EXAMPLE 5 Standardized Test Practice SOLUTION

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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

EXAMPLE 5 Standardized Test Practice Use the result to write f (x) as a product of two factors. Then factor completely. f (x) = x3 – 2x2 – 23x + 60 = (x – 3)(x2 + x – 20) = (x – 3)(x + 5)(x – 4) The zeros are 3, –5, and 4. The correct answer is A. ANSWER

EXAMPLE 6 Use a polynomial model BUSINESS The profit P (in millions of dollars) for a shoe manufacturer can be modeled by P = –21x3 + 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?

EXAMPLE 6 Use a polynomial model SOLUTION 25 = –21x3 + 46x Substitute 25 for P in P = –21x3 + 46x. 0 = 21x3 – 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 21x3 – 46x + 25. Use synthetic division to find the other factors. 1 21 0 –46 25 21 21 –25 21 21 –25 0

EXAMPLE 6 Use a polynomial model So, (x – 1)(21x2 + 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

GUIDED PRACTICE for Examples 5 and 6 Find the other zeros of f given that f (–2) = 0. 7. f (x) = x3 + 2x2 – 9x – 18 8. f (x) = x3 + 8x2 + 5x – 14 ANSWER 3 and –3 ANSWER 1 and –7 9. What if? In Example 6, how does the answer change if the profit for the shoe manufacturer is modeled by P = –15x3 + 40x? ANSWER The company could still make the same profit producing about 900,000 shoes.