Question

Using the given zero, find one other zero of f(x). Explain the process you used to find your solution.

1 - 6i is a zero of f(x) = x^4 - 2x^3 + 38x^2 - 2x + 37.

Answer 1

Because 1-6i is a zero, 1+6i also must be a 0

Answer 2

I know how to do the 1-6i part, and I got the x^2+x+37, but how do I find the other roots?

Answer 3

quadratic

Answer 4

I know that... But is there any way to find the (x-a) roots?

Answer 5

no, 37 is prime

Answer 6

unless you did the 1-6i part wrong lol

Answer 7

(x-(1-6i)) and (x+(1-6i)) right?

Answer 8

I mean when you find all the zeros for one of these, you get (x-a), (x-a), (x-(1-6i)), and (x-(1+6i))..

Answer 9

Im not sure I follow you

Answer 10

its not necessarily both (x-a)

Answer 11

I know that.. that's just the formula. Normally when they give you a problem like this, they want you to do the opposite, and they give you all of the roots. For example, earlier I did a problem, and the roots wer (x-3), (x+13), (x+(5+4i)), and (x-(5-4i)). See what I mean?

Answer 12

@Isaiah.Feynman

Answer 13

Shoot! I forget this..

Answer 14

:c Think harddd.

Answer 15

here we go again
it is actually very easy

Answer 16

\[5+4i\] is a zero so lets work backwards
\[x=5+4i\\
x-5=4i\\
(x-5)^2=(4i)^2=-16\\
x^2-10x+25=-16\\
x^2-10x+41=0\]

Answer 17

easiest of all is to memorize that if \(a+bi\) is the zero of a quadratic, then it is
\[x^2-2ax+(a^2+b^2)=0\]

Answer 18

in this case you have \(1-6i\) and \(1+6i\) so your quadratic is
\[x^2-2x+(1^2+6^2)=x^2-2x+17\]

Answer 19

oops that was wrong
\[x^2-2x+(1^1+6^2)=x^2-2x+37\]

Answer 20

Why didn't I get the 2x in the middle, or why did you?

Answer 21

ok i see that you almost had that
not quite, but almost
so you have
\[(x^2-2x+37)(ax^2+bx+c)=x^4 - 2x^3 + 38x^2 - 2x + 37\] and it should be clear that \(a=1\) and \(b=1\) making it
\[(x^2-2x+37)(x^2+bc+1)=x^4 - 2x^3 + 38x^2 - 2x + 37\]

Answer 22

i got it by memorization, but working backwards also give it to you
\[x=1+6i\\
x-1=6i\\
(x-1)^2=(6i)^2=-36\\
x^2-2x+1=-36\\
x^2-2x+37=0\]

Answer 23

assuming you get that, are you good to this step
\[(x^2-2x+37)(x^2+bc+1)=x^4 - 2x^3 + 38x^2 - 2x + 37\] ? because we are still not done

Answer 24

Oh, I see what I did. Nevermind. Continue.

Answer 25

On the second group of three, how did you know the last number is a 1?

Answer 26

first of all a made a stupid typo, let me correct it\[(x^2-2x+37)(x^2+bx+1)=x^4 - 2x^3 + 38x^2 - 2x + 37\]

Answer 27

take a look and see if it is obvious or not that the last number in the second trinomial has to be 1

Answer 28

Haha. You're good. And I see the 1 now, because we have to keep that 37.

Answer 29

ok good
just like we know the first term is \(x^2\) and not for example \(3x^2\)

Answer 30

Exactly. Let's continue.

Answer 31

now how about \(b\) ?

Answer 32

we have a couple choices lets do an easy one
on the right hand side of the equal sign you have \(-2x\)
where is the \(x\) term going to come from on the left hand side?

Answer 33

b would be a 2, right? So that you would multiply it by the x^2, and get 2x^3.

Answer 34

when you multiply out on the left, there will be two place to get an \(x\) term:
\(37\times bx\) and \(1\times (-2x)\)

Answer 35

that means \(37bx-2x\) has to be \(-2x\) i.e.
\[37b-2=-2\] making \(b=0\)

Answer 36

So lost me there. Hang on...

Answer 37

there are a couple places on the left to get an \(x^3\) term, not just one

Answer 38

when you multiply \((x^2-2x+37)(x^2+bx+1)\) there will be two product that will give you \(x^3\)
they come form \(x^2\times bx\) and also \(-2x\times x^2\)

Answer 39

this means \(bx^3-2x^3=-2x^3\) (because that is what you have on the right) so once again \(b=0\)

Answer 40

Where did you get a -2?

Answer 41

on the right hand side or on the left hand side? the \(-2x^3\) is on the right hand side because it is

Answer 42

lets look at this again
\[(x^2-2x+37)(x^2+bx+1)\]

Answer 43

I wrote it down wrong. Ugh. Okay, I'm caught up now.

Answer 44

ok so you see that if you multiply out on the left, there will be two products to give an \(x^3\) term
one is \(x^2\times bx\), the other is \(-2x\times x^2\)
when you go to combine like terms you will have
\[bx^3-2x^2\] but this must match on the right with \(-2x^3\) making \(b=0\)

Answer 45

typo again , i meant \(bx^3-2x^3\) sorry

Answer 46

Yes, so b=0. Got it.

Answer 47

right, and we didn't have to pick \(x^3\) term to look at, we could have done it as i did at first matching up the \(x\) terms left and right

Answer 48

way above i wrote \(37bx-2x\) which has to be \(-2x\) once again making \(b=0\)

Answer 49

you can pick whichever one you like, it has to work with all of them

Answer 50

Okay, well now we got back to the original problem, but how do we find the zeros?

Answer 51

now we know \(b=0\) it is easy

Answer 52

\[(x^2-2x+37)(x^2+1)=0\]

Answer 53

we already have the zero of \(x^2-2x+37\) as \(1\pm 6i\) now set \[x^2+1=0\] and solve in your head

Answer 54

Oh, I knew that. Blonde moment. Okay, so they are going to be imaginary numbers.

Answer 55

yes, two rather famous ones

Answer 56

So they would be 1+6i, 1-6i, and +i, and -i?

Answer 57

yes

Answer 58

Awesome! Thanks so much! God bless you! :D

Answer 59

lets check it
www.wolframalpha.com/input/?i=+x^4+-+2x^3+%2B+38x^2+-+2x+%2B+37

Answer 60

yw, my pleasure

Answer 61

I have that site on my desktop from one of the last times you helped me. :P Lol. It gave me the same, (x^2+1). We're good. :)

Answer 62

I'll be back tomorrow. I have lots of quizzes to take. Ha. :p