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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 Lesson Menu

Graph f (x) = (x – 2)3 + 3. A. B. C. D. 5–Minute Check 1

Falls left & Rises Right 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, . Falls left & Rises Right 5–Minute Check 2

Determine all of the real zeros of f (x) = 4x 6 – 16x 4. B. 0, −2, 2, 4 C. −2, 2 D. −4, 4 5–Minute Check 3

You factored quadratic expressions to solve equations. (Lesson 0–3) Divide polynomials using long division and synthetic division. Use the Remainder and Factor Theorems. Then/Now

synthetic substitution synthetic division depressed polynomial synthetic substitution Vocabulary

THE REMAINDER & FACTOR THEOREMS Long Division DAY 1 HW: Pg. 115, #’s 1-18

Use Long Division to Factor Polynomials Factor 6x 3 + 17x 2 – 104x + 60 completely using long division if (2x – 5) is a factor. ←Multiply divisor by 3x 2 because = 3x 2. (–)6x 3 – 15x 2 32x 2 – 104x ←Subtract and bring down next term. ←Multiply divisor by 16x because = 16x. (–)32x 2 – 80x ←Subtract and bring down next term. –24x + 60 ←Multiply divisor by –12 because = –12 (–)–24x + 60 ←Subtract. Notice that the remainder is 0. Example 1

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) Example 1

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) Example 1

  Key Concept 1

Divide 6x 3 – 5x 2 + 9x + 6 by 2x – 1. (–)6x 3 – 3x 2 –2x 2 + 9x 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 Example 2

You can write the result as . Long Division with Nonzero Remainder You can write the result as . Answer: 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  Example 2

Divide 4x 4 – 2x 3 + 8x – 10 by x + 1. A. B. 4x 3 + 2x 2 + 2x + 10 C. Example 2

Divide x 3 – x 2 – 14x + 4 by x 2 – 5x + 6. (–)x 3 – 5x 2 + 6x 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 Example 3

You can write this result as . Division by Polynomial of Degree 2 or Higher You can write this result as . Answer: Example 3

Divide 2x 4 + 9x 3 + x2 – x + 26 by x 2 + 6x + 9. A. B. C. D. Example 3

THE REMAINDER & FACTOR THEOREMS Synthetic Division LESSON 2 HW: Pg. 115, #’s 19-28

Key Concept 2

Factor 6x 3 + 17x 2 – 104x + 60 completely using Synthetic division if (2x – 5) is a factor.

= Multiply by c, and write the product. 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 = add terms. 6 6 9 9 12 2 2 3 3 4 4 = Multiply by c, and write the product. coefficients of depressed quotient remainder Example 4

The quotient has degree one less than that of its dividend, so Synthetic Division The quotient has degree one less than that of its dividend, so Answer: Example 4

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. Example 4

So, . Perform the synthetic division. Example 4

Synthetic Division So, . Answer: Example 4

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 Example 4

THE REMAINDER & FACTOR THEOREMS Remainder Theorem

If f(x)= (6x 4 – 2x 3 + 8x 2 – 9x – 3) and is divided by (x – 1) , then f(1) will be the remainder. Key Concept 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. Example 5

Use the Remainder Theorem 5 –150 1000 480,000 –750 1250 – 150 250 481,250 The remainder is 481,250, so f (5) = 481,250. Therefore, the revenue will be $481,250 when the rent is decreased by $50. Example 5

Check You can check your answer using direct substitution. Use the Remainder Theorem Answer: $481,250 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. Example 5

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 Example 5

THE REMAINDER & FACTOR THEOREMS Factor Theorem

Key Concept 4

Use synthetic division to test each factor, (x – 5) and (x + 5). Use the Factor Theorem 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 synthetic division to test each factor, (x – 5) and (x + 5). 5 1 –18 60 25 5 –65 –25 1 –13 –5 0 Example 6

Answer: f (x) = (x – 5)(x 2 – 13x – 5) Use the Factor Theorem –5 1 –18 60 25 –5 115 –875 1 –23 175 –850 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). Answer: f (x) = (x – 5)(x 2 – 13x – 5) Example 6

Use synthetic division to test the factor (x – 5). Use the Factor Theorem 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 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). Example 6

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). Answer: f (x) = (x – 5)(x + 2)(x + 1) Example 6

A. yes, yes; f(x) = (x + 2)(x – 3)(–14x + 5) 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) Example 6

Key Concept 5

End of the Lesson