Question

please help. i'll post the picture in a bit

Answer 1

#6

Answer 2

\[x ^{5} + 9x ^{3} = x ^{3}(x^{2}+9)\]

we are looking for zeros, for which x is the given expression equal to zero
first zero:
\[x^{3} = 0\]\[x = 0\]

second and third zero:
\[x^2 + 9 = 0\]\[x^2 = - 9\]\[x = \pm \sqrt{-9}\]\[x = \pm \sqrt{9}i = \pm 3i\]

Answer 3

for the top part it can still reduce to x^(3)(x-3)(x+3)

Answer 4

@e.mccormick can you explain this to me?

Answer 5

Which part were you trying to reduce more?

Answer 6

That us a sum of squares, which does not factor. You are thinking about the difference of squares.

Answer 7

the fist part.

Answer 8

first part

Answer 9

That is why stijena only factored like this:

\(x ^{5} + 9x ^{3} =0\implies x ^{3}(x^{2}+9)=0\)

Let me go over where you made a small mistake:
\((x^{2}+9)\ne (x-3)(x+3)\)
\((x^{2}-9)= (x-3)(x+3)\)

See that now?

Answer 10

but isnt the answer just x^3(x^(2)+9)=0
howcome you get (x^(2)+9) and (x^(2)-9) ???

Answer 11

You said you could factor \((x^2+9)\) more. I am just trying to point out why you can't factor it that way.

Answer 12

ohh i see

Answer 13

Right. And if you have \(x^3(x^2+9)=0\) then you set each factor to 0 and solve for x.

Answer 14

so would it be x^(3)=0
and (x^(2)+9)=0

Answer 15

\(x^3(x^2+9)=0\) Yes, in factoring that, there is:
\(x^3=0\) and \((x^2+9)=0\)
Now, we could factor this more:
\(x=0\), \(x=0\), \(x=0\) and \((x^2+9)=0\)

That is why stijena skipped to just \(x=0\) for one answer.

Answer 16

so there is actually four answers?

Answer 17

Five, it is a 5th degree poly, so there are going to be 5 answers. But in this case, 3 of them are the same, so it can be shoretened as 3 answeres.

That leaves you solving: \((x^2+9)=0\) for any other two answers.

Answer 18

x^2+9=0
x^(2)=-9
x=+- 3i?

Answer 19

Yah, that was that part of the explanation. Do you get how the i part works? How stijena got that out of the root? It has to do with root multiplication rules.

Answer 20

yeah when there is a negative in a squareroot you use an i

Answer 21

Kk. Yah, it is because \(\sqrt{-9}=\sqrt{9}\cdot \sqrt{-1}\) and \(\sqrt{-1}=i\) so as long as you have that part, it is easy to do the rest like you did.

Answer 22

what about for #9?

Answer 23

You do any U substitutions?

Answer 24

i got x^(2)(x^(2)+2+1)
x^(2)(x^(2)+3)

Answer 25

substitution?

Answer 26

Becuase of the +1, it does not work that way. See:
\(x^2(x^2+2+1)\) means I would multiply through by \(x^2\) so:
\(x^4+2x^2+1x^2\)

That is not the original.

Answer 27

Let me show you the start of a substitution.

\(x^4+2x^2+1\)
Let \(u=x^2\)

\(u^2+u+1=0\)

Factor that, then put \(x^2\) back for any Us, then factor more as needed.

Answer 28

so what do we do in this case? do we just leave it as it is because 1 doesn't have an x

Answer 29

That is what the sub. lets you address.

Answer 30

...

Answer 31

?

Answer 32

im confuse

Answer 33

You can't just factor out the \(x^2\) because of the 1, so you can't solve it that way.

Answer 34

so i leave it like that?

Answer 35

Do you see my u posting? Lets look at that.

Answer 36

which one :O

Answer 37

This one:
Let me show you the start of a substitution.

\(x^4+2x^2+1\)
Let \(u=x^2\)

\(u^2+u+1=0\)

Factor that, then put \(x^2\) back for any Us, then factor more as needed.

Answer 38

(u+1)(u-1)=0
u=-1, 1

Answer 39

Don't go quite that far.

Answer 40

\((u+1)(u-1)=0\)
Now, put \(x^2=u\) back in there. So wherever there is a u, replace it with \(x^2\).

Answer 41

Hmmm.... and is that the right factoring?

Answer 42

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

Answer 43

Oh, I see.... I missed a 2 in what I posted.

Answer 44

\(x^4+2x^2+1\)
Let \(u=x^2\)

\(u^2+2u+1=0\)
So:
\((u+1)(u+1)\)

Answer 45

okay

Answer 46

Now, when you put the \(x^2\) back into that, where do things go?

Answer 47

(x^(2)+1)(x^(2)+1)

Answer 48

Right. And you can set those to 0. Either one, since they are duplicates.

Answer 49

answer for a: + - i
b: (x-i)^(2)(x+i)^(2)

Answer 50

that was in the back of the book

Answer 51

Which makes sense, since solving the \(x^2=1\) gets you there.

Answer 52

but i still dont get how they get those answers v.v

Answer 53

Well, you have the \(x=\pm i\) but there is a square in there we sort of ignored. Might be because of that.

Answer 54

???????????????????????

Answer 55

Because there were 2 that were the same.

Answer 56

AH. Factor completely. OK.

Answer 57

So that is why the b looked a little odd. I was just thinking about the zeros.

Answer 58

i think i just got more confuse

Answer 59

Do you see where what we did leads to this?
\(x^2+1=0\implies x^2=-1\implies x=\sqrt{-1}\)

Answer 60

The process again, just to make it clear how we got here:

\(x^4+2x^2+1\)
Let \(u=x^2\)
\(u^2+2u+1=0\)
\((u+1)(u+1)=0\)
Replace \(x^2\) for u
\((x^2+1)(x^2+1)=0\)
\((x^2+1)^2=0\)
\((x^2+1)=\sqrt{0}\)
\((x^2+1)=0\)
That is where we drop an answer out, which shows up in b, but not a.

Answer 61

than there is no answer for a?

Answer 62

No, no. That got us to where it is set to 0.

Answer 63

\(x^2+1=0\)
\(x^2+1=0\)
\(x^2=-1\)
\(x=\sqrt{-1}\)
\(x=\pm i\)

Answer 64

okay

Answer 65

Look at that book again. Is b this:
\((x-i)^2(x+i)^2\)

Hmmm... lets multiplyt that out to see what it does.
\((x-i)(x-i)(x+i)(x+i)\)
\((x-i)(x+i)(x+i)(x-i)\)
\((x^2-i^2)(x^2-i^2)\)
\((x^2+1)(x^2+1)\)

Answer 66

it say its (x-i)^(2)(x+1)^(2)

Answer 67

(x+1)?? Earlier you said (x+i).

Answer 68

It is a 4th degree problem, so it will have four factors. Whenever a variable is still to a power, it is not fully factored. So while using \(x^2=-1\) foes get you the unique zeros, it does not get you the fully factored form.

That is where the \((x^2+1)(x^2+1)\) comes back in. They are factoring that, which makes 4 terms with single x values. But two of them are with a +i and two with a -i. That is why they regrouped them the way they did.

Answer 69

sorry i meant i

Answer 70

So, you see how the answer to b is related to the middle of getting the answer for a?

Answer 71

yeah

Answer 72

this si still confusing but thanks for trying to teach me

Answer 73

Well, it is i. A complex number. While that replates to the whole way they are used, it also says it for the topic!

Answer 74

This might help a little with seeing the overall topic.
http://www.math.hmc.edu/funfacts/ffiles/10005.1.shtml