Question

Using the given zero, find all other zeros of f(x)

Answer 1

2+6i is a zero of \[f(x)=x^4-16x^3+124x^2-624x+1440\]

Answer 2

k so you know another zero right?

Answer 3

No, 2+6i is the only known zero

Answer 4

what i mean is this: if \(2+6i\) is a zero, then also you know an other zero

Answer 5

or not?

Answer 6

Complex zeroes always come in conjugate pairs. So if x = i is a zero, then so is x = -i. This means (x+i)(x-i) = x^2 + 1 can be factored from the original polynomial.

So use polynomial long division (or synthetic division) to divide the original polynomial by x^2 + 1. You'll be left with a quadratic. Set it equal to 0 and use the quadratic formula to find the other two zeroes

Answer 7

if \(a+bi\) is a zero, then so is its conjugate \(a-bi\) so you know one, you know two

Answer 8

unfortunately your zero is no \(i\) it is \(2+6i\) so you have some more work to do

Answer 9

is it clear that the other zero is \(2-6i\) ?

Answer 10

yeah, I remember that now. so, what do I do from here with the two zeros?

Answer 11

ok now it gets annoying, but we can do it

Answer 12

you know two factors
\[(x-(2+6i))(x-(2-6i))\] what we want is an easy way to multiply that out, to see what the quadratic is
there are a couple more or less snap ways to do this

Answer 13

one way is to work backwards to find the quadratic
put
\[x=2+6i\] then subtract \(2\) get
\[x-2=6i\] square and get
\[(x-2)^2=-36\] or
\[x^2-4x+4=-36\] then add \(36\) to get
\[x^2-4x+40\] as your quadratic

Answer 14

there is in fact an even easier way
memorize that if \(a+bi\) is a solution of a quadratic, then it is
\[x^2-2ax+(a^2+b^2)\] which in your case is
\[x^2-2\times 2x+2^2+6^2=x^2-4x+40\] this is very easy, but it requires memorizing it

Answer 15

did i lose you yet?

Answer 16

nope, I follow it!

Answer 17

ok good
we still have work to do

Answer 18

you have to factor
\[x^4-16x^3+124x^2-624x+1440=(x^2-4x+40)(something)\]

Answer 19

now you could use polynomial long division if you like
if you are good at that fine
i hate it

Answer 20

the best way for me to do this is just to think what the "something" has to be
it is a quadratic and it must look like \(ax^2+bx+c\)
fortunately \(a\) and \(c\) are pretty obvious

Answer 21

pretty clear that the first term has to be \(x^2\) right? because you have to get \(x^4\) when you multiply out

Answer 22

how about \(c\)? is it obvious what \(c\) is?

Answer 23

I'm not quite sure...

Answer 24

\[(x^2-4x+40)(x^2+bx+c)=x^4-16x^3+124x^2-624x+1440\]right?

Answer 25

where is the constant \(1440\) going to come from?

Answer 26

So would you divide 1440 by 40?

Answer 27

yup

Answer 28

To get 36

Answer 29

yes
now comes the only real thinking part

Answer 30

\[(x^2-4x+40)(x^2+bx+36)=x^4-16x^3+124x^2-624x+1440\] we still don't know \(b\)

Answer 31

but we have a choice of ways to find it
where is \(-16x^3\) going to come from?

Answer 32

if it is not clear, let me know
there are two terms that will give you \(x^3\)

Answer 33

I know I should know this but I can't think of what it is

Answer 34

the \(x^2\) from the first factor \(x^2-4x+40\) times the \(bx\) from the \(x^2+x+36\) is one of them

Answer 35

you see what i mean? that will give you an \(x^3\) term, namely \(bx^3\)

Answer 36

there is another one as well
do you see the other one?

Answer 37

i made a typo there, i means the \(bx\) term from \(x^2+bx+36\) sorry

Answer 38

let me know if you see the other two terms that will give \(x^3\) when you multiply

Answer 39

\[(\color{red}{x^2}\color{blue}{-4x}+40)(\color{blue}{x^2}+\color{red}{bx}+36)=x^4-16x^3+124x^2-624x+1440\]

Answer 40

so would it just be 16 or am I overlooking something?

Answer 41

the two terms that give \(x^3\) are
\[\color{blue}{-4x^3}+\color{red}{bx^3}\]

Answer 42

and on the right you have \(-16x^3\) telling you
\[-4+b=-16\] and so
\[b=-12\]

Answer 43

final answer will therefore be
\[(x^2-4x+40)(x^2-12x+36)=x^4-16x^3+124x^2-624x+1440\]

Answer 44

this is not the way it is usually done,
usually you are taught polynomial long division, but that is really easy to screw up and very cumbersome, just like regular long division
although you probably have not seen this, lets recap what we did to get
\[x^2-12x+36\] because it was not that hard

the first term must be \(x^2\) because you have to get \(x^4\) when you multiply
the constant must be \(36\) because \(40\times 36=1440\)
and finally \(-4+b=-16\iff b=-12\)

Answer 45

btw not to draw this out, but you could have used the \(x\) term instead of the \(x^3\) term
just the numbers are more annoying
\[-4\times 36x+40bx=-624x\]

Answer 46

I totally get it now! Thank you so much!! :)

Answer 47

yw
you can always use division if you like (ick)
have fun

Answer 48

division is hard, i try to avoid it haha

Answer 49

this problem was hard
look at all the steps!

Answer 50

I wrote them all down as we went :)