Question

Can someone help me with this please? I don't want the answers, I want someone to help me understand how to solve the question.

A function is shown below:

f(x) = x3 + 2x2 - x - 2

Part A: What are the factors of f(x)? Show your work.

Part B: What are the zeros of f(x)? Show your work.

Part C: What are the steps you would follow to graph f(x)? Describe the end behavior of the graph of f(x).

Thanks!

Answer 1

@zepdrix @mathmale @mathstudent55 @triciaal @ganeshie8 @Directrix

Answer 2

Figure out part A? :)
Looks like grouping again.

Answer 3

Yeah, I was thinking of doing that later. I thot I took it out of the question...

Eh I'll just do it here

Answer 4

Well you need part A for the rest of the problem XD
so make sure you can do that

Answer 5

Alright

Answer 6

f(x) = (x3 + 2x2 - x - 2)

First group:

f(x) = (x3 + 2x2) - (x - 2)

Find GCF

First Binomial x2

Second Binomial -1

Factor out:

f(x) = x2(x + 2) -1(x + 2)

(x2 - 1)(x + 2)

Is this correct @zepdrix I think I might have done something wrong...

Answer 7

Looks good so far :)
We can go a little bit further.
Here is something to think about:

Answer 8

wait wait wait wait

Answer 9

No wait >:U\[\large\rm x^2-1\quad=\quad x^2-1^2\]

Answer 10

See your next step? :)

Answer 11

but theres something incorrect

f(x) = x2(x + 2) -1(x + 2)

Factoring out -1 would give

f(x) = x2(x + 2) -1(-x + 2)

Not

f(x) = x2(x + 2) -1(x + 2)

Answer 12

D:

Answer 13

oh i missed that boo boo.

You should've grouped like this:

Answer 14

\[\large\rm (x^3+2x^2)+(-x-2)\]You have to input a plus sign to put the brackets the way you would like.
You can't just drop brackets like this:\[\large\rm (x^3+2x^2)-(x-2)\]I don't think that's what you did, just a note.

Answer 15

\[\large\rm x^3+2x^2-x-2\]\[\large\rm (x^3+2x^2)+(-x-2)\]\[\large\rm x^2(x+2)-1(x+2)\]\[\large\rm (x^2-1)(x+2)\]These are the correct steps.
Confused by any of this?

Answer 16

@zepdrix Alright, I get it so far. It makes sense.

Answer 17

So for A, the factors are (x2−1)(x+2) correct @zepdrix ???

Answer 18

No, we would like `linear factors`, so we factor further.

Answer 19

0-0

Answer 20

\[\large\rm x^2-1\quad=\quad x^2-1^2\]Remember your `difference of squares` formula?

Answer 21

\[\large\rm a^2-b^2=(a-b)(a+b)\]

Answer 22

so how would i use that in this???

Answer 23

btw thx zep, ur awesome. :)

Answer 24

@zepdrix hmmm...x^2−1=x2−1^2. I don't really get how you suddenly got 1^2 oh 1 * 1 is still 1 nvm.

Alright that makes sesnse so far.

Equation is

a2−b2=(a−b)(a+b)

Im gonna go on a limb here and say a2 is x2 and b2 is 1^2.

So x^2 - 1^2 = (x - 1)(x + 1)

Therefore, the fully factored form would be

(x - 1)(x + 1)

Correct @zepdrix ?

Answer 25

@zepdrix So the factors of f(x) would be (x - 1)(x + 1), correct?

Answer 26

Good good good.
That gives us our fully factored expression for part A:
(x-1)(x+1)(x+2)

Answer 27

Alright, the fully factored form is (x-1)(x+1)(x+2) for Part A?

Answer 28

\[\large\rm (x^2-1)(x+2)\]\[\large\rm (x-1)(x+1)(x+2)\]Yes.

Answer 29

got it, thanks!

Answer 30

Alright, so to calculate the x intercepts you have to make the equation equal to 0, so i think to calculate the 0's you have to make x = 0 first, and then y = 0? Or is that not true...

Answer 31

\[\large\rm y=(x-1)(x+1)(x+2)\]Looking for zeros, or x-intercepts,\[\large\rm 0=(x-1)(x+1)(x+2)\]Applying your `Zero-Factor Property`:\[\large\rm 0=(x-1),\qquad\qquad\qquad 0=(x+1),\qquad\qquad\qquad 0=(x+2)\]And solve for x in each case.

Answer 32

Setting x=0 is a separate issue.
That gives us the `y-intercept`.
it doesn't appear that they want information though,
so we don't need to do that.

Answer 33

Alright, so the 0s are just the x intercepts, correct @zepdrix ?

Answer 34

Yes, these words mean all the same thing:
Zeros, x-intercepts, solutions.

They'll use those three different words a lot,
so get comfortable with the fact that they mean the same thing.

Answer 35

\[0=(x−1) 0=(x+1) 0=(x+2)\]

Alright

In the first equation x = 1

Second equation x = -1

Third equation x = -2

Answer 36

Good :) You've got your zeroes.

Answer 37

HOORAY! On to part C

Answer 38

Part C: What are the steps you would follow to graph f(x)? Describe the end behavior of the graph of f(x).

Answer 39

To graph it?
You would label your x-intercepts to start.

Answer 40

f(x) = x3 + 2x2 - x - 2

Right right, that is a good start lol

Answer 41

Wait, this has 3 intercepts, does that mean 2 vertexes?

Answer 42

Yes, and without the use of calculus, we can't graph those very easily.
So I don't think they want you to worry about that too much.

Answer 43

Hold on for a minute please

Answer 44

\(\large\rm f(x) = x^3 + 2x^2 - x - 2\)
If you let x=0, you have:
\(\large\rm f(x) = 0 + 0 - 0 - 2\)
So you can also graph your y-intercept if you like.
y=-2

Answer 45

@zepdrix Alright I'm back. I'm fairly sure I somehow found the vertex in my other question tho

Answer 46

Forgot how...

Answer 47

@zepdrix Dont I need to plot all the intercepts to be able to graph f(x)? Cause then I can just draw a line in between, right?

Answer 48

No, not in this type of problem you didn't.
The vertices aren't even rational numbers for this one. :)
Really weird decimal values.

Answer 49

Yes, you can connect things somewhat effectively after you do that.

Answer 50

But you also need to know the `end behavior` of the function.

Answer 51

but how would i graph without the vertexes?

Answer 52

You would just sort of... "guess" how high the vertices go.
You wouldn't be able to do the same thing you did with a parabola,
vertex form is nice and easy with a parabola :))
not with this bad boy.

Answer 53

Sec, lemme get some graph paper for us.

Answer 54

@zepdrix Sorry, bathroom break.

Alright I see the 0

Answer 55

@zepdrix Did you get banned?

Answer 56

Lol, what?
Why would I ever get banned? XD

Answer 57

I don't know, conversing with an OS fugitive whose account has been banned?

Answer 58

lol

Answer 59

Alright, I still don't understand how you would be able to graph this without knowing the vertexes @zepdrix .

Answer 60

So if we graph the intercepts, that gives us a basic idea.