Factoring the quadratic expression yields f (x) = (x – 1)(x – 6)(x + 5)(x + 1). Thus, the rational zeros of g are 1, 6, −5, and −1. 5. h(x) = 6x4 + 13x3 – 67x2 – 156x – 60 SOLUTION: The leading coefficient is 6 and the constant term is −60. The possible rational zeros are eSolutions Manual - Powered by Cognero Page 1 = ±1, ±2, ±3, ±4, ±5, ±6, ±10, ±12, ±15, ±20, ±30, ±60, ± , ± , ± , ± ,± ,± ,± ,± ,± ,± , ± , and ± . 2-4 Because (x − 1) and (x − 6) are factors of g(x), we can use the final quotient to write a factored form of g(x) as g 2 (x) = (x −of1)(x 5). Factoring the quadratic expression yields f (x) = (x – 1)(x – 6)(x + 5)(x + 1). Thus, − 6)(x + 6x +Functions Zeros Polynomial the rational zeros of g are 1, 6, −5, and −1. 5. h(x) = 6x4 + 13x3 – 67x2 – 156x – 60 SOLUTION: The leading coefficient is 6 and the constant term is −60. The possible rational zeros are = ±1, ±2, ±3, ±4, ±5, ±6, ±10, ±12, ±15, ±20, ±30, ±60, ± , ± , ± , ± By using synthetic division, it can be determined that ,± ,± ,± ,± ,± and , ± , and ± . is a rational zero. By using synthetic division on the depressed polynomial, it can be determined that Because ,± is a rational zero. are factors of h(x), we can use the final quotient to write a factored form of h(x) as Factoring the quadratic expression yields 2 Because the factor (x − 12) yields no rational zeros, the rational zeros of h are Solve each equation. 11. x4 + 9x3 + 23x2 + 3x − 36 = 0 SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −36. Therefore, the possible rational zeros are . By using synthetic division, it can be determined that 1 is a rational zero. By using synthetic division on the depressed polynomial, it can be determined that −4 is a rational zero. eSolutions Manual - Powered by Cognero Page 2 Because (x − 1) and (x + 4) are factors of the equation, we can use the final quotient to write a factored form as (x 2 2 − 1)(x + 4)(x + 6x + 9) = 0. Factoring the quadratic expression yields (x − 1)(x + 4)(x + 3) = 0. Thus, the solutions Factoring the quadratic expression yields 2 2-4 Because Zeros of thePolynomial factor (x − 12)Functions yields no rational zeros, the rational zeros of h are Solve each equation. 11. x4 + 9x3 + 23x2 + 3x − 36 = 0 SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −36. Therefore, the possible rational zeros are . By using synthetic division, it can be determined that 1 is a rational zero. By using synthetic division on the depressed polynomial, it can be determined that −4 is a rational zero. Because (x − 1) and (x + 4) are factors of the equation, we can use the final quotient to write a factored form as (x 2 2 − 1)(x + 4)(x + 6x + 9) = 0. Factoring the quadratic expression yields (x − 1)(x + 4)(x + 3) = 0. Thus, the solutions are 1, −4, and −3 (multiplicity: 2). 13. x4 – 3x3 – 20x2 + 84x – 80 = 0 SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −80. Therefore, the possible rational zeros are ±1, ±2, ±4, ±5, ±8, ±10, ±16, ±20, ±40, and ±80. By using synthetic division, it can be determined that 4 is a rational zero. By using synthetic division on the depressed polynomial, it can be determined that −5 is a rational zero. Because (x − 4) and (x + 5) are factors of the equation, we can use the final quotient to write a factored form as (x 2 2 − 4)(x + 5)(x − 4x + 4) = 0. Factoring the quadratic expression yields (x − 4)(x + 5)(x − 2) = 0. Thus, the solutions are 4, −5, and 2 (multiplicity: 2). 15. 6x4 + 41x3 + 42x2 – 96x + 6 = –26 SOLUTION: 4 3 2 The equation can be written as 6x + 41x + 42x − 96x + 32 = 0. Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the constant term is 32. The possible rational eSolutions Manual - Powered by Cognero zeros are Page 3 = ±1, ±2, ±4, ±8, ±16, ±32, ± , ± , ± , ± ,± , ± ,± , and ± . Because (x − 4) and (x + 5) are factors of the equation, we can use the final quotient to write a factored form as (x 2 2 4)(x + 5)(x Factoring the quadratic expression yields (x − 4)(x + 5)(x − 2) − 4x + 4) = 0. Functions 2-4 −Zeros of Polynomial = 0. Thus, the solutions are 4, −5, and 2 (multiplicity: 2). 15. 6x4 + 41x3 + 42x2 – 96x + 6 = –26 SOLUTION: 4 3 2 The equation can be written as 6x + 41x + 42x − 96x + 32 = 0. Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the constant term is 32. The possible rational = ±1, ±2, ±4, ±8, ±16, ±32, ± , ± , ± , ± ,± , ± zeros are ,± , and ± . By using synthetic division, it can be determined that is a rational zero. By using synthetic division on the depressed polynomial, it can be determined that Because and is a rational zero. are factors of the equation, we can use the final quotient to write a factored form as . Factoring the quadratic expression yields .Thus, the solutions are , , and −4 (multiplicity: 2). 17. SALES The sales S(x) in thousands of dollars that a store makes during one month can be approximated by S(x) = 3 2 2x – 2x + 4x, where x is the number of days after the first day of the month. How many days will it take the store to make $16,000? SOLUTION: 3 2 Substitute S(x) = 16 into S(x) = 2x – 2x + 4x and apply the Rational Zeros Theorem to find possible rational zeros of the function. 3 2 The equation can be written as 2(x − x + 2x − 8) = 0. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −8. Therefore, the possible rational zeros are ±1, ±2, ±4, and ±8. By using synthetic division, it can be determined that x = 2 is a rational zero. eSolutions Manual - Powered by Cognero 2 The depressed polynomial x + x + 4 has no real zeros. Thus, x = 2. The store will make $16,000 in 2 days. Page 4 Factoring the quadratic expression yields 2-4 Zeros of Polynomial Functions .Thus, the solutions are , , and −4 (multiplicity: 2). 17. SALES The sales S(x) in thousands of dollars that a store makes during one month can be approximated by S(x) = 3 2 2x – 2x + 4x, where x is the number of days after the first day of the month. How many days will it take the store to make $16,000? SOLUTION: 3 2 Substitute S(x) = 16 into S(x) = 2x – 2x + 4x and apply the Rational Zeros Theorem to find possible rational zeros of the function. 3 2 The equation can be written as 2(x − x + 2x − 8) = 0. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −8. Therefore, the possible rational zeros are ±1, ±2, ±4, and ±8. By using synthetic division, it can be determined that x = 2 is a rational zero. 2 The depressed polynomial x + x + 4 has no real zeros. Thus, x = 2. The store will make $16,000 in 2 days. Determine an interval in which all real zeros of each function must lie. Explain your reasoning using the upper and lower bound tests. Then find all the real zeros. 19. f (x) = 2x4 – x3 – 29x2 + 34x + 24 SOLUTION: Graph f (x) using a graphing calculator. From this graph, it appears that the real zeros of this function lie in the interval [−6, 5]. Test a lower bound of c = −6 and an upper bound of c = 5. Every number in the last line is alternately nonnegative and nonpositive, so −6 is a lower bound. Every number in the last line is nonnegative, so 5 is an upper bound. Manual - Powered by Cognero eSolutions Page 5 Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 2 and the constant term is 24. 2-4 Every Zerosnumber of Polynomial in the last line Functions is nonnegative, so 5 is an upper bound. Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 2 and the constant term is 24. or ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24, The possible rational zeros are , and . Because the real zeros are in the interval [−6, 5], narrow this list to just ±1, ±2, ±3, ±4, −6, , and . From the are reasonable. graph, it appears that By using synthetic division, it can be determined that −4 is a rational zero. By using synthetic division on the depressed polynomial, it can be determined that Because (x + 4) and is a rational zero. are factors of the equation, we can use the final quotient to write a factored form as . Factoring the quadratic expression yields . Thus, the solutions are 21. g(x) = 6x4 – 33x3 – 6x2 + 123x – 90 SOLUTION: Graph g(x) using a graphing calculator. From this graph, it appears that the real zeros of this function lie in the interval [−4, 7]. Test a lower bound of c = −4 and an upper bound of c = 7. eSolutions Manual - Powered by Cognero Page 6 2-4 Zeros of Polynomial Functions Test a lower bound of c = −4 and an upper bound of c = 7. The numbers in the last line alternate signs, , so −4 is a lower bound. Every number in the last line is nonnegative, so 7 is an upper bound. Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the constant term is 90. or ±1,±2, ±3, ±5, ±6, ±9, ±10, ±15, The possible rational zeros are ±18, ±30, ±45, ±90, ± , ± , ± , ± , ± ,± ,± ,± ,± ,± , ± , and ± . Because the real zeros are in the interval [−4, 7], narrow this list to just ±1,±2, ±3, 5, 6, ± , ± , ± , ± , ± , ± , ± , ± , ± , and ± . From the graph, it appears that −2, 1, , and 5 are reasonable. By using synthetic division, it can be determined that −2 is a rational zero. By using synthetic division on the depressed polynomial, it can be determined that 1 is a rational zero. Because (x + 2) and (x − 1) are factors of the equation, we can use the final quotient to write a factored form as 2 g(x) = (x + 2)(x − 1)(6x − 39x + 45). Factoring the quadratic expression yields g(x) = 3(x + 2)(x − 1)(2x − 3)(x − 5). Thus, the solutions are −2, 1, , and 5. 23. f (x) = 2x4 – 13x3 + 21x2 + 9x – 27 SOLUTION: Graph f (x) using a graphing calculator. From this graph, it appears that the real zeros of this function lie in the eSolutions Manual - Powered interval [−2, 7]. by Cognero Page 7 5. 4 3 2 23. fZeros (x) = 2xof– Polynomial 13x + 21x + 9xFunctions – 27 2-4 SOLUTION: Graph f (x) using a graphing calculator. From this graph, it appears that the real zeros of this function lie in the interval [−2, 7]. Test a lower bound of c = −2 and an upper bound of c = 7. The numbers in the last line alternate signs, so −2 is a lower bound. Every number in the last line is nonnegative, so 7 is an upper bound. Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 2 and the constant term is −27. The possible rational zeros are = ±1, ±3, ±9, ±27, ± ,± ,± and ± . Because the real zeros are in the interval [−2, 7], narrow this list to just ±1, 3, ± From the graph, it appears that −1 and ,± , and . are reasonable. By using synthetic division, it can be determined that −1 is a rational zero. By using synthetic division on the depressed polynomial, it can be determined that eSolutions Manual - Powered by Cognero Because (x + 1) and is a rational zero. Page 8 are factors of the equation, we can use the final quotient to write a factored form as 2-4 Zeros of Polynomial Functions Because (x + 1) and are factors of the equation, we can use the final quotient to write a factored form as . . Thus, the solutions are −1, Factoring the quadratic expression yields , and 3 (multiplicity: 2). 25. h(x) = 4x5 – 20x4 + 5x3 + 80x2 – 75x + 18 SOLUTION: Graph h(x) using a graphing calculator. From this graph, it appears that the real zeros of this function lie in the interval [−3, 5]. Test a lower bound of c = −3 and an upper bound of c = 5. The numbers in the last line alternate signs, so −3 is a lower bound. Every number in the last line is nonnegative, so 5 is an upper bound. Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 4 and the constant term is 18. The possible rational zeros are or ±1, ±3, ±6, ±9, ±18, ± ,± ,± ,± ,± , and ± . Because the real zeros are in the interval [−3, 5], narrow this list to just ±1, ±3, ± From the graph, it appears that −2, ,± , ,± ,± , and ± . and 3 are reasonable. By using synthetic division, it can be determined that −2 is a rational zero. eSolutions Manual - Powered by Cognero Page 9 Because the real zeros are in the interval [−3, 5], narrow this list to just ±1, ±3, ± ,± , ,± ,± , and ± . theof graph, it appears that −2, and 3 are reasonable. 2-4 From Zeros Polynomial Functions By using synthetic division, it can be determined that −2 is a rational zero. By using synthetic division on the depressed polynomial, it can be determined that is a rational zero. By using synthetic division on the new depressed polynomial, it can be determined that 3 is a rational zero. Because form as are factors of the equation, we can use the final quotient to write a factored . Factoring the quadratic expression yields , which simplifies to Thus, the solutions are −2, eSolutions Manual - Powered by Cognero . (multiplicity: 2), and 3 (multiplicity: 2). Page 10