Download Presentation
## Bell-ringer

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**Bell-ringer**• Geometry Text**7.3.2 Products and Factors of Polynomials**• Objectives: 1).Divide one polynomial by another using long and synthetic division 2).Use the Remainder Theorem to solve problems**Definitons**• Divisor-expression that you divide by • Dividend – the expression being divided • Quotient - the answer after division is complete • Remainder - what is left over after dividing. • Refer to the example**20**31 Long Division With Numbers Find the quotient: (23828) ÷ (31) 7 6 8 2 31 238 2 8 8 - 217 768 20/31 or 768 + 20/31 21 - 186 26 - 248 20**Practice with variables**• 2x2 ÷ x 9y3÷ 3y • 16w2 ÷4w 36x3÷2x**- 16**x – 4 x2 + 2x – 14 – 16 x – 4 Example Find the quotient: (x3 – 2x2 – 22x + 40) ÷ (x – 4) x2 + 2x – 14 x – 4 x3 – 2x2 – 22x + 40 - (x3 – 4x2) 2x2 – 22x - (2x2 – 8x) –14x + 40 - (–14x + 56) – 16**Example**Find the quotient: (x3 –x2 – 4) ÷ (x2 + x + 2) x - 2 x2 + x + 2 x3 –x2 + 0x - 4 - (x3 + x2 +2x) -2x2 – 2x - 4 - (-2x2 – 2x – 4) 0 Quotient: x – 2**Now You**• Find the quotient: (x3 + 3x2 – 13x - 15) ÷ (x2 – 2x – 3) Quotient: x + 5**Synthetic Divsion**Use synthetic division to find the quotient: (x3 + x2 – 9x - 9) ÷ (x -3) Step 1: Write the coefficients of the polynomial, and the r-value of the divisor on the left 1 -9 -9 1 3**Example 2**Use synthetic division to find the quotient: (x3 + x2 – 9x - 9) ÷ (x -3) Step 2: Draw a line and write the first coefficient under the line. 1 -9 -9 1 1 3 1**Example 2**Use synthetic division to find the quotient: (x3 + x2 – 9x - 9) ÷ (x - 3) Step 3: Multiply the r-value, 3, by the number below the line and write the product below the next coefficient. 1 -9 -9 1 3 3 1**Example 2**Use synthetic division to find the quotient: (x3 + x2 – 9x - 9) ÷ (x - 3) Step 4: Write the sum of 1 and 3 below the line. 1 -9 -9 1 3 3 1 4**Example 2**Use synthetic division to find the quotient: (x3 + x2 – 9x - 9) ÷ (x - 3) Repeat steps 3 and 4. 1 -9 -9 1 3 3 12 1 4 3**Example 2**Use synthetic division to find the quotient: (x3 + x2 – 9x - 9) ÷ (x - 3) Repeat steps 3 and 4. 1 -9 -9 1 3 3 12 9 1 4 3 0**Example 2**Use synthetic division to find the quotient: (x3 + x2 – 9x - 9) ÷ (x - 3) The remainder is 0 and the resulting numbers are the coefficients of the quotient. 1 -9 -9 1 3 3 12 9 1 4 3 0 x2 + 4x + 3**Synthetic Division v. long division***Synthetic Division can only be used when dividing by a linear binomial of the form x –r. **Otherwise long division must be used.**Example**Use synthetic division to find the quotient: (x4 – 3x + 2x3 – 6) ÷ (x - 2) -6 2 0 -3 1 2 26 2 8 16 1 4 8 13 20 x3 + 4x2 + 8x + 13 + 20/(x-2)**Now You**Use synthetic division to find the quotient: (6x2 – 5x - 6) ÷ (x + 3) -5 -6 6 -3 -18 69 6 -23 63 6x -23 + 63/(x+3).**2**Example 3 Given that 2 is a zero of P(x) = x3 – 3x2 + 4, use division to factor x3 – 3x2 + 4. Since 2 is a zero, x = 2 , so x – 2 = 0 , which means x – 2 is a factor of x3 – 3x2 + 4. (x3 – 3x2 + 4) ÷ (x – 2) Method 2 Method 1 x2 - x – 2 1 -3 0 4 x – 2 x3 – 3x2 + 0x + 4 2 -2 -4 - (x3 – 2x2) 1 -1 -2 0 -x2 + 0x - (-x2 + 2x) x3 – 3x2 + 4 = (x – 2)(x2 – x – 2) –2x + 4 - (–2x + 4) 0**Practice**Given that -3 is a zero of P(x) = x3 – 13x - 12, use division to factor x3 – 13x – 12. x3 -13x -12 = (x + 3)(x2 -3x -4)**Remainder Theorem**If the polynomial expression that defines the function of P is divided by x – a, then the remainder is the number P(a).**6**Example 4 Given P(x) = 3x3 – 4x2 + 9x + 5 find P(6) by using both synthetic division and substitution. Method 2 Method 1 3 -4 9 5 P(6) = 3(6)3 – 4(6)2 + 9(6) + 5 = 3(216) – 4(36) + 54 + 5 18 84 558 3 14 93 563 = 648 – 144 + 54 + 5 = 563**Practice**Given P(x) = 3x3 + 2x2 + 3x + 1 find P(-2) using synthetic division and substitution.**Homework**p.446 #72 – 96 by 3’s