1. Chapter 5 Section 5

3. EXAMPLE 1 Use polynomial long division Divide f (x) = 3x 4 – 5x 3 + 4x – 6 by x 2 – 3x + 5. SOLUTION Write polynomial division in the same format you use when dividing numbers. Include a “0” as the coefficient of x 2 in the dividend. At each stage, divide the term with the highest power in what is left of the dividend by the first term of the divisor. This gives the next term of the quotient.

4. EXAMPLE 1 Use polynomial long division Multiply divisor by 3x 4 /x 2 = 3x 2 3x 4 – 9x 3 + 15x 2 4x 3 – 15x 2 + 4x Subtract. Bring down next term. Multiply divisor by 4x 3 /x 2 = 4x 4x 3 – 12x 2 + 20x –3x 2 – 16x – 6 Subtract. Bring down next term. Multiply divisor by – 3x 2 /x 2 = – 3 –3x 2 + 9x – 15 –25x + 9 remainder 3x 2 + 4x – 3 x 2 – 3x + 5 3x 4 – 5x 3 + 0x 2 + 4x – 6 ) quotient

5. EXAMPLE 1 Use polynomial long division You can check the result of a division problem by multiplying the quotient by the divisor and adding the remainder. The result should be the dividend. (3x 2 + 4x – 3)(x 2 – 3x + 5) + (–25x + 9) = 3x 2 (x 2 – 3x + 5) + 4x(x 2 – 3x + 5) – 3(x 2 – 3x + 5) – 25x + 9 CHECK = 3x 4 – 9x 3 + 15x 2 + 4x 3 – 12x 2 + 20x – 3x 2 + 9x – 15 – 25x + 9 = 3x 4 – 5x 3 + 4x – 6 3x 4 – 5x 3 + 4x – 6 x 2 – 3x + 5 = 3x 2 + 4x – 3 + –25x + 9 x 2 – 3x + 5 ANSWER

6. EXAMPLE 2 Use polynomial long division with a linear divisor Divide f(x) = x 3 + 5x 2 – 7x + 2 by x – 2. x 2 + 7x + 7 x – 2 x 3 + 5x 2 – 7x + 2 ) quotient x 3 – 2x 2 Multiply divisor by x 3 /x = x 2. 7x 2 – 7x Subtract. Multiply divisor by 7x 2 /x = 7x. 7x 2 – 14x 7x + 2 Subtract. Multiply divisor by 7x/x = 7. 16 remainder 7x – 14 ANSWER x 3 + 5x 2 – 7x +2 x – 2 = x 2 + 7x + 7 + 16 x – 2

7. GUIDED PRACTICE for Examples 1 and 2 Divide using polynomial long division. 1. (2x 4 + x 3 + x – 1) (x 2 + 2x – 1) (2x 2 – 3x + 8) + –18x + 7 x 2 + 2x – 1 ANSWER (x 2 – 3x + 10) + –30 x + 2 ANSWER 2. (x 3 – x 2 + 4x – 10)  (x + 2)

9. EXAMPLE 3 Use synthetic division Divide f (x)= 2x 3 + x 2 – 8x + 5 by x + 3 using synthetic division. –3 2 1 –8 5 –6 15 –21 2 –5 7 –16 2x 3 + x 2 – 8x + 5 x + 3 = 2x 2 – 5x + 7 – 16 x + 3 ANSWER SOLUTION

11. EXAMPLE 4 Factor a polynomial Factor f (x) = 3x 3 – 4x 2 – 28x – 16 completely given that x + 2 is a factor. SOLUTION Because x + 2 is a factor of f (x), you know that f (–2) = 0. Use synthetic division to find the other factors. –2 3 –4 –28 –16 –6 20 16 3 –10 –8 0

12. EXAMPLE 4 Factor a polynomial Use the result to write f (x) as a product of two factors and then factor completely. f (x) = 3x 3 – 4x 2 – 28x – 16 Write original polynomial. = (x + 2)(3x 2 – 10x – 8 ) Write as a product of two factors. = (x + 2)(3x + 2)(x – 4) Factor trinomial.

13. GUIDED PRACTICE for Examples 3 and 4 Divide using synthetic division. 3. (x 3 + 4x 2 – x – 1)  (x + 3) x 2 + x – 4 + 11 x + 3 ANSWER 4. (4x 3 + x 2 – 3x + 7)  (x – 1) 4x 2 + 5x + 2 + 9 x – 1 ANSWER

14. GUIDED PRACTICE for Examples 3 and 4 Factor the polynomial completely given that x – 4 is a factor. 5. f (x) = x 3 – 6x 2 + 5x + 12 (x – 4)(x –3)(x + 1) 6. f (x) = x 3 – x 2 – 22x + 40 (x – 4)(x –2)(x +5) ANSWER

15. 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

16. 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)

17. 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?

18. 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

19. 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

20. 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 3 and –3 8. f (x) = x 3 + 8x 2 + 5x – 14 1 and –7 The company could still make the same profit producing about 900,000 shoes. ANSWER 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 ? ANSWER