Question

PLEASE HELP ME WITH THIS ASSIGNMENT.

1.A mysterious box is delivered to the dinner party you are attending. The label on the box says that the volume of a box is the function f(x) = x3 + 3x2 – 10x – 24. To open the box, you need to identify the correct factors of f(x). Party-goers offer up solutions, and it is your job to find the right ones.


Their suggestions are:
•(x – 1)
•(x + 2)
•(x – 3)
•(x + 4)
•(x + 6)
•(x – 12)


List the correct factors. Then justify your selections with complete sentences.

2.Three party-goers are in the corner of the ballroom having an intense argument. You walk over to settle the debate. They are discussing a function g(x). You take out your notepad and jot down their statements.

•Professor McCoy: She says that 2 is a zero of g(x) because long division with (x + 2) results in a remainder of 0.
•Ms. Guerra: She says that 2 is a zero of g(x) because g(2) = 0.
•Mr. Romano: He says that 2 is a zero of g(x) because synthetic division with 2 results in a remainder of 0.


Correct the reasoning of any inaccurate reasoning by the party-goers in full and complete sentences. Make sure you reference any theorems that support your justifications.

3.Dr. Collier summons you over to his table. He wants to demonstrate the graph of a fourth-degree polynomial function, but the batteries in his graphing calculator have run out of juice. Explain to Dr. Collier how to create a rough sketch of a graph of a fourth-degree polynomial function.

4.Mrs. Collins is at the table with you and states that the fourth-degree graphs she has seen have four real zeros. She asks you if it is possible to create a fourth-degree polynomial with only two real zeros. Demonstrate how to do this and explain your steps.

Answer 1

the first one is number 3

Answer 2

How do you know that?

Answer 3

i did this before

Answer 4

im not lieng

Answer 5

Not what I'm looking for.. I'm looking for how you got that..

Answer 6

oh ok

Answer 7

I'm awful at this, so, I need all the help I can get.

Answer 8

Oh, @ranga, I see you!
Help please! :)

Answer 9

To check if (x-a) is a factor put x = a in f(x) and see if it becomes zero.

For example, the first choice is (x-1). Is (x-1) a factor?
Put x = 1 in f(x)
f(x) = x3 + 3x2 – 10x – 24
f(1) = 1^3 + 3(1)^2 - 10(1) - 24 = 1 + 3 - 10 - 24 = -30.
f(1) is not 0 and therefore (x-1) is NOT a factor.

Try each of the other choices in the same manner and you will find out which ones are the factors.

Answer 10

Oh, okay, I see how you did that. I'll try that. Gimme a sec. (Sorry for the slow reply, I had to step away for a few moments)

Answer 11

(x + 2)
f(x) = x3 + 3x2 - 10x - 24
f(2) 2^3 + 3(2)2 - 10(2) - 24 = -34
Not zero, so it's not a factor.

Did I do that one right?

Answer 12

(x - 3)
f(x) = x3 + 3x^22 - 10x - 24
f(3) = 3^3 + 3(3)^2 - 10(3) - 24 = 0
A factor!

(x + 4)
f(x) = x3 + 3x^2 - 10x - 24
f(4) = 4^3 + 3(4)^2 - 10(4) - 24 = 48
Not zero, so not a factor.

Okay, I'm thinking all the others won't be factors, either, because they just get bigger.

Answer 13

poor thing

Answer 14

lol

Answer 15

Poor thing? lol

Answer 16

For (x+2) and (x+4), you want to plug in -2 and -4, respectively, because those are the numbers that make the monomial 0. Likewise, like you did earlier, for (x-1), you plug in 1 into the equation because 1 makes (x-1)=0.

Answer 17

Wait -- So I'm supposed to put in the opposite sign given? I thought no sign at all. Ugghhh I hate math!

Answer 18

Yes you have to put the opposite sign.

Answer 19

Ohh, alright, let me try those again, then.

Answer 20

f(x) = x3 + 3x^2 - 10x - 24

(x + 2)
f(-2) = -2^3 + 3(-2)^2 - 10(-2) - 24 = 0
Factor!

(x - 3)
f(3) = 3^3 + 3(3)^3 - 10(3) - 24 = 54
Not a factor

(x + 4)
f(-4) = -4^3 + 3(-4)^3 - 10(-4) - 24 = -240
Not a factor

(x + 6)
f(-6) = -6^3 + 3(-6)^3 - 10(-6) - 24 = -828
Not a factor

(x - 12)
f(12) = 12^3 + 3(12)^3 - 10(12) - 24 = 6768
Not a factor

I do that right??

Answer 21

The second term is 3x^2 and NOT 3x^3. In all your calculations you have used 3x^3.

(x - 3)
f(3) = 3^3 + 3(3)^2 - 10(3) - 24 = 27 + 27 - 30 - 24 = 54 - 54 = 0
Factor!

Answer 22

OMG are you kidding me? I USED THE WRONG EXPONENT?! *headwall* OMG I'll fix it. >.<

Answer 23

But you got (x+2) as a factor.
I showed (x-3) as a factor.
There is just one more to be found because this is a cubic equation and at most it will have three roots (meaning three x values that will make f(x) zero).

Answer 24

I can't believe I did that..

(x + 4)
f(-4) = -4^3 + 3(-4)^2 - 10(-4) - 24 = 0
Factor!

(x + 6)
f(-6) = -6^3 + 3(-6)^2 - 10(-6) - 24 = -72
Not a factor

(x - 12)
f(12) = 12^3 + 3(12)^2 - 10(12) - 24 = 2016
Not a factor

I do that right this time?? xDD

Answer 25

Yes. So the factors are:
x3 + 3x^2 - 10x - 24 = (x-3)(x+2)(x+4)

"justify your selections with complete sentences."

(x-3) is a factor because x =3 makes f(x) zero and therefore (x-3) must be a factor.
(x-3) is a factor means x3 + 3x^2 - 10x - 24 = (x-3){some other factor)
And if we put x = 3 on the expression on the right (x-3) = 0 and the product of (x-3){some other factor) will be 0.

Answer 26

Sweet! Okay, that takes care of that. What about number 2?

2.Three party-goers are in the corner of the ballroom having an intense argument. You walk over to settle the debate. They are discussing a function g(x). You take out your notepad and jot down their statements.

•Professor McCoy: She says that 2 is a zero of g(x) because long division with (x + 2) results in a remainder of 0.
•Ms. Guerra: She says that 2 is a zero of g(x) because g(2) = 0.
•Mr. Romano: He says that 2 is a zero of g(x) because synthetic division with 2 results in a remainder of 0.


Correct the reasoning of any inaccurate reasoning by the party-goers in full and complete sentences. Make sure you reference any theorems that support your justifications.

Answer 27

I see you there, @Ranga.. LOL

Answer 28

Oh, I typed an answer but it automatically refreshed the page, lost my reply and came back to this page. So even though you saw me here I was not on this page. I was answering another question.

Answer 29

Oh dear.. OS has been such a pain today, it seems. I wonder what's doing on with it? Anyway, all help is dearly appreciated.. Sorry to be such a bother!

Answer 30

2 is a zero to f(x) means that (x-2) is a factor of f(x).
That means if you divide f(x) by (x-2) the remainder must be 0.
But Professor McCoy says that 2 is a zero of g(x) because long division with (x + 2) results in a remainder of 0.
That is wrong. Should not divide by (x+2).
2 is a zero means long division by (x-2) will yield a remainder of 0.

Ms. Guerra says that 2 is a zero of g(x) because g(2) = 0.
That is a correct statement. That is the definition of a root (or a zero). What is the value of x that will make g(x) = 0 and that value is called the zero or the root of g(x).

Mr. Romano says that 2 is a zero of g(x) because synthetic division with 2 results in a remainder of 0.
That is also a correct statement. If 2 is a root (or zero) then synthetic division with 2 will yield a remainder of 0.

Answer 31

Oh, wow, okay, thank you. Um.. What about number 3, @Ranga?


3.Dr. Collier summons you over to his table. He wants to demonstrate the graph of a fourth-degree polynomial function, but the batteries in his graphing calculator have run out of juice. Explain to Dr. Collier how to create a rough sketch of a graph of a fourth-degree polynomial function.

Answer 32

First look at the sign of the leading coefficient. For a fourth-degree polynomial function, the leading term will be ax^4.
If a is positive, the graph will open upwards.
If a is negative, the graph will open downwards.
Since this is a fourth-degree polynomial (an even degree), the ends of the graph will behave much like the parabola ax^2. If a > 0, both ends will be up.
If a < 0 both ends will be down.
A fourth-degree polynomial will have four roots.
If all four roots are real, the graph will cross or touch the x axis a maximum of 4 times.
Complex roots always occur in pairs. So a fourth-degree polynomial can have 4 complex roots, 2 complex roots or 0 complex roots.
So the possibilities of roots (or zeros) of the fourth-degree polynomial are:
4 real roots 0 complex roots ==> touches /crosses x axis at most four times
2 real roots 2 complex roots ==> touches /crosses x axis at most twice
0 real roots 4 complex roots ==> does not touch /cross the x axis at all.

Finally you can put x = 0 and evaluate the function to find out the y-intercept and that is where the function will cross the y axis.

Answer 33

Hnnn, Ranga, I love you for helping me so much! How about the last one, number 4?

4.Mrs. Collins is at the table with you and states that the fourth-degree graphs she has seen have four real zeros. She asks you if it is possible to create a fourth-degree polynomial with only two real zeros. Demonstrate how to do this and explain your steps.

Answer 34

Yes, Mrs. Collins: It is possible to create a fourth-degree polynomial with ONLY two real zeros.

A a fourth-degree polynomial with have a total of 4 roots. If only 2 roots are real then the other 2 must be complex.

How to create such a polynomial?

First find a quadratic function with two complex roots and then find another quadratic function with two real roots. Multiply the two quadratics together and you will have a fourth-degree polynomial with 2 real roots and 2 complex roots.

Answer 35

Ahhh, thank you, Ranga! Thank you so much! I wish I could give you a boatload of medals for this. I'm so grateful!

Answer 36

Example: (x-1)(x-2) will have two real roots 1 and 2 because if you put either value for x, (x-1)(x-2) will be 0.
Multiply it out and you have the quadratic x^2 - 3x + 2 with two real roots

Take another quadratic, for example, x^2 + 2x + 3. This has two complex roots. How do I know that? Because the discriminant (b^2 - 4ac) is negative for this function.

To get the 4th degree polynomial multiply the above two quadratics and you will end up with a fourth-degree polynomial with 2 real roots and 2 complex roots.

(x^2 - 3x + 2)(x^2 + 2x + 3)

IDK if they actually want a fourth-degree polynomial or just the steps as to how you would go about creating one. But if they do actually want one just multiply the above two quadratics and simplify it.

Answer 37

You are welcome.

Answer 38

This material is a quiz on Florida Virtual school. Providing the answers is a violation of the code of conduct.

Answer 39

oh F off. @dkeating

Answer 40

The information included in this question is the intellectual property of Florida Virtual School. Posting questions from any FLVS course and receiving answers is considered cheating.

Answer 41

Shut the F off please. Thank you @dashreeve

Answer 42

stfu