Question

s

Answer 1

@zepdrix ?

Answer 2

degree of 3 means it will have 3 zeros

complex zeros always come in pairs, so the third zero must be 1+i

Answer 3

The pair for the complex zeroes, pointed out by dumb cow, is called the complex conjugate. It is always the opposite function of the first number. For example, the conjugate of 2+i is 2-i. All we switch is the function, addition or subtraction.

Answer 4

Ahhhh, ok, thanks!

What if it's trying to find the polynomial?

Degree 4 zero -3 -4i; -1 multiplicity 2?

Answer 5

So how many zeroes should there be?

Answer 6

yea, 4. you just forgot to put a variable in for your complex values. It's structured the same as the real ones. (x-____) or (x+____)

Answer 7

then instead of "totaling them up" you need to Multiply them out using FOIL

Answer 8

alright, so just to be sure, then answer would be x^2 + 2x + 26, correct?

Answer 9

no, if it is a fourth degree poly, it has to have degree 4. That is a second degree

Answer 10

So, write out the factored form to start

Answer 11

with all of the x's

Answer 12

oh crap, you're right

so it should have been (x+1)(x+1)(x-3-4i)(x+3-4i)

Answer 13

another approach for dealing with the complex zeros is this:

\[x = 3 \pm 4i\]
\[x-3 = 4i\]
\[(x-3)^2 = (4i)^2\]
\[x^2 -6x+9 = -16\]
\[x^2 -6x +25\]

Then the 4th degree polynomial is:
\[(x+1)^2 (x^2-6x+25)\]

Answer 14

yes that is one way ... if you foil it out you get same thing, but foiling with the imaginary part can be tedious and lead to mistakes

Answer 15

Actually, there are 3 zeros. If one of the zeros is 1-i, then another is the complex conjugate of 1-i, which is 1+i. The third zero is 2. Therefore, one factor of this polynomial is (x-2). Find the factors that stem from x= 1+i and x=1-i.

Therefore

Answer 16

There is just one "remaining zero." What is it? (Everything has already been typed out, above.)

Answer 17

@umulas
You are making an error in the second problem, still:

if the zero is \(r\), then the factor is \((x-r)\)

for the complex conjugate pair \(a\pm bi\) you will have factors \[(x-(a+bi))(x-(a-bi))= (x-a-bi)(x-a+bi)\]

but in your proposed polynomial you did the equivalent of \[(x-a-bi)(x+a-bi)\]

I will expand your polynomial:

\[(x+1)(x+1)(x-3-4i)(x+3-4i)\]\[ = (x+1)^2(x^2+3x-4ix-3x-9+12i-4ix-12i+16i^2)\]\[=(x+1)^2(x^2-8ix-9+16i^2) \]\[= (x+1)^2(x^2-8ix-9+16(-1))\]\[=(x+1)^2(x^2-8ix-25)\]Notice that you've got an \(i\) stuck in there, and nothing is going to get rid of it. That's because you changed the wrong sign when constructing the complex conjugate.

Answer 18

@UMULAS Do you understand your question now or do you need more help? :)