1. Warm – up #3 Divide

x + 5 4 𝑥2+23 𝑥−164x

4 𝑥2+20x

R

+ 3

3x - 16 3x

-

+15-31

-

- -

= 4x + 3-

Homework Log

Tues

11/24

Lesson 5 – 4

Learning Objective: To divide polynomials by synthetic division

Hw: Pg. 308 #21-29 ODD, 35-39 ODD, 41cc, 55-63 ODD

11/24/15 Lesson 5 – 4 Dividing Polynomials Day

2Algebra II

To divide polynomials by synthetic division

To apply the Remainder Theorem

To understand the Factor Theorem

Learning Objective

1.

Divide using Synthetic Division

=

x = 44𝑥3 𝑥2 𝑥¿2 −13 26 −24

2

8

−5−20624

0𝑥2 𝑥¿ 𝑅

2.

Divide using Synthetic Division

=

x = -2

-2𝑥3 𝑥2 𝑥¿02 −4 12

−8

16

18−36−4080

92𝑥2 𝑥¿ 𝑅

𝑥4

4

4𝑥3

−8

3.

Divide using Synthetic Division

=

x = 77𝑥2 𝑥¿3 −29 56

3

21

−8−560

𝑥¿ 𝑅

4a) Divide by long division

3x + 1 +0𝑥3

3 𝑥4 +

- 2

-6 + -6

-

-23

-

- -

+ 7

+

3 + - -

6 + 0

+ 2

6x + 2- -

- 2

-

4b)

Divide using Synthetic Division

*** Need to divide by 3 b/c div by 3x+1

x = 𝑥3 𝑥2 𝑥¿−5 17 0

−6

2

3−16

−2

−2𝑥2 𝑥¿ 𝑅

𝑥4

3

3𝑥3

−1

=

If you divide a polynomial P(x) by x – a, then the remainder is P(a)

5. Given , what is ? (Find remainder)

a) Method 1: Plug in (Using Remainder Theorem)

= 243 – 2(27) – 9 + 2

= 182

Remainder Theorem

5. Given , what is ?

b) Method 2: Synthetic Substitution

Remainder Theorem

3𝑥3 𝑥2 𝑥¿−2 −1 0 2

721206060

180182

1

13

𝑅

𝑥4

0

3

𝑥5

9

𝑃 (3 )=182

6. , what is ?Remainder Theorem

-4𝑥3 𝑥2 𝑥¿−28 0 5 20

00005

−200

1

1−4

𝑅

𝑥4

−3

−7

𝑥5

28

0

If you divide a polynomial P(x) by x – a, and the remainder is

ZERO, then x – a is a factor of P(x).

Ex (from #6): a factor of

Factor Theorem

7. Is a factor of ? If it is, find the remaining factors.

Factor Theorem

-5𝑥3 𝑥2 𝑥¿17 −38 0

1

−5

2−10−48240

0𝑥2 𝑥¿ 𝑅

R 0, so x + 5 is a factor of P(x)!

+ 2x - 48 Factor more!

#7 continued… Factors.. =

= (x + 5)(x – 6)(x + 8)

Factor Theorem

Assignment:

modified for period 3:didn’t get through all examples/notes.

Pg. 308 #21-27 ODD, 41cc, 57-63 ODD

8a) is the volume of a box in cubic inches, and the length is (x + 1) inches. What are the dimensions of the box?

Word Problem

-1𝑥3 𝑥2 𝑥¿16 116

1

−1

5−56−6

0𝑥2 𝑥¿ 𝑅 + 5x + 6

Factor more!

8. V = = (x + 1)( + 5x + 6) = (x + 1)(x + 3)(x + 2) = (length)(width)(height)

Length = (x + 1) inches (given)Width = (x + 3) inchesHeight = (x + 2) inches

Word Problem

8b) What are the actual dimensions if the length is 10 inches?Length = x + 110 = x + 1x = 9Width = (x + 3) = (9 + 3) = 12 in.Height = (x + 2) = (9 + 2) = 11 in. 10” x 12” x 11”

Word Problem

is the volume in cubic inches of a box, and the length is (x + 7) inches. What are the other dimensions of the box?

What are the actual dimensions if the length is 12 inches?

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