1. Warm-Up 2/24 1. 12
𝐶=𝑑𝜋=6 3 𝜋
6 3 6 B

3. 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

4. 2-3 The Remainder and Factor Theorems

6. 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 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, , and −

7. 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

8. Example 3: Divide the polynomial.
2 𝑥 4 −4 𝑥 𝑥 2 +3𝑥−11; 𝑥 2 −2𝑥+7
2 𝑥 2
− 1
𝑥 2 −2𝑥+7
2 𝑥 4 −4 𝑥 𝑥 2 +3𝑥−11
−2 𝑥 4 +4 𝑥 3 −14 𝑥 2
2 𝑥 4 −4 𝑥 𝑥 2
− 𝑥 2 +3𝑥−11
+ 𝑥 2 −2𝑥+7
− 𝑥 2 +2𝑥−7
𝑥 −4
2 𝑥 4 −4 𝑥 𝑥 2 +3𝑥−11 𝑥 2 −2𝑥+7 =2 𝑥 2 −1+ 𝑥−4 𝑥 2 −2𝑥+7

10. 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

11. 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 − 𝑥 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− =5 𝑥 2 +𝑥+4− 2 2𝑥−3

13. 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 𝑥 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)

14. 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)

16. −1
math!
2-3 Assignment: TX p115, 4-44 EOE