Question

PLEASE HELP ME!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
I WILL AWARD MEDAL!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Part 1: Find the polynomial f(x) that has the roots of –3, 5 of multiplicity 2. (4 points)

Part 2: Explain how you would verify the zeros of f(x). (4 points)
@Loser66 @primeralph @satellite73

Answer 1

(x+3)(x-5)
Plug in -3 and/or 5 and the result should yield 0. That's how you check.

Answer 2

I am totally lost.

Answer 3

maps.google.com

Answer 4

HA.
I do not think they have the blueprints to my house.

Answer 5

@Loser66 Please post your idea here so we can talk about it.

Answer 6

you can handle everything by yourself, right? no need to put many recipe into one meal, right?

Answer 7

You always need a little Happy Feet.

Answer 8

see primeralph's smartscore!! it scared me, hehhehehe

Answer 9

@Loser66 Your idea could be right too.

Answer 10

@magbak , any options given?

Answer 11

No @primeralph I would have provided them if their were.

Answer 12

@magbak There are two possibilities:
(x+3)(x-5) or (x+3)^2(x-5)^2

Answer 13

Depending on how the question points.

Answer 14

@satellite73

Answer 15

@satellite73 @primeralph @KingGeorge @radar

Answer 16

@zepdrix

Answer 17

@some_someone

Answer 18

@amistre64 @ash2326

Answer 19

@magbak Do you understand the question? Have you started solving it?

Answer 20

No I do not even understand the problem.

Answer 21

ok, I'll explain. We are give the roots of a polynomial. Do you understand roots/zeros of a polynomial?

Answer 22

Please explain it to me.

Answer 23

Suppose I have a simple polynomial
\[x-2\]
The root or zero is the value of x for which the polynomial becomes zero. Can you tell what's the root here?

Answer 24

@magbak

Answer 25

Yes I am sorry I was afk.

Answer 26

Um I do not know.

Answer 27

x-2
What value of x will make (x-2) zero?
so
\[x-2=0\]
Can you find value of x from here?

Answer 28

x=2

Answer 29

Good, so x=2 is the root of (x-2). Do you get this?

Answer 30

yes Ido. It is that sipl.

Answer 31

Sorry typo. Yes I do. It is that Simple.

Answer 32

How do you become a moderator?

Answer 33

yes, now suppose we have a quadratic equation
\[x^2+3x+2\]
There would be two values of x for which this polynomial would become zero. Those are x=-1 and x=-2
How do we find that? that is not required for this problem

Answer 34

Oh um I can tell you wait one sec please.
But first can I get my final answer to my original question.

Answer 35

If we have the roots, we can find the polynomail
x=-1 and x=-2 are the roots, then the polynomial is
\[(x-(-1) )(x-(-2))\]
or
\[(x+1)(x+2)\]
Multiply and you'll get x^2+3x+2

Answer 36

We'll work on the question, do you understand till here?

Answer 37

Yes.

Answer 38

Suppose a root has a multiplicity of 2, let's say x=-2, then the polynomial can be found by
\[ (x+2)(x+2)\]
Can you multiply these and find the polynomial?

Answer 39

Yes.

Answer 40

Can you post it here, after solving

Answer 41

Yes ok.

Answer 42

x^2 +4x +4

Answer 43

Good, our question says
Find the polynomial f(x) that has the roots of –3, 5 of multiplicity 2. (4 points)

Answer 44

So can you please sum up the final answer into one post PELASE.

Answer 45

Can you try to frame the polynomial from this?

Answer 46

What do you mean by frame.

Answer 47

I mean find

Answer 48

Oh ok.

Answer 49

No I can not.

Answer 50

Well I do not know how to.

Answer 51

Suppose a root has a multiplicity of 2, let's say x=-2, then the polynomial can be found by
(x+2)(x+2)

now the roots are differen x=-3 and x=5

Answer 52

Ok now what do I do.

Answer 53

just a min, I'll call someone to help you

Answer 54

@ash2326 See the problem too?

Answer 55

Yes, I have seen it. @satellite73 Sir could you please help? I have to bounce

Answer 56

OK I have been here for like 1 hour and no answer that is bad.

Answer 57

Is the polynomial x^2 +4x +4

Answer 58

Either that or (x+3)^2(x-5)^2

Answer 59

It must only be one it can not be 2

Answer 60

@KingGeorge @satellite73 @UnkleRhaukus @dumbcow @kropot72

Answer 61

@satellite73 @primeralph I REALY NEED HELP I AM GOING TO FRY

Answer 62

@ajprincess @timo86m

Answer 63

@magbak I'm sorry, but the question is ambiguous.

Answer 64

@Noura11

Answer 65

@Luigi0210

Answer 66

Is the question :
Find the polynomial f(x) that has the roots of –3, 5 of multiplicity 2. (4 points)

Answer 67

How did this question get solong?

Answer 68

*so long

Answer 69

I do not know.

Answer 70

@Noura11

Answer 71

I thought @primeralph had this no problem, huh.

Answer 72

Well he said it is ambiguous

Answer 73

@Noura11 this is the second part
Part 2: Explain how you would verify the zeros of f(x).

Answer 74

because the multiplicity of each root is 2, then the polynomial can be like this :
\[f(x)=(x-(-3))^\color{red}2(x-1)^\color{red}2\]
so :
\[f(x)=(x+3)^2(x-1)^2\]

Answer 75

Are you in the FLVS?

Answer 76

Yes I am.

Answer 77

is august 1 the last day for you?

Answer 78

Yes it is.

Answer 79

And I am stuck on this and like 20 more lessons of algebra 2 and If I do not finish it will look bad on my freshmen record.@Sloth7

Answer 80

tell your school maybe they can do an exception

Answer 81

@Noura11 so it is not x^2 +4x +4

Answer 82

No, it's not !

Answer 83

how about x^2 - 3x - 10

Answer 84

I already asked. @Noura11 can you sum up the entire answer for part one and 2 in one post please I have been studying since 12 in the morning and it is now 1 at night please help me.

Answer 85

f(x) = (x+3)(x-5)^2 then expand
f(-3) = 0
f(5) = 0

Answer 86

I asked for an extension that is what I meant.

Answer 87

. . . . . .. . . . . . . . . . . ,.-‘”. . . . . . . . . .``~.,
. . . . . . . .. . . . . .,.-”. . . . . . . . . . . . . . . . . .“-.,
. . . . .. . . . . . ..,/. . . . . . . . . . . . . . . . . . . . . . . ”:,
. . . . . . . .. .,?. . . . . . . . . . . . . . . . . . . . . . . . . . .\,
. . . . . . . . . /. . . . . . . . . . . . . . . . . . . . . . . . . . . . ,}
. . . . . . . . ./. . . . . . . . . . . . . . . . . . . . . . . . . . ,:`^`.}
. . . . . . . ./. . . . . . . . . . . . . . . . . . . . . . . . . ,:”. . . ./
. . . . . . .?. . . __. . . . . . . . . . . . . . . . . . . . :`. . . ./
. . . . . . . /__.(. . .“~-,_. . . . . . . . . . . . . . ,:`. . . .. ./
. . . . . . /(_. . ”~,_. . . ..“~,_. . . . . . . . . .,:`. . . . _/
. . . .. .{.._$;_. . .”=,_. . . .“-,_. . . ,.-~-,}, .~”; /. .. .}
. . .. . .((. . .*~_. . . .”=-._. . .“;,,./`. . /” . . . ./. .. ../
. . . .. . .\`~,. . ..“~.,. . . . . . . . . ..`. . .}. . . . . . ../
. . . . . .(. ..`=-,,. . . .`. . . . . . . . . . . ..(. . . ;_,,-”
. . . . . ../.`~,. . ..`-.. . . . . . . . . . . . . . ..\. . /\
. . . . . . \`~.*-,. . . . . . . . . . . . . . . . . ..|,./.....\,__
,,_. . . . . }.>-._\. . . . . . . . . . . . . . . . . .|. . . . . . ..`=~-,
. .. `=~-,_\_. . . `\,. . . . . . . . . . . . . . . . .\
. . . . . . . . . .`=~-,,.\,. . . . . . . . . . . . . . . .\
. . . . . . . . . . . . . . . . `:,, . . . . . . . . . . . . . `\. . . . . . ..__
. . . . . . . . . . . . . . . . . . .`=-,. . . . . . . . . .,%`>--==``
. . . . . . . . . . . . . . . . . . . . _\. . . . . ._,-%. . . ..`

Answer 88

@Noura11

Answer 89

For the part 2 we can verify the roots of f(x) by calculating f(-1) and f(3).
And if we find them equal to zero, then -1 and 3 are certainly roots of f(x)

Answer 90

What about part 1

Answer 91

@AccessDenied @Noura11 @watermelon14 @tylerj37
@RANE

Answer 92

I am so sorry but I haven't learn this yet :(