1. 8.4 Dividing Polynomials CORD Math Mrs. Spitz Fall 2006

2. Objective Divide polynomials by binomials

3. Upcoming 8.4 Monday 10/23 8.5 Tuesday/Wednesday – Skip 8.6 8.7 Thursday 10/26 8.8 Friday 10/27 8.9 Monday 10/30 8.10 Tuesday/Wed Chapter 8 Review Wed/Thur Chapter 8 Test Friday

4. Assignment Pg. 320 #3-31 all

5. Introduction To divide a polynomial by a polynomial, you can use a long division process similar to that used in arithmetic. For example, you can divide x 2 + 8x +15 by x + 5 as shown on the next couple of slides.

6. Step 1 To find the first term of the quotient, divide the first term of the dividend (x 2 ) by the first term of the divisor (x). x 3x + 15

7. Step 2 To find the next term of the quotient, divide the first term of the partial dividend (3x) by the first term of the divisor (x). x + 3 3x + 15 3x - 15 0 Therefore, x 2 + 8x + 15 divided by x + 5 is x + 3. Since the remainder is 0, the divisor is a factor of the divident. This means that (x + 5)(x + 3) = x 2 + 8x + 15.

8. What happens if it doesn’t go evenly? If the divisor is NOT a factor of the dividend, there will be a non-zero remainder. The quotient can be expressed as follows: Quotient = partial quotient +

9. Ex. 1: Find (2x 2 -11x – 20)  (2x + 3). - 14x - 20 + 14x+ 21 x - 7 + 1 ← Multiply by x(2x+3) ← Subtract, then bring down - 20 ← Multiply -7(2x+3) ← Subtract. The remainder is 1 The quotient is x – 7 with a remainder of 1. Thus, (2x 2 -11x – 20)  (2x + 3) = x – 7 +

10. Other note... In an expression like s 3 +9, there is no s 2 term and no s term. In such situations, rename the expression using 0 as a coefficient of these terms as follows: s 3 + 9 = s 3 + 0s 2 + 0s + 9

11. Ex. 2: Find 3s 2 + 0s s 2 +3s + 9 + 36 ← Insert 0s 2 and 0s. Why? ← Subtract, then bring down 0s ← Multiply 3s(s - 3) ← Subtract. The remainder is 36 The quotient is s 2 +3s+9 with a remainder of 36. Thus, (s 3 + 9)  (s - 3) = s2 + 3s + 9 + 3s 2 - 9s 9s + 9 -9s + 27 ← Multiply by s 2