Warm-Up Exercises 1. Use the quadratic formula to solve 2x 2 – 3x – 1 = 0. Round the nearest hundredth. 2. Use synthetic substitution to evaluate f (x) = x 3 + x 2 – 3x – 10 when x = 2. ANSWER 1.78, –0.28 ANSWER –4–4

Warm-Up Exercises 3. A company’s income is modeled by the function P = 22x 2 – 571x. What is the value of P when x = 200 ? ANSWER 765,800

Warm-Up Exercises 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.

Warm-Up Exercises EXAMPLE 1 Use polynomial long division Multiply divisor by 3x 4 /x 2 = 3x 2 3x 4 – 9x x 2 4x 3 – 15x 2 + 4x Subtract. Bring down next term. Multiply divisor by 4x 3 /x 2 = 4x 4x 3 – 12x x – 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

Warm-Up Exercises 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 x 2 + 4x 3 – 12x x – 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

Warm-Up Exercises 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 = remainder 7x – 14 ANSWER x 3 + 5x 2 – 7x +2 x – 2 = x 2 + 7x x – 2

Warm-Up Exercises GUIDED PRACTICE for Examples 1 and 2 Divide using polynomial long division. 1. (2x 4 + x 3 + x – 1) (x 2 + 2x – 1) 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.

Warm-Up Exercises GUIDED PRACTICE for Examples 1 and 2 Multiply divisor by 2x 4 /x 2 = – 2x 2. 2x 4 – 4x 3 – 2x 2 3x 3 – 2x 2 + x Subtract. Bring down next term. Multiply divisor by –3x 3 /x 2 = –3. – 3x 3 – 6x 2 + 3x 8x 2 – 2x – 1 Subtract. Bring down next term. Multiply divisor by 4x 2 /x 2 = 8. 8x 2 –16x – 8 – 18x + 7 remainder 2x 2 – 3x + 8 x 2 + 2x – 1 2x 4 + x 3 + 0x 2 + x – 1 ) quotient

Warm-Up Exercises GUIDED PRACTICE for Examples 1 and 2 2x 4 + 5x 3 + x – 1 x 2 + 2x – 1 = (2x 2 – 3x + 8)+ – 18x + 7 x 2 + 2x – 1 ANSWER

Warm-Up Exercises GUIDED PRACTICE for Examples 1 and 2 2. (x 3 – x 2 + 4x – 10)  (x + 2) 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.

Warm-Up Exercises GUIDED PRACTICE for Examples 1 and 2 Multiply divisor by x 3 /x = x 2. x 3 + 2x 2 –3x 2 + 4x Subtract. Bring down next term. Multiply divisor by –3x 2 /x = –3x. – 3x 2 – 6x 10x – 1 Subtract. Bring down next term. Multiply divisor by 10x/x = x + 20 – 30 remainder x 2 – 3x + 10 x + 2 x 3 – x 2 + 4x – 10 ) quotient

Warm-Up Exercises GUIDED PRACTICE for Examples 1 and 2 x 3 – x 2 +4x – 10 x + 2 = (x 2 – 3x +10)+ – 30 x + 2 ANSWER

Warm-Up Exercises EXAMPLE 3 Use synthetic division Divide f (x)= 2x 3 + x 2 – 8x + 5 by x + 3 using synthetic division. – – 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

Warm-Up Exercises 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 – – 10 – 8 0

Warm-Up Exercises 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.

Warm-Up Exercises GUIDED PRACTICE for Examples 3 and 4 Divide using synthetic division. 3. (x 3 + 4x 2 – x – 1)  (x + 3) SOLUTION (x 3 + 4x 2 – x – 1)  (x + 3) – – 1 – 1 – 3 – – 4 11 x x 2 – x – 1 x + 3 = x 2 + x – x + 3 ANSWER

Warm-Up Exercises GUIDED PRACTICE for Examples 3 and 4 4. (4x 3 + x 2 – 3x + 7)  (x – 1) SOLUTION (4x 3 + x 2 – 3x + 7)  (x – 1) – x 3 + x 2 – 3x + 1 x – 1 = 4x 2 + 5x x – 1 ANSWER

Warm-Up Exercises 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 SOLUTION Because x – 4 is a factor of f (x), you know that f (4) = 0. Use synthetic division to find the other factors. 4 1 – – 8 –12 1 – 2 – 3 0

Warm-Up Exercises GUIDED PRACTICE for Examples 3 and 4 Use the result to write f (x) as a product of two factors and then factor completely. f (x) = x 3 – 6x 2 + 5x + 12 Write original polynomial. = (x – 4)(x 2 – 2x – 3 ) Write as a product of two factors. = (x – 4)(x –3)(x + 1) Factor trinomial.

Warm-Up Exercises GUIDED PRACTICE for Examples 3 and 4 6. f (x) = x 3 – x 2 – 22x + 40 SOLUTION Because x – 4 is a factor of f (x), you know that f (4) = 0. Use synthetic division to find the other factors – – – 10 0

Warm-Up Exercises GUIDED PRACTICE for Examples 3 and 4 Use the result to write f (x) as a product of two factors and then factor completely. f (x) = x 3 – x 2 – 22x + 40 Write original polynomial. = (x – 4)(x 2 + 3x – 10 ) Write as a product of two factors. = (x – 4)(x –2)(x +5) Factor trinomial.

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

Warm-Up Exercises 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)

Warm-Up Exercises EXAMPLE 6 Use a polynomial model BUSINESS The profit P (in millions of dollars ) for a shoe manufacturer can be modeled by P = – 21x x 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?

Warm-Up Exercises EXAMPLE 6 Use a polynomial model SOLUTION 25 = – 21x x Substitute 25 for P in P = – 21x x. 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 Use synthetic division to find the other factors – – – 25 0

Warm-Up Exercises EXAMPLE 6 Use a polynomial model So, (x – 1)(21x x – 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

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

Warm-Up Exercises 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 )

Warm-Up Exercises 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. – – 14 – 2 – – 7 0

Warm-Up Exercises 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 )

Warm-Up Exercises 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 x ? SOLUTION 25 = – 15x x Substitute 25 for P in P = – 15x x. 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 Use synthetic division to find the other factors – – – 25 0

Warm-Up Exercises GUIDED PRACTICE for Examples 5 and 6 So, (x – 1)(15x x – 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

Warm-Up Exercises Daily Homework Quiz 2. Use synthetic division to divide f(x) = x 3 – 3x 2 – 5x – 25 by x – Divide 6x 4 – x 3 – x x – 18 by 2x 2 + x – 3. ANSWER 3x 2 – 2x x 2 + x – 3 – 3 ANSWER x 2 + 2x + 5

Warm-Up Exercises Daily Homework Quiz ANSWER 5 About 4300 or about One zero of f(x) = x 3 – x 2 – 17x – 15 is x = – One of the costs to print a novel can be modeled by C = x 3 – 10x x, where x is the number of novels printed in thousands. The company now prints 5000 novels at a cost of $15,000. What other numbers of novels would cost about the same amount?