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Example 1 Divide a Polynomial by a Monomial Answer: a – 3b 2 + 2a 2 b 3 Sum of quotients Divide. = a – 3b 2 + 2a 2 b 3 a 1 – 1 = a 0 or 1 and b 1 – 1 = b 0 or 1

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A.A B.B C.C D.D Example 1 A.2x 3 y – 3x 5 y 2 B.1 + 2x 3 y – 3x 5 y 2 C.6x 4 y 2 + 9x 7 y 3 – 6x 9 y 4 D.1 + 2x 7 y 3 – 3x 9 y 4

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

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

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

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Concept

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Example 4 Synthetic Division Use synthetic division to find (x 3 – 4x 2 + 6x – 4) ÷ (x – 2). Step 1Write the terms of the dividend so that the degrees of the terms are in descending order. Then write just the coefficients as shown. Step 2Write the constant r of the divisor x – r to the left. In this case, r = 2. Bring the first coefficient, 1, down as shown. x 3 – 4x 2 + 6x – 4 1 –46–4 1

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

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Example 4 Synthetic Division Step 5Multiply the sum, 2, by r : 2 ● 2 = 4. Write the product under the next coefficient and add: –4 + 4 = 0. The remainder is 0. 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. 4 0 2 –2 1 2 –4 1–46–4 Answer: The quotient is x 2 – 2x + 2.

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A.A B.B C.C D.D Example 4 A.x + 9 B.x + 7 C.x + 8 D.x + 7 Use synthetic division to find (x 2 + 8x + 7) ÷ (x + 1).

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Example 5 Divisor with First Coefficient Other than 1 Use synthetic division to find (4y 3 – 6y 2 + 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.

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

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Example 5 Divisor with First Coefficient Other than 1 Answer:The solution is. Check:Divide using long division. The result is. 2y22y2 4y 3 – 2y 2 0 –4y 2 + 4y –2y –4y 2 + 2y 2y – 1 + 1 2y – 1

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A.A B.B C.C D.D Example 5 Use synthetic division to find (8y 3 – 12y 2 + 4y + 10) ÷ (2y + 1). A. B. C. D.

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End of the Lesson

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