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

Vocabulary synthetic division depressed polynomial synthetic substitution

Example 1 Use Long Division to Factor Polynomials Factor 6x x 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.

Example 1 Use Long Division to Factor Polynomials From this division, you can write 6x x 2 – 104x + 60 = (2x – 5)(3x x – 12). Factoring the quadratic expression yields 6x x 2 – 104x + 60 = (2x – 5)(3x – 2)(x + 6). Answer: (2x – 5)(3x – 2)(x + 6)

Example 1 Factor 6x 3 + x 2 – 117x 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)

Key Concept 1

Example 2 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 Long Division with Nonzero Remainder Answer: You can write the result as. Check Multiply to check this result. (2x – 1)(3x 2 – x + 4) + 10= 6x 3 – 5x 2 + 9x + 6 6x 3 – 2x 2 + 8x – 3x 2 + x – = 6x 3 – 5x 2 + 9x + 6 6x 3 – 5x 2 + 9x + 6= 6x 3 – 5x 2 + 9x + 6

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

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

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

Key Concept 2

Example 4 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 – coefficients of depressed quotient remainder = add terms. =Multiply by c, and write the product

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

Example 4 Synthetic Division B. Find (8x x 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 Synthetic Division 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.6x 3 – 8x D.6x 3 + 4x x + 3

Key Concept 3

Example 5 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 x + 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 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– ,000 – – ,250

Example 5 Answer:$481,250 Use the Remainder Theorem Check You can check your answer using direct substitution. R(x) = –150x x + 480,000 Original function R(5) = –150(5) (5) + 480,000 Substitute 5 for x. R(5) = – ,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

Key Concept 4

Example 6 A. Use the Factor Theorem to determine if (x – 5) and (x + 5) are factors of f (x) = x 3 – 18x x 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). 51– –65–25 1–13–50

Example 6 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). –51– –5115–875 1–23175–850

Example 6 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). 51–2–13– 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 x 2 + 3x + 2. –2132–2 110 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).

Example 6 Use the Factor Theorem to determine if the binomials (x + 2) and (x – 3) are factors of f (x) = 4x 3 – 9x 2 – 19x 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)(4x 2 – 17x – 15) D.no, yes; f(x) = (x – 3)(4x 2 + 3x + 10)

Key Concept 5

End of the Lesson