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Five-Minute Check (over Lesson 5–1) CCSS Then/Now New Vocabulary Example 1: Divide a Polynomial by a Monomial Example 2: Division Algorithm Example 3: Standardized Test Example: Divide Polynomials Key Concept: Synthetic Division Example 4: Synthetic Division Example 5: Divisor with First Coeffecient Other than 1 Lesson Menu

Simplify b2 ● b5 ● b3. A. b5 B. b8 C. b10 D. b30 5-Minute Check 1

A. B. C. D. 5-Minute Check 2

Simplify (10a2 – 6ab + b2) – (5a2 – 2b2). A. 15a2 + 8ab + 3b2 B. 10a2 – 6ab – b2 C. 5a2 + 6ab – 3b2 D. 5a2 – 6ab + 3b2 5-Minute Check 3

Simplify 7w(2w2 + 8w – 5). A. 14w3 + 56w2 – 35w B. 14w2 + 15w – 35 C. 9w2 + 15w – 12 D. 2w2 + 15w – 5 5-Minute Check 4

State the degree of 6xy2 – 12x3y2 + y4 – 26. B. 7 C. 5 D. 4 5-Minute Check 5

Find the product of 3y(2y2 – 1)(y + 4). A. 18y5 + 72y4 – 9y3 – 36y2 B. 6y4 + 24y3 – 3y2 – 12y C. –18y3 – 3y2 + 12y D. 6y3 – 2y + 4 5-Minute Check 6

Mathematical Practices 6 Attend to precision. Content Standards A.APR.6 Rewrite simple rational expressions in different forms; write a(x)/b(x) in the form q(x) + r(x)/b(x), where a(x), b(x), q(x), and r(x) are polynomials with the degree of r(x) less than the degree of b(x), using inspection, long division, or, for the more complicated examples, a computer algebra system. Mathematical Practices 6 Attend to precision. CCSS

Divide polynomials using long division. You divided monomials. Divide polynomials using long division. Divide polynomials using synthetic division. Then/Now

synthetic division Vocabulary

= a – 3b2 + 2a2b3 a1 – 1 = a0 or 1 and b1 – 1 = b0 or 1 Divide a Polynomial by a Monomial Sum of quotients Divide. = a – 3b2 + 2a2b3 a1 – 1 = a0 or 1 and b1 – 1 = b0 or 1 Answer: a – 3b2 + 2a2b3 Example 1

A. 2x3y – 3x5y2 B. 1 + 2x3y – 3x5y2 C. 6x4y2 + 9x7y3 – 6x9y4 D. 1 + 2x7y3 – 3x9y4 Example 1

Use long division to find (x2 – 2x – 15) ÷ (x – 5). Division Algorithm Use long division to find (x2 – 2x – 15) ÷ (x – 5). 3(x – 5) = 3x – 15 x(x – 5) = x2 – 5x –2x – (–5x) = 3x Answer: The quotient is x + 3. The remainder is 0. Example 2

Use long division to find (x2 + 5x + 6) ÷ (x + 3). A. x + 2 B. x + 3 C. x + 2x D. x + 8 Example 2

Which expression is equal to (a2 – 5a + 3)(2 – a)–1? Divide Polynomials Which expression is equal to (a2 – 5a + 3)(2 – a)–1? A a + 3 B C D Example 3

Divide Polynomials Read the Test Item Since the second factor has an exponent of –1, this is a division problem. Solve the Test Item 3(–a + 2) = –3a + 6 –a(–a + 2) = a2 – 2a Rewrite 2 – a as –a + 2. –5a – (–2a) = –3a Subtract. 3 – 6 = –3 Example 3

The quotient is –a + 3 and the remainder is –3. Divide Polynomials The quotient is –a + 3 and the remainder is –3. Therefore, . Answer: The answer is D. Example 3

Which expression is equal to (x2 – x – 7)(x – 3)–1? B. C. D. Example 3

Concept

Use synthetic division to find (x3 – 4x2 + 6x – 4) ÷ (x – 2). Step 1 Write the terms of the dividend so that the degrees of the terms are in descending order. Then write just the coefficients as shown. x3 – 4x2 + 6x – 4     1 –4 6 –4 Step 2 Write the constant r of the divisor x – r to the left. In this case, r = 2. Bring the first coefficient, 1, down as shown. 1 –4 6 –4 1 Example 4

Synthetic Division Step 3 Multiply the first coefficient by r : 1 ● 2 = 2. Write the product under the second coefficient. Then add the product and the second coefficient: –4 + 2 = –2. 1 1 –4 6 –4 2 –2 2 –2 1 1 –4 6 –4 Step 4 Multiply the sum, –2, by r : –2 ● 2 = –4. Write the product under the next coefficient and add: 6 + (–4) = 2. –4 2 Example 4

Answer: The quotient is x2 – 2x + 2. Synthetic Division 2 –2 1 –4 1 –4 6 –4 Step 5 Multiply the sum, 2, by r : 2 ● 2 = 4. Write the product under the next coefficient and add: –4 + 4 = 0. The remainder is 0. 4 The numbers along the bottom are the coefficients of the quotient. Start with the power of x that is one less than the degree of the dividend. Answer: The quotient is x2 – 2x + 2. Example 4

Use synthetic division to find (x2 + 8x + 7) ÷ (x + 1). A. x + 9 B. x + 7 C. x + 8 D. x + 7 Example 4

Use synthetic division to find (4y3 – 6y2 + 4y – 1) ÷ (2y – 1). Divisor with First Coefficient Other than 1 Use synthetic division to find (4y3 – 6y2 + 4y – 1) ÷ (2y – 1). Rewrite the divisor so it has a leading coefficient of 1. Divide numerator and denominator by 2. Simplify the numerator and denominator. Example 5

Divisor with First Coefficient Other than 1 The result is . Example 5

Answer: The solution is . Divisor with First Coefficient Other than 1 Answer: The solution is . Check: Divide using long division. 2y2 –2y + 1 4y3 – 2y2 –4y2 + 4y –4y2 + 2y 2y – 1 2y – 1 The result is .  Example 5

Use synthetic division to find (8y3 – 12y2 + 4y + 10) ÷ (2y + 1). A. B. C. D. Example 5

End of the Lesson