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3-4 Notes Part 1

3-4 Notes Part 1

Assessment

Presentation

Mathematics

9th - 12th Grade

Medium

Created by

Alyson Foley

Used 2+ times

FREE Resource

30 Slides • 13 Questions

1

3-4 Notes Part 1

Please make sure you take notes throughout the activity

Slide image

2

Multiple Choice

If one number is divisible by another, what should be the remainder if you were to divide them?

1

1

2

0

3

undefined

4

2

3

This also applies when dividing polynomials.

4

Multiple Select

How can you tell if (x-5) is a factor of x3-125? Select all that apply.

1

Divide. If the remainder is 0 then (x-5) is a factor.

2

Divide. If the remainder is 1 then (x-5) is a factor.

3

Use the remainder theorem by plugging in 5. If the remainder if 0 then (x-5) is a factor.

4

Use the remainder theorem by plugging in -5. If the remainder if 0 then (x-5) is a factor.

5

Multiple Choice

Is (x-5) a factor of x3-125?

1

yes

2

no

6

Slide image

7

We can use division to factor polynomials of higher degrees.


8

In the previous example, (x3-125)÷(x-5) = x2+5x+25.

Working backwards, that means multiplying (x2+5x+25) and (x-5) is equal to x3-125.

9

Slide image

The factored form of x3-125 is (x-5)(x2+5x+25).

10

If x2+5x+25 was factorable, we would keep factoring.

Since it's not, the expression is factored completely.

11

Multiple Choice

One factor is x-3. Find the other factor of

 p(x)=x3+2x213x6p\left(x\right)=x^3+2x^2-13x-6  


To do this problem, use synthetic division to divide p(x) by x-3.  What is the result?

1

 x3x210x+24x^3-x^2-10x+24  

2

 x3+5x2+2xx^3+5x^2+2x  

3

 x2x10+24x3x^2-x-10+\frac{24}{x-3}  

4

 x2+5x+2x^2+5x+2  

12

Since you were told x-3 is a factor, you know the remainder has to be 0. If it isn't, you made a mistake somewhere.

13

Multiple Choice

Select the factored form of

 p(x)=x3+2x213x6p\left(x\right)=x^3+2x^2-13x-6  


1

 (x3)(x3x210x+24)\left(x-3\right)\left(x^3-x^2-10x+24\right)  

2

 (x3)(x3+5x2+2x)\left(x-3\right)\left(x^3+5x^2+2x\right)  

3

 (x3)(x2x10+24x3)\left(x-3\right)\left(x^2-x-10+\frac{24}{x-3}\right)  

4

 (x3)(x2+5x+2)\left(x-3\right)\left(x^2+5x+2\right)  

14

Fill in the Blank

Review:

Factor. Type your answer with NO SPACES.

x2-8x-20

15

Multiple Select

Review:

Now solve by factoring.

x2-8x-20=0

1

10

2

-2

3

-10

4

2

16

Remember, to solve by factoring, set each factor equal to 0 and solve for x.

  • x-10=0

  • Add 10 to get x=10

  • x+2=0

  • Subtract 2 to get x=-2

17

Slide image

Read the problem. I will go through this one step at a time.

18

Slide image

Read the problem again and determine the year that the pond will dry up. You will select an answer on the next slide.

19

Multiple Choice

Question image

What year will the pond dry up?

1

2016

2

2010

3

2030

4

2006

20

The problem states that the water level of the pond has been measured since 2006. That means the y-intercept is the water level in 2006. The pond's depth reaches 0 feet at an x-value of 10. Since x is years, 10 years since 2006 would be the year 2016.


21

Slide image

22

 d(x)=x3+16x274x+140d\left(x\right)=-x^3+16x^2-74x+140  

  • Remember the first step in factoring is to take out what each term has in common

  • These terms don't have anything in common but the polynomial starts with a negative

  • I will factor out -1

23

Slide image

24

 d(x)=1(x316x2+74x140)d\left(x\right)=-1\left(x^3-16x^2+74x-140\right)  

  • Now I will use the graph to continue factoring

  • Remember our x-intercept is x=10

  • If the x-intercept is x=10, that means it should have come from a factor of (x-10)

  • When x-10=0, x=10

25

Slide image

That means we can divide x3-16x2+74x-140 by x-10 and get a remainder of 0.

26

Multiple Choice

Use synthetic division to divide

 (x316x2+74x140)÷(x10)\left(x^3-16x^2+74x-140\right)\div\left(x-10\right)  

1

 x26x+14x^2-6x+14  

2

 x36x2+14xx^3-6x^2+14x  

3

 x226x+3343480x10x^2-26x+334-\frac{3480}{x-10}  

4

 x226x+3343480x+10x^2-26x+334-\frac{3480}{x+10}  

27

Slide image


28

Slide image

This is the factored form of d(x).

29

Slide image

The amount of voters on a single day is represented by the graph, where x is hours and y is number of voters.

30

Slide image

We can tell a lot about this scenario just by looking at the graph.

31

Slide image

Approximately how many voters arrived when the voting location opened? Go to the next slide to answer.

32

Multiple Choice

Approximately how many voters arrived when the voting location opened?

1

56

2

4

3

8

4

26

33

Slide image

Since we can't have negative time, 0 hours is when the voting location opens. The y-value is 56.

34

Slide image

Approximately how many voters were there 2.6 hours after opening? Go to the next slide to answer.

35

Multiple Choice

Approximately how many voters were there 2.6 hours after opening?

1

4

2

10

3

26

4

34

36

Slide image

At about 2.6 hours, the y-value is approximately 4.

37

Slide image

How long is this polling place open? Go to the next slide to answer.

38

Multiple Choice

How long is this polling place open?

1

8 hours

2

2.6 hours

3

6.1 hours

4

4.7 hours

39

Slide image

The number of people becomes negative after 8 hours, which doesn't make sense.

40

Slide image

The equation of the graph is p(x)=x3+13x247x+56p\left(x\right)=-x^3+13x^2-47x+56  

41

Multiple Choice

 p(x)=x3+13x247x+56p\left(x\right)=-x^3+13x^2-47x+56  

Use the graph to factor p(x).

1

 (x25x+7)(x8)\left(x^2-5x+7\right)\left(x-8\right)  

2

 1(x25x+7)(x8)-1\left(x^2-5x+7\right)\left(x-8\right)  

3

 1(x+8)(x3+21x2215x+1776)-1\left(x+8\right)\left(-x^3+21x^2-215x+1776\right)  

4

 1(x8)(x2+21x215)-1\left(x-8\right)\left(x^2+21x-215\right)  

42

Slide image

43

Make sure you also complete 3-4 Notes Part 2!

3-4 Notes Part 1

Please make sure you take notes throughout the activity

Slide image

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