x + 5 4x +20x r + 3 3x - 16 3x - +15 -31 - - -. tues 11/24 lesson 5 – 4 learning objective: to...
TRANSCRIPT
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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