Warm-Up 2/241.

B

12

6

6

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.

2-3 The Remainder and Factor Theorems

−3 𝑥+9

Example 1: Use long division to factor polynomial.6 𝑥3−25 𝑥2+18 𝑥+9 ; (𝑥−3 )

6 𝑥3−25 𝑥2+18 𝑥+9𝑥−36 𝑥2

6 𝑥3−18𝑥2−6 𝑥3+18 𝑥2

−7 𝑥2+18 𝑥+9

−7 𝑥

−7 𝑥2+21𝑥+7 𝑥2−21𝑥

−3

−3 𝑥+9+3 𝑥−90

(𝑥−3 )(6 𝑥2−7𝑥−3)(𝑥−3 )(2 𝑥−3)(3𝑥+1)So there are real zeros at x = 3, , and .

3 𝑥−3

Example 2: Divide the polynomial.9 𝑥3−𝑥−3 ; (3 𝑥+2 )

9 𝑥3+0 𝑥2−𝑥−33 𝑥+23 𝑥2

9 𝑥3+6 𝑥2−9𝑥3−6 𝑥2

−6 𝑥2−𝑥−3

−2 𝑥

−6 𝑥2−4 𝑥+6 𝑥2+4 𝑥

+1

3 𝑥+2−3 𝑥−2−5

9𝑥3−𝑥−33 𝑥+2

=3 𝑥2−2𝑥+1+−5

3𝑥+2,𝑥 ≠− 2

39𝑥3−𝑥−3

3 𝑥+2=3 𝑥2−2𝑥+1− 5

3𝑥+2,𝑥≠− 2

3

𝑥−4

Example 3: Divide the polynomial.2 𝑥4−4 𝑥3+13 𝑥2+3 𝑥−11 ; (𝑥2−2𝑥+7 )

2 𝑥4−4 𝑥3+13 𝑥2+3 𝑥−11𝑥2−2 𝑥+72 𝑥2

2 𝑥4−4 𝑥3+14 𝑥2−2 𝑥4+4 𝑥3−14 𝑥2

−𝑥2+3 𝑥−11

−1

−𝑥2+2𝑥−7+𝑥2−2𝑥+7

2𝑥4−4 𝑥3+13 𝑥2+3 𝑥−11𝑥2−2 𝑥+7

=2 𝑥2−1+𝑥−4

𝑥2−2 𝑥+7

Example 4a: Divide the polynomial using synthetic division.(2 𝑥4−5 𝑥2+5 𝑥−2)÷ (𝑥+2 )

– 5

– 6

– 4

2 0 5

↓

– 2

– 4 8

32 – 1

2

0

– 2

2𝑥4−5 𝑥2+5𝑥−2𝑥+2

=2𝑥3−4 𝑥2+3 𝑥−1

2 𝑥3−4 𝑥2+3 𝑥−1

Example 4b: Divide the polynomial using synthetic division.(10 𝑥3−13𝑥2+5 𝑥−14 )÷ (2 𝑥−3 )

52

6

1

5 − 132 – 7

↓ 152

32

45 – 1

32

5 𝑥2+𝑥+4− 1

𝑥− 32

=5 𝑥2+𝑥+4− 22𝑥−3

(10 𝑥3−13𝑥2+5 𝑥−14 )(2 𝑥−3)

(10 𝑥3−13𝑥2+5 𝑥−14 )÷2(2 𝑥−3)÷2

=5 𝑥3− 13

2 𝑥2+52 𝑥−7

𝑥− 32

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 𝑥4+21𝑥3+25 𝑥2−5𝑥+3 ; (𝑥−1) , (𝑥+3 )

25

50

25

4 21 – 5

↓

3

4 25

504 45

45

48

1

, so is not a factor.

25

6

9

4 21 – 5

↓

3

– 12 – 27

– 2 4 1

– 3

0

– 3

, so is a factor.

𝑓 (𝑥 )=(𝑥+3 )(4 𝑥3+9𝑥2−2𝑥+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 )=2𝑥3−𝑥2−41𝑥−20 ;(𝑥+4) , (𝑥−5 )

– 41

20

– 9

2 – 1 – 20

↓ – 8 36

– 5 2 0

– 4

, so is a factor.

𝑓 (𝑥 )=(𝑥+4 )(𝑥−5)(2 𝑥+1)

– 5

1

2 – 9

↓ 10 5

02

5

, so is a factor.

√−1math!

2-3 Assignment: TX p115, 4-44 EOE