Warm-Up 2/24 1. 12 𝐶=𝑑𝜋=6 3 𝜋 6 3 6 B
Rigor: You will learn how to divide polynomials and use the Remainder and Factor Theorems. Relevance: You will be able to use graphs and equations of polynomial functions to solve real world problems. MA.912. A.2.11
2-3 The Remainder and Factor Theorems
Example 1: Use long division to factor polynomial. 6 𝑥 3 −25 𝑥 2 +18𝑥+9; 𝑥−3 6 𝑥 2 − 7𝑥 − 3 𝑥−3 6 𝑥 3 −25 𝑥 2 +18𝑥+9 −6 𝑥 3 +18 𝑥 2 6 𝑥 3 −18 𝑥 2 −7 𝑥 2 +18𝑥+9 +7 𝑥 2 −21𝑥 −7 𝑥 2 +21𝑥 −3𝑥+9 +3𝑥−9 −3𝑥+9 𝑥−3 (6 𝑥 2 −7𝑥−3) 𝑥−3 (2𝑥−3)(3𝑥+1) So there are real zeros at x = 3, 3 2 , and − 1 3 .
Example 2: Divide the polynomial. 9 𝑥 3 −𝑥−3; 3𝑥+2 3 𝑥 2 − 2𝑥 + 1 3𝑥+2 9 𝑥 3 +0 𝑥 2 −𝑥−3 −9 𝑥 3 −6 𝑥 2 9 𝑥 3 +6 𝑥 2 −6 𝑥 2 −𝑥−3 −6 𝑥 2 −4𝑥 +6 𝑥 2 +4𝑥 3𝑥 −3 −3𝑥−2 3𝑥+2 −5 9 𝑥 3 −𝑥−3 3𝑥+2 =3 𝑥 2 −2𝑥+1+ −5 3𝑥+2 ,𝑥≠ − 2 3 9 𝑥 3 −𝑥−3 3𝑥+2 =3 𝑥 2 −2𝑥+1− 5 3𝑥+2 ,𝑥≠ − 2 3
Example 3: Divide the polynomial. 2 𝑥 4 −4 𝑥 3 +13 𝑥 2 +3𝑥−11; 𝑥 2 −2𝑥+7 2 𝑥 2 − 1 𝑥 2 −2𝑥+7 2 𝑥 4 −4 𝑥 3 +13 𝑥 2 +3𝑥−11 −2 𝑥 4 +4 𝑥 3 −14 𝑥 2 2 𝑥 4 −4 𝑥 3 +14 𝑥 2 − 𝑥 2 +3𝑥−11 + 𝑥 2 −2𝑥+7 − 𝑥 2 +2𝑥−7 𝑥 −4 2 𝑥 4 −4 𝑥 3 +13 𝑥 2 +3𝑥−11 𝑥 2 −2𝑥+7 =2 𝑥 2 −1+ 𝑥−4 𝑥 2 −2𝑥+7
Example 4a: Divide the polynomial using synthetic division. (2 𝑥 4 −5 𝑥 2 +5𝑥−2)÷ 𝑥+2 – 2 2 – 5 5 – 2 ↓ – 4 8 – 6 2 2 – 4 3 – 1 2 𝑥 3 −4 𝑥 2 +3𝑥−1 2 𝑥 4 −5 𝑥 2 +5𝑥−2 𝑥+2 =2 𝑥 3 −4 𝑥 2 +3𝑥−1
Example 4b: Divide the polynomial using synthetic division. (10 𝑥 3 −13 𝑥 2 +5𝑥−14)÷ 2𝑥−3 (10 𝑥 3 −13 𝑥 2 +5𝑥−14)÷2 (2𝑥−3)÷2 = 5 𝑥 3 − 13 2 𝑥 2 + 5 2 𝑥−7 𝑥− 3 2 (10 𝑥 3 −13 𝑥 2 +5𝑥−14) (2𝑥−3) 3 2 5 − 13 2 5 2 – 7 ↓ 15 2 3 2 6 5 1 4 – 1 5 𝑥 2 +𝑥+4− 1 𝑥− 3 2 =5 𝑥 2 +𝑥+4− 2 2𝑥−3
Example 6a: Use the Factor Theorem to determine if the binomials are factors of f(x). Write f(x) in factor form if possible. 𝑓 𝑥 =4 𝑥 4 +21 𝑥 3 +25 𝑥 2 −5𝑥+3;(𝑥−1), 𝑥+3 1 4 21 25 – 5 3 – 3 4 21 25 – 5 3 ↓ 4 25 50 45 ↓ – 12 – 27 6 – 3 4 25 50 45 48 4 9 – 2 1 𝑓 1 =48, so (𝑥−1 ) is not a factor. 𝑓 −3 =0, so (𝑥+3 ) is a factor. 𝑓 𝑥 = 𝑥+3 (4 𝑥 3 +9 𝑥 2 −2𝑥+1)
Example 6b: Use the Factor Theorem to determine if the binomials are factors of f(x). Write f(x) in factor form if possible. 𝑓 𝑥 =2 𝑥 3 − 𝑥 2 −41𝑥−20;(𝑥+4), 𝑥−5 – 4 2 – 1 – 41 – 20 ↓ – 8 36 20 𝑓 −4 =0, so (𝑥+4 ) is a factor. 2 – 9 – 5 5 2 – 9 – 5 ↓ 10 5 𝑓 5 =0, so (𝑥−5 ) is a factor. 2 1 𝑓 𝑥 = 𝑥+4 (𝑥−5)(2𝑥+1)
−1 math! 2-3 Assignment: TX p115, 4-44 EOE