Question

f(x)=-2x^3-7x^2-166x+85
Find the complex zeros of f. Repeat any zeros if their multiplicity is greater than 1.

Answer 1

@satellite73

Answer 2

@jim_thompson5910 @zepdrix

Answer 3

@ganeshie8 @myininaya

Answer 4

yikes a third degree polynomial

Answer 5

i have no idea how you are supposed to do this, but apparently the one real zero is at \(\frac{1}{2}\) which means it factors as
\[-(2x-1)(\text{something})\]

Answer 6

the "something" will be a quadratic with two complex zeros
they have to be conjugates so they are different, making the multiplicity of each zero as 1

Answer 7

me, i would cheat because this is donkey work
http://www.wolframalpha.com/input/?i=-2x^3-7x^2-166x%2B85

Answer 8

I also have to write it in factored form...

Answer 9

yeah it is a thankless task to find the zeros of a third degree polynomial

Answer 10

since on zero is \(\frac{1}{2}\) you can factor it as
\[-(2 x-1) (x^2+4 x+85)\]

Answer 11

* one zero

Answer 12

the other two complex ones you find either using the quadratic formula to solve
\[x^2+4x+85=0\] or, what is easier because 4 is even, completing the square

Answer 13

I have an example
So for the equation -2x^3-27x^2-116x+65 it was factored down to (-2x+1)(x-(-7+4i))(x-(-7-4i))

Answer 14

similarly you can factor this one once you find the complex zeros

Answer 15

so\[x^2+4x+85=0\]
\[x^2+4x+4=-85+4\]
\[(x+2)^2=-81\]
\[x+2=\pm9i\]
\[x=-2\pm9i\]

Answer 16

So\[x=\frac{ 1 }{ 2 },-2+9i, -2-9i\]

Answer 17

\[-2x^3-7x^2-166x+85=(2x-1)(x-(-2+9i))(x-(-2-9i))\]

Answer 18

@satellite73

Answer 19

I am not sure if I wrote it correctly in factored form. Is it (2x-1)(x-(-2+9i))(x-(-2-9i))

Answer 20

Hmm your roots look good!

Factored? mmm yesss good job.

Answer 21

Hard to read through all those brackets sometimes lol

Answer 22

Well it is telling me it is wrong...

Answer 23

Hmm :d

Answer 24

wait satellite said -(2x-1) so I guess I missed the - sign?

Answer 25

I got it correct?

Answer 26

*.. not a ?

Answer 27

The negative floating out front shouldn't affect your roots 0_o that's very strange....

Answer 28

I got it correct it should have been a (-2x+1) not (2x-1)

Answer 29

Those are the same :U But whatever lol.
They're being silly I think.

Answer 30

No actually I forgot to put the - sign so I just distributed it in to it...

Answer 31

I mean....about half way through the problem, after you find your 1/2 root,
you get something of this form:\[\Large\rm 0=(2x-1)(-x^2-4x-85)\]Pulling a negative out of the bracketed part:\[\Large\rm 0=-(2x-1)(x^2+4x+85)\]And then just divide both sides by -1.
I don't see why they want you to hold on to that lonely negative sign >.<
So weird!

Answer 32

Thanks again both of you @satellite73 and @zepdrix