Warm-Up Exercises
1. Use the quadratic formula to solve 2x2 – 3x – 1 = 0. Round the nearest hundredth.
2. Use synthetic substitution to evaluate f (x) = x3 + x2 – 3x – 10 when x = 2.
ANSWER 1.78, –0.28
ANSWER –4
Warm-Up Exercises
3. A company’s income is modeled by the function P = 22x2 – 571x. What is the value of P when x = 200?
ANSWER 765,800
Warm-Up ExercisesEXAMPLE 1 Use polynomial long division
Divide f (x) = 3x4 – 5x3 + 4x – 6 by x2 – 3x + 5.
SOLUTION
Write polynomial division in the same format you use when dividing numbers. Include a “0” as the coefficient of x2 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 ExercisesEXAMPLE 1 Use polynomial long division
Multiply divisor by 3x4/x2 = 3x2
3x4 – 9x3 + 15x2
4x3 – 15x2 + 4x Subtract. Bring down next term.
Multiply divisor by 4x3/x2 = 4x4x3 – 12x2 + 20x
– 3x2 – 16x – 6 Subtract. Bring down next term.
Multiply divisor by – 3x2/x2 = – 3–3x2 + 9x – 15
– 25x + 9 remainder
3x2 + 4x – 3x2 – 3x + 5 3x4 – 5x3 + 0x2 + 4x – 6)
quotient
Warm-Up ExercisesEXAMPLE 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.
(3x2 + 4x – 3)(x2 – 3x + 5) + (– 25x + 9)
= 3x2(x2 – 3x + 5) + 4x(x2 – 3x + 5) – 3(x2 – 3x + 5) – 25x + 9
CHECK
= 3x4 – 9x3 + 15x2 + 4x3 – 12x2 + 20x – 3x2 + 9x – 15 – 25x + 9
= 3x4 – 5x3 + 4x – 6
3x4 – 5x3 + 4x – 6x2 – 3x + 5
= 3x2 + 4x – 3 + – 25x + 9x2 – 3x + 5
ANSWER
Warm-Up ExercisesEXAMPLE 2 Use polynomial long division with a linear divisor
Divide f (x) = x3 + 5x2 – 7x + 2 by x – 2.
x2 + 7x + 7x – 2 x3 + 5x2 – 7x + 2)
quotient
x3 – 2x2 Multiply divisor by x3/x = x2.
7x2 – 7x Subtract.
Multiply divisor by 7x2/x = 7x.7x2 – 14x
7x + 2 Subtract.
Multiply divisor by 7x/x = 7.16 remainder
7x – 14
ANSWER x3 + 5x2 – 7x +2x – 2
= x2 + 7x + 7 + 16x – 2
Warm-Up ExercisesGUIDED PRACTICE for Examples 1 and 2
Divide using polynomial long division.
1. (2x4 + x3 + x – 1) (x2 + 2x – 1)
SOLUTION
Write polynomial division in the same format you use when dividing numbers. Include a “0” as the coefficient of x2 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 ExercisesGUIDED PRACTICE for Examples 1 and 2
Multiply divisor by 2x4/x2
= –2x2.2x4 – 4x3 – 2x2
3x3 – 2x2 + x Subtract. Bring down next term.
Multiply divisor by –3x3/x2
= –3.– 3x3 – 6x2 + 3x
8x2 – 2x – 1 Subtract. Bring down next term.
Multiply divisor by 4x2/x2 = 8.
8x2 –16x – 8
– 18x + 7 remainder
2x2 – 3x + 8x2 + 2x – 1 2x4 + x3 + 0x2 + x – 1)
quotient
Warm-Up ExercisesGUIDED PRACTICE for Examples 1 and 2
2x4 + 5x3 + x – 1x2 + 2x – 1
= (2x2 – 3x + 8)+ – 18x + 7x2 + 2x – 1
ANSWER
Warm-Up ExercisesGUIDED PRACTICE for Examples 1 and 2
2. (x3 – x2 + 4x – 10) (x + 2)
SOLUTION
Write polynomial division in the same format you use when dividing numbers. Include a “0” as the coefficient of x2 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 ExercisesGUIDED PRACTICE for Examples 1 and 2
Multiply divisor by x3/x = x2.x3 + 2x2
–3x2 + 4x Subtract. Bring down next term.
Multiply divisor by –3x2/x
= –3x.– 3x2 – 6x
10x – 1 Subtract. Bring down next term.
Multiply divisor by 10x/x = 10.
10x + 20
– 30 remainder
x2 – 3x + 10 x + 2 x3 – x2 + 4x – 10)
quotient
Warm-Up ExercisesGUIDED PRACTICE for Examples 1 and 2
x3 – x2 +4x – 10x + 2
= (x2 – 3x +10)+ – 30 x + 2
ANSWER
Warm-Up ExercisesEXAMPLE 3Use synthetic division
Divide f (x)= 2x3 + x2 – 8x + 5 by x + 3 using synthetic division.
– 3 2 1 – 8 5
– 6 15 – 21
2 – 5 7 – 16
2x3 + x2 – 8x + 5x + 3
= 2x2 – 5x + 7 –16
x + 3ANSWER
SOLUTION
Warm-Up ExercisesEXAMPLE 4 Factor a polynomial
Factor f (x) = 3x3 – 4x2 – 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
Warm-Up ExercisesEXAMPLE 4 Factor a polynomial
Use the result to write f (x) as a product of two factors and then factor completely.
f (x) = 3x3 – 4x2 – 28x – 16 Write original polynomial.
= (x + 2)(3x2 – 10x – 8) Write as a product of two factors.
= (x + 2)(3x + 2)(x – 4) Factor trinomial.
Warm-Up ExercisesGUIDED PRACTICE for Examples 3 and 4
Divide using synthetic division.
3. (x3 + 4x2 – x – 1) (x + 3)
SOLUTION
(x3 + 4x2 – x – 1) (x + 3)
– 3 1 4 – 1 – 1
– 3 – 3 12
3 1 – 4 11
x3 + 4 x2 – x – 1x + 3
= x2 + x – 4 +11
x + 3ANSWER
Warm-Up ExercisesGUIDED PRACTICE for Examples 3 and 4
4. (4x3 + x2 – 3x + 7) (x – 1)
SOLUTION
(4x3 + x2 – 3x + 7) (x – 1)
1 4 1 – 3 7
4 5 2
4 5 2 9
4x3 + x2 – 3x + 1x – 1
= 4x2 + 5x + 2 +9
x – 1ANSWER
Warm-Up ExercisesGUIDED PRACTICE for Examples 3 and 4
Factor the polynomial completely given that x – 4 is a factor.
5. f (x) = x3 – 6x2 + 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 – 6 5 12
4 – 8 –12
1 – 2 – 3 0
Warm-Up ExercisesGUIDED 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) = x3 – 6x2 + 5x + 12 Write original polynomial.
= (x – 4)(x2 – 2x – 3) Write as a product of two factors.
= (x – 4)(x –3)(x + 1) Factor trinomial.
Warm-Up ExercisesGUIDED PRACTICE for Examples 3 and 4
6. f (x) = x3 – x2 – 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.
4 1 4 – 22 40
4 12 –40
1 3 – 10 0
Warm-Up ExercisesGUIDED 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) = x3 – x2 – 22x + 40 Write original polynomial.
= (x – 4)(x2 + 3x – 10) Write as a product of two factors.
= (x – 4)(x –2)(x +5) Factor trinomial.
Warm-Up ExercisesEXAMPLE 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
Warm-Up ExercisesEXAMPLE 5
Use the result to write f (x) as a product of two factors. Then factor completely.
f (x) = x3 – 2x2 – 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)(x2 + x – 20)
Warm-Up ExercisesEXAMPLE 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 numberof shoes produced (in millions). The companynow 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 ExercisesEXAMPLE 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
Warm-Up ExercisesEXAMPLE 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
Warm-Up ExercisesGUIDED PRACTICE for Examples 5 and 6
Find the other zeros of f given that f (– 2) = 0.
7. f (x) = x3 + 2x2 – 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
Warm-Up ExercisesGUIDED PRACTICE for Examples 5 and 6
Use the result to write f (x) as a product of two factors. Then factor completely.
f (x) = x3 + 2x2 – 9x – 18
The zeros are 3, – 3, and – 2.
= (x + 2)(x + 3)(x – 3)
= (x + 2)(x2 – 92)
Warm-Up ExercisesGUIDED PRACTICE for Examples 5 and 6
8. f (x) = x3 + 8x2 + 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
Warm-Up ExercisesGUIDED PRACTICE for Examples 5 and 6
Use the result to write f (x) as a product of two factors. Then factor completely.
f (x) = x3 + 8x2 + 5x – 14
The zeros are 1, – 7, and – 2.
= (x + 2)(x + 7)(x – 1)
= (x + 2)(x2 + 6x – 7 )
Warm-Up ExercisesGUIDED 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 = – 15x3 + 40x?
SOLUTION
25 = – 15x3 + 40x Substitute 25 for P in P = – 15x3 + 40x.0 = 15x3 – 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 15x3 – 40x + 25. Use synthetic division to find the other factors.
1 15 0 – 40 25
15 15 –25
15 15 – 25 0
Warm-Up ExercisesGUIDED PRACTICE for Examples 5 and 6
So, (x – 1)(15x2 + 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
Warm-Up ExercisesDaily Homework Quiz
2. Use synthetic division to divide f(x) = x3 – 3x2 – 5x – 25 by x – 5.
1. Divide 6x4 – x3 – x2 + 11x – 18 by 2x2 + x – 3.
ANSWER
3x2 – 2x + 5 + 2x2 + x – 3– 3
ANSWER
x2 + 2x + 5
Warm-Up ExercisesDaily Homework Quiz
ANSWER
5
ANSWER
About 4300 or about 700
3. One zero of f(x) = x3 – x2 – 17x – 15 is x = – 1.
4. One of the costs to print a novel can be modeled by C = x3 – 10x2 + 28x, 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?