Five-Minute Check (over Lesson 2-2)

Then/Now

New Vocabulary

Example 1:Use Long Division to Factor Polynomials

Key Concept:Polynomial Division

Example 2:Long Division with Nonzero Remainder

Example 3:Division by Polynomial of Degree 2 or Higher

Key Concept:Synthetic Division Algorithm

Example 4:Synthetic Division

Key Concept:Remainder Theorem

Example 5:Real-World Example: Use the Remainder Theorem

Key Concept:Factor Theorem

Example 6:Use the Factor Theorem

Concept Summary: Synthetic Division and Remainders

Over Lesson 2-2

Graph f

(x) = (x – 2)3 + 3.

A.

B.

C.

D.

Over Lesson 2-2

Describe the end behavior of the graph of f (x) = 2x3 – 4x + 1 using limits. Explain your reasoning using the leading term

test.

A. The degree is 3 and the leading coefficient 2. Because the degree is odd and the leading coefficient is positive,

.

B. The degree is 3 and the leading coefficient 2. Because the degree is odd and the leading coefficient is positive,

.

C. The degree is 3, and the leading coefficient is 2. Because the degree is odd and the leading coefficient is

positive,

.

D. The degree is 3 and the leading coefficient is 2. Because the degree is odd and the leading coefficient is

positive,

.

Over Lesson 2-2

Determine all of the real zeros of f

(x) = 4x

6 – 16x

4.

A. 0, −2, 2

B. 0, −2, 2, 4

C. −2, 2

D. −4, 4

You factored quadratic expressions to solve equations. (Lesson 0–3)

• Divide polynomials using long division and synthetic division.

• Use the Remainder and Factor Theorems.

• synthetic division

• depressed polynomial

• synthetic substitution

Use Long Division to Factor Polynomials

Factor 6x 3 + 17x

2 – 104x + 60 completely using long division if (2x – 5) is a factor.

(–)6x

3 – 15x

2←Multiply divisor by 3x 2 because

= 3x 2.

←Subtract and bring down next term.32x

2 – 104x

(–)32x

2 – 80x ←Multiply divisor by 16x because

= 16x.

–24x + 60 ←Subtract and bring down next term.

(–)–24x + 60 ←Multiply divisor by –12 because

= –12

0 ←Subtract. Notice that the remainder is 0.

Use Long Division to Factor Polynomials

From this division, you can write

6x3 + 17x2 – 104x + 60 = (2x – 5)(3x2 + 16x – 12).

Factoring the quadratic expression yields

6x3 + 17x2 – 104x + 60 = (2x – 5)(3x – 2)(x + 6).

Answer: (2x – 5)(3x – 2)(x + 6)

Factor 6x

3 + x

2 – 117x + 140 completely using long division if (3x – 4) is a factor.

A. (3x – 4)(x – 5)(2x + 7)

B. (3x – 4)(x + 5)(2x – 7)

C. (3x – 4)(2x

2 + 3x – 35)

D. (3x – 4)(2x + 5)(x – 7)

Long Division with Nonzero Remainder

Divide 6x 3 – 5x

2 + 9x + 6 by 2x – 1.

(–)6x

3 – 3x

2

–2x

2 + 9x

(–)–2x

2 + x

8x + 6

(–)8x – 4

10

Long Division with Nonzero Remainder

Answer:

You can write the result as

.

Check Multiply to check this result.

(2x – 1)(3x2 – x + 4) + 10 = 6x3 – 5x2 + 9x + 6

6x3 – 2x2 + 8x – 3x2 + x – 4 + 10 = 6x3 – 5x2 + 9x + 6

6x3 – 5x2 + 9x + 6 = 6x3 – 5x2 + 9x + 6

Divide 4x

4 – 2x

3 + 8x – 10 by x + 1.

A.

B. 4x

3 + 2x

2 + 2x + 10

C.

D.

Division by Polynomial of Degree 2 or Higher

Divide x 3 – x

2 – 14x + 4 by x

2 – 5x + 6.

(–)x

3 – 5x

2 + 6x

4x

2 – 20x + 4

(–)4x

2 – 20x + 24

–20

Division by Polynomial of Degree 2 or Higher

Answer:

You can write this result as

.

Divide 2x

4 + 9x

3 + x2 – x + 26 by x

2 + 6x + 9.

A.

B.

C.

D.

Synthetic Division

A. Find (2x 5 – 4x

4 – 3x

3 – 6x

2 – 5x – 8) ÷ (x – 3) using synthetic division.

Because x – 3 is x – (3), c = 3. Set up the synthetic division as follows. Then follow the

synthetic division procedure.

3 2 –4 –3 –6 –5 –8

4 2

2 3 3 4

coefficients of depressed quotient remainder

= add terms.

= Multiply by c, and write

the product.

6 6 9 9 12

Synthetic Division

Answer:

The quotient has degree one less than that of its dividend, so

Synthetic Division

B. Find (8x 4 + 38x

3 + 5x

2 + 3x + 3) ÷ (4x + 1) using synthetic division.

Rewrite the division expression so that the divisor is of the form x – c.

Synthetic Division

So, . Perform the synthetic division.

Synthetic Division

So, .

Answer:

Find (6x

4 – 2x

3 + 8x

2 – 9x – 3) ÷ (x – 1) using synthetic division.

A.

B.

C. 6x3 – 8x2 + 3

D. 6x3 + 4x2 + 12x + 3

Use the Remainder Theorem

REAL ESTATE Suppose 800 units of beachfront property have tenants paying $600 per

week. Research indicates that for each $10 decrease in rent, 15 more units would be

rented. The weekly revenue from the rentals is given by

R

(x) = –150x

2 + 1000x + 480,000, where x is the number of $10 decreases the property

manager is willing to take. Use the Remainder Theorem to find the revenue from the

properties if the property manager decreases the rent by $50.

To find the revenue from the properties, use synthetic substitution to evaluate f

(x) for x = 5

since $50 is

5 times $10.

Use the Remainder Theorem

The remainder is 481,250, so f

(5) = 481,250. Therefore, the revenue will be $481,250 when

the rent is decreased by $50.

5 –150 1000 480,000

–750 1250

– 150 250 481,250

Answer: $481,250

Use the Remainder Theorem

Check You can check your answer using direct substitution.

R(x) = –150x2 + 1000x + 480,000

Original function

R(5) = –150(5)2 + 1000(5) + 480,000 Substitute 5 for x.

R(5) = –3750 + 5000 + 480,000 or 481,250 Simplify.

REAL ESTATE Use the equation for R(x) from Example 5 and the Remainder Theorem

to find the revenue from the properties if the property manager decreases the rent by

$100.

A. $380,000

B. $450,000

C. $475,000

D. $479,900

A. Use the Factor Theorem to determine if (x – 5) and (x + 5) are factors of f (x) = x

3 –

18x

2 + 60x + 25. Use the binomials that are factors to write a factored form of f

(x).

Use the Factor Theorem

Use synthetic division to test each factor, (x – 5) and (x + 5).

5 1 –18 60 25

5 –65 –25

1 –13 –5 0

Answer: f

(x) = (x – 5)(x

2 – 13x – 5)

Use the Factor Theorem

Because the remainder when f

(x) is divided by (x – 5) is 0, f(5) = 0, and (x – 5) is a factor.

Because the remainder when f

(x) is divided by (x + 5) is –850, f

(–5) = –850 and (x + 5) is not a

factor.

Because (x – 5) is a factor of f

(x), we can use the quotient of f

(x) ÷ (x – 5) to write a factored

form of f(x).

–5 1 –18 60 25

–5 115 –875

1 –23 175 –850

B. Use the Factor Theorem to determine if (x – 5) and (x + 2) are factors of f (x) = x

3 – 2x

2 – 13x – 10. Use the binomials that are factors to write a factored form of f

(x).

Use the Factor Theorem

Use synthetic division to test the factor (x – 5).

5 1 –2 –13 –10

5 15 10

1 3 2 0

Because the remainder when f

(x) is divided by (x – 5) is 0, f

(5) = 0 and (x – 5) is a factor of f

(x).

Answer: f

(x) = (x – 5)(x + 2)(x + 1)

Use the Factor Theorem

Next, test the second factor (x + 2), with the depressed polynomial x2 + 3x + 2.

–2 1 3 2

–2 –2

1 1 0

Because the remainder when the quotient of

f

(x) ÷ (x – 5) is divided by (x + 2) is 0, f(–2) = 0 and

(x + 2) is a factor of f

(x).

Because (x – 5) and (x + 2) are factors of f

(x), we can use the final quotient to write a

factored form of f

(x).

Use the Factor Theorem to determine if the binomials (x + 2) and (x – 3) are factors of

f

(x) = 4x

3 – 9x

2 – 19x + 30. Use the binomials that are factors to write a factored form

of f

(x).

A. yes, yes; f(x) = (x + 2)(x – 3)(–14x + 5)

B. yes, yes; f(x) = (x + 2)(x – 3)(4x – 5)

C. yes, no; f(x) = (x + 2)(4x2 – 17x – 15)

D. no, yes; f(x) = (x – 3)(4x2 + 3x + 10)