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EXAMPLE 5 Standardized Test Practice SOLUTION Because f (3) = 0, x – 3 is a factor of f (x). Use synthetic division. 3 1 – 2 – 23 60 3 3 – 60 1 1 – 20 0

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EXAMPLE 5 Use the result to write f (x) as a product of two factors. Then factor completely. f (x) = x 3 – 2x 2 – 23x + 60 The zeros are 3, – 5, and 4. Standardized Test Practice The correct answer is A. ANSWER = (x – 3)(x + 5)(x – 4) = (x – 3)(x 2 + x – 20)

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EXAMPLE 6 Use a polynomial model BUSINESS The profit P (in millions of dollars ) for a shoe manufacturer can be modeled by P = – 21x 3 + 46x where x is the number of shoes produced (in millions). The company now produces 1 million shoes and makes a profit of $25,000,000, but would like to cut back production. What lesser number of shoes could the company produce and still make the same profit?

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EXAMPLE 6 Use a polynomial model SOLUTION 25 = – 21x 3 + 46x Substitute 25 for P in P = – 21x 3 + 46x. 0 = 21x 3 – 46x + 25 Write in standard form. You know that x = 1 is one solution of the equation. This implies that x – 1 is a factor of 21x 3 – 46x + 25. Use synthetic division to find the other factors. 1 21 0 – 46 25 21 21 –25 21 21 – 25 0

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EXAMPLE 6 Use a polynomial model So, (x – 1)(21x 2 + 21x – 25) = 0. Use the quadratic formula to find that x 0.7 is the other positive solution. The company could still make the same profit producing about 700,000 shoes. ANSWER

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GUIDED PRACTICE for Examples 5 and 6 Find the other zeros of f given that f (– 2) = 0. 7. f (x) = x 3 + 2x 2 – 9x – 18 SOLUTION Because f (– 2 ) = 0, x + 2 is a factor of f (x). Use synthetic division. – 2 1 2 – 9 – 18 – 2 0 18 1 0 – 9 0

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GUIDED PRACTICE for Examples 5 and 6 Use the result to write f (x) as a product of two factors. Then factor completely. f (x) = x 3 + 2x 2 – 9x – 18 The zeros are 3, – 3, and – 2. = (x + 2)(x + 3)(x – 3) = (x + 2)(x 2 – 9 2 )

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GUIDED PRACTICE for Examples 5 and 6 8. f (x) = x 3 + 8x 2 + 5x – 14 SOLUTION Because f (– 2 ) = 0, x + 2 is a factor of f (x). Use synthetic division. – 2 1 8 5 – 14 – 2 –12 14 1 6 – 7 0

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GUIDED PRACTICE for Examples 5 and 6 Use the result to write f (x) as a product of two factors. Then factor completely. f (x) = x 3 + 8x 2 + 5x – 14 The zeros are 1, – 7, and – 2. = (x + 2)(x + 7)(x – 1) = (x + 2)(x 2 + 6x – 7 )

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GUIDED PRACTICE for Examples 5 and 6 9. What if? In Example 6, how does the answer change if the profit for the shoe manufacturer is modeled by P = – 15x 3 + 40x ? SOLUTION 25 = – 15x 3 + 40x Substitute 25 for P in P = – 15x 3 + 40x. 0 = 15x 3 – 40x + 25 Write in standard form. You know that x = 1 is one solution of the equation. This implies that x – 1 is a factor of 15x 3 – 40x + 25. Use synthetic division to find the other factors. 1 15 0 – 40 25 15 15 –25 15 15 – 25 0

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GUIDED PRACTICE for Examples 5 and 6 So, (x – 1)(15x 2 + 15x – 25) = 0. Use the quadratic formula to find that x 0.9 is the other positive solution. The company could still make the same profit producing about 900,000 shoes. ANSWER

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