6 3 Dividing Polynomials Warm Up Divide using

6 -3 Dividing Polynomials Warm Up Divide using long division. 1. 161 ÷ 7 23 2. 18 ÷ 2. 1 5. 8 Divide. 3. 6 x – 15 y 3 2 – ab 7 a 4. a Holt Algebra 2 2 x + 5 y 7 a – b

6 -3 Dividing Polynomials Objective Use long division and synthetic division to divide polynomials. Holt Algebra 2

6 -3 Dividing Polynomials Vocabulary synthetic division Holt Algebra 2

6 -3 Dividing Polynomials Polynomial long division is a method for dividing a polynomial by another polynomials of a lower degree. It is very similar to dividing numbers. Holt Algebra 2

6 -3 Dividing Polynomials Example 1: Using Long Division to Divide a Polynomial Divide using long division. (–y 2 + 2 y 3 + 25) ÷ (y – 3) Step 1 Write the dividend in standard form, including terms with a coefficient of 0. 2 y 3 – y 2 + 0 y + 25 Step 2 Write division in the same way you would when dividing numbers. y – 3 2 y 3 – y 2 + 0 y + 25 Holt Algebra 2

6 -3 Dividing Polynomials Example 1 Continued Step 3 Divide. 2 y 2+ 5 y + 15 y – 3 2 y 3 – y 2 + 0 y + 25 –(2 y 3 – 6 y 2) 5 y 2 + 0 y –(5 y 2 – 15 y) 15 y + 25 –(15 y – 45) 70 Holt Algebra 2 Notice that y times 2 y 2 is 2 y 3. Write 2 y 2 above 2 y 3. Multiply y – 3 by 2 y 2. Then subtract. Bring down the next term. Divide 5 y 2 by y. Multiply y – 3 by 5 y. Then subtract. Bring down the next term. Divide 15 y by y. Multiply y – 3 by 15. Then subtract. Find the remainder.

6 -3 Dividing Polynomials Example 1 Continued Step 4 Write the final answer. –y 2 + 2 y 3 + 25 y– 3 Holt Algebra 2 = 2 y 2 70 + 5 y + 15 + y – 3

6 -3 Dividing Polynomials Check It Out! Example 1 a Divide using long division. (15 x 2 + 8 x – 12) ÷ (3 x + 1) Step 1 Write the dividend in standard form, including terms with a coefficient of 0. 15 x 2 + 8 x – 12 Step 2 Write division in the same way you would when dividing numbers. 3 x + 1 15 x 2 + 8 x – 12 Holt Algebra 2

6 -3 Dividing Polynomials Check It Out! Example 1 a Continued Step 3 Divide. 5 x + 1 3 x + 1 15 x 2 + 8 x – 12 –(15 x 2 + 5 x) 3 x – 12 –(3 x + 1) – 13 Holt Algebra 2 Notice that 3 x times 5 x is 15 x 2. Write 5 x above 15 x 2. Multiply 3 x + 1 by 5 x. Then subtract. Bring down the next term. Divide 3 x by 3 x. Multiply 3 x + 1 by 1. Then subtract. Find the remainder.

6 -3 Dividing Polynomials Check It Out! Example 1 a Continued Step 4 Write the final answer. 15 x 2 + 8 x – 12 13 = 5 x + 1 – 3 x + 1 Holt Algebra 2

6 -3 Dividing Polynomials Synthetic division is a shorthand method of dividing a polynomial by a linear binomial by using only the coefficients. For synthetic division to work, the polynomial must be written in standard form, using 0 and a coefficient for any missing terms, and the divisor must be in the form (x – a). Holt Algebra 2

6 -3 Dividing Polynomials Holt Algebra 2

6 -3 Dividing Polynomials Example 2 A: Using Synthetic Division to Divide by a Linear Binomial Divide using synthetic division. (3 x 2 + 9 x – 2) ÷ (x – 1 ) 3 Step 1 Find a. Then write the coefficients and a in the synthetic division format. a= 1 3 3 Holt Algebra 2 For (x – 1 ), a = 31. 3 9 – 2 Write the coefficients of 3 x 2 + 9 x – 2.

6 -3 Dividing Polynomials Example 2 A Continued Step 2 Bring down the first coefficient. Then multiply and add for each column. 1 3 3 9 – 2 1 31 3 3 10 1 1 3 Draw a box around the remainder, 1 31. Step 3 Write the quotient. 11 3 3 x + 10 + 1 x– 3 Holt Algebra 2

6 -3 Dividing Polynomials Example 2 B: Using Synthetic Division to Divide by a Linear Binomial Divide using synthetic division. (3 x 4 – x 3 + 5 x – 1) ÷ (x + 2) Step 1 Find a. a = – 2 For (x + 2), a = – 2. Step 2 Write the coefficients and a in the synthetic division format. – 2 3 – 1 Holt Algebra 2 0 5 – 1 Use 0 for the coefficient of x 2.

6 -3 Dividing Polynomials Example 2 B Continued Step 3 Bring down the first coefficient. Then multiply and add for each column. – 2 3 – 1 0 5 – 1 – 6 14 – 28 46 Draw a box around the remainder, 45. 3 – 7 14 – 23 45 Step 4 Write the quotient. 3 x 3 – 7 x 2 + 14 x – 23 + Holt Algebra 2 45 x+2 Write the remainder over the divisor.

6 -3 Dividing Polynomials You can use synthetic division to evaluate polynomials. This process is called synthetic substitution. The process of synthetic substitution is exactly the same as the process of synthetic division, but the final answer is interpreted differently, as described by the Remainder Theorem. Holt Algebra 2

6 -3 Dividing Polynomials Example 3 A: Using Synthetic Substitution Use synthetic substitution to evaluate the polynomial for the given value. P(x) = 2 x 3 + 5 x 2 – x + 7 for x = 2. 2 2 5 – 1 2 4 9 7 18 34 17 41 Write the coefficients of the dividend. Use a = 2. P(2) = 41 Check Substitute 2 for x in P(x) = 2 x 3 + 5 x 2 – x + 7. P(2) = 2(2)3 + 5(2)2 – (2) + 7 P(2) = 41 Holt Algebra 2

6 -3 Dividing Polynomials Example 3 B: Using Synthetic Substitution Use synthetic substitution to evaluate the polynomial for the given value. 1 P(x) = 6 x 4 – 25 x 3 – 3 x + 5 for x = –. 3 1 Write the coefficients of 6 – 25 0 – 3 5 – 3 the dividend. Use 0 for – 2 9 – 3 2 the coefficient of x 2. Use 1 6 – 27 9 – 6 7 a=. 3 1 P( ) = 7 3 Holt Algebra 2