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).

2. Holt Algebra 2 6-3 Dividing Polynomials

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

4. Holt Algebra 2 6-3 Dividing Polynomials Example 1 Continued Draw a box around the remainder, 45. 3 –1 0 5 –1–2 Step 3 Bring down the first coefficient. Then multiply and add for each column. –6 345 Step 4 Write the quotient. 3x 3 – 7x 2 + 14x – 23 + 45 x + 2 Write the remainder over the divisor. 46–2814 –2314–7

5. Holt Algebra 2 6-3 Dividing Polynomials Example 2 Divide using synthetic division. (6x 2 – 5x – 6) ÷ (x + 3) Step 1 Find a. Write the coefficients of 6x 2 – 5x – 6. For (x + 3), a = –3. a = –3 –3 6 –5 –6 Step 2 Write the coefficients and a in the synthetic division format.

6. Holt Algebra 2 6-3 Dividing Polynomials Example 2 Continued Draw a box around the remainder, 63. 6 –5 –6–3 Step 3 Bring down the first coefficient. Then multiply and add for each column. –18 663 Step 4 Write the quotient. 6x – 23 + 63 x + 3 Write the remainder over the divisor. –23 69

7. Holt Algebra 2 6-3 Dividing Polynomials Example 3 Divide using synthetic division. (x 2 – 3x – 18) ÷ (x – 6) Step 1 Find a. Write the coefficients of x 2 – 3x – 18. For (x – 6), a = 6. a = 6 6 1 –3 –18 Step 2 Write the coefficients and a in the synthetic division format.

8. Holt Algebra 2 6-3 Dividing Polynomials Example 3 Continued There is no remainder. 1 –3 –18 6 Step 3 Bring down the first coefficient. Then multiply and add for each column. 6 10 Step 4 Write the quotient. x + 3 18 3

9. Holt Algebra 2 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.

10. Holt Algebra 2 6-3 Dividing Polynomials Example 4: Using Synthetic Substitution Use synthetic substitution to evaluate the polynomial for the given value. P(x) = 2x 3 + 5x 2 – x + 7 for x = 2. Write the coefficients of the dividend. Use a = 2. 2 5 –1 7 2 4 241 P(2) = 41 Check Substitute 2 for x in P(x) = 2x 3 + 5x 2 – x + 7. P(2) = 2(2) 3 + 5(2) 2 – (2) + 7 P(2) = 41 3418 179

11. Holt Algebra 2 6-3 Dividing Polynomials Example 5: Using Synthetic Substitution Use synthetic substitution to evaluate the polynomial for the given value. P(x) = 6x 4 – 25x 3 – 3x + 5 for x = –. 6 –25 0 –3 5 –2 67 1 3 – 1 3 Write the coefficients of the dividend. Use 0 for the coefficient of x 2. Use a =. 1 3 P( ) = 7 1 3 2–39 –69–27

12. Holt Algebra 2 6-3 Dividing Polynomials Example 6 Use synthetic substitution to evaluate the polynomial for the given value. P(x) = x 3 + 3x 2 + 4 for x = –3. Write the coefficients of the dividend. Use 0 for the coefficient of x 2 Use a = –3. 1 3 0 4 –3 14 P(–3) = 4 Check Substitute –3 for x in P(x) = x 3 + 3x 2 + 4. P(–3) = (–3) 3 + 3(–3) 2 + 4 P(–3) = 4 00 00