2-4 Zeros of Polynomial Functions

Transcription

1 List all possible rational zeros of each function Then determine which, if any, are zeros 1 g(x) = x 4 6x 3 31x x 180 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 By using synthetic division on the depressed polynomial, it can be determined that x = 5 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 (x) = (x 1)(x 5)(x 2 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) = x 4 x 3 31x 2 + x + 30 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 By using synthetic division, it can be determined that x = 1 By using synthetic division on the depressed polynomial, it can be determined that x = 6 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 (x) = (x 1)(x 6)(x 2 +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 esolutions Manual - Powered by Cognero Page 1

2 5 h(x) = 6x x 3 67x 2 156x 60 The leading coefficient is 6 and the constant term is 60 The possible rational zeros are or By using synthetic division, it can be determined that By using synthetic division on the depressed polynomial, it can be determined that Because and 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 Because the factor (x 2 12) yields no rational zeros, the rational zeros of h are esolutions Manual - Powered by Cognero Page 2

3 7 h(x) = x 5 11x x 3 147x x 432 Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 432 Therefore, the possible rational zeros of g are By using synthetic division, it can be determined that x = 3 By using synthetic division on the depressed polynomial, it can be determined that x = 4 By using synthetic division on the new depressed polynomial, it can be determined that x = 4 is a repeated rational zero Because (x 3) and (x 4) 2 are factors of h(x), we can use the final quotient to write a factored form of h(x) as h (x) = (x 3)(x 4) 2 (x 2 + 9) Because the factor (x 2 + 9) yields no real zeros, the rational zeros of h are 3 and 4 (multiplicity: 2) esolutions Manual - Powered by Cognero Page 3

4 9 MANUFACTURING The specifications for the dimensions of a new cardboard container are shown If the volume of the container is modeled by V(h) = 2h 3 9h 2 + 4h and it will hold 45 cubic inches of merchandise, what are the container's dimensions? Substitute V(h) = 45 into V(h) = 2h 3 9h 2 + 4h and apply the Rational Zeros Theorem to find possible rational zeros of the function The leading coefficient is 2 and the constant term is 45 The possible rational zeros are or By using synthetic division, it can be determined that h = 5 The depressed polynomial 2x 2 + x + 9 has no real zeros Thus, h = 5 The dimensions of the container are 5, 5 4 or 1, and 2(5) 1 or 9 Solve each equation 11 x 4 + 9x x 2 + 3x 36 = 0 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 x = 1 By using synthetic division on the depressed polynomial, it can be determined that x = 4 Because (x 1) and (x + 4) are factors of the equation, we can use the final quotient to write a factored form as 0 = (x 1)(x + 4)(x 2 + 9x + 9) Factoring the quadratic expression yields 0 = (x 1)(x + 4)(x + 3) 2 Thus, the solutions are 1, 4, and 3 (multiplicity: 2) esolutions Manual - Powered by Cognero Page 4

5 13 x 4 3x 3 20x x 80 = 0 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 By using synthetic division, it can be determined that x = 4 By using synthetic division on the depressed polynomial, it can be determined that x = 5 Because (x 4) and (x + 5) are factors of the equation, we can use the final quotient to write a factored form as 0 = (x 4)(x + 5)(x 2 4x + 4) Factoring the quadratic expression yields 0 = (x 4)(x + 5)(x 2) 2 Thus, the solutions are 4, 5, and 2 (multiplicity: 2) esolutions Manual - Powered by Cognero Page 5

6 15 6x x x 2 96x + 6 = 26 The equation can be written as 6x x x 2 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 zeros are or By using synthetic division, it can be determined that x = By using synthetic division on the depressed polynomial, it can be determined that x = Because and 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) esolutions Manual - Powered by Cognero Page 6

7 17 SALES The sales S(x) in thousands of dollars that a store makes during one month can be approximated by S(x) = 2x 3 2x 2 + 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? Substitute S(x) = 16 into S(x) = 2x 3 2x 2 + 4x and apply the Rational Zeros Theorem to find possible rational zeros of the function The equation can be written as 2(x 3 x 2 + 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 The depressed polynomial x 2 + x + 4 has no real zeros Thus, x = 2 The store will make $16,000 in 2 days esolutions Manual - Powered by Cognero Page 7