Find a polynomial function with real coefficients that has the given zeros. (There are many correct answers.)
2, −2i

Question

Find a polynomial function with real coefficients that has the given zeros. (There are many correct answers.)
2, −2i

zepdrix · Answer

oh oh oh thinking

zepdrix · Answer

I'm not sure of how to do this systematically.. I was just trying to think of something.. then it said it had to be a polynomial.. hmm

But finally, :O i think this works, lemme know what you think.

f(x) = (x-2)(x^2 + 4)

unklerhaukus · Answer

$$(x+ 2i)(x-2i)=0$$$$x^2+4=0$$

anonymous · Answer

I attempted that, webassign didn't like it. I got the question wrong, but my test is tuesday and I'd really like to figure this out.

anonymous · Answer

unkle how does that work?! where would you then get the positive two from, I can see the -2i...

zepdrix · Answer

that's what im wondering :D heh

unklerhaukus · Answer

$$f(x)=x^2+4=0$$
$$f(-2i)=(-2i)^2+4=-4+4=0$$$$f(2i)=(2i)^2+4=-4+4=0$$

oh i thought the roots were $2i,-2i$

zepdrix · Answer

Oh double, try this maybe? :o
$$x^3 - 8 = 0$$

anonymous · Answer

but its 2 and -2i, only one has an imaginary number.

zepdrix · Answer

x^3 - 8 should have 3 roots, 1 real and 2 imaginary
2, 2i, -2i

I think at least :o

anonymous · Answer

Zep's first polynomial works just fine. (Second one doesn't though. (2i)^3 - 8 = -8i - 8.

You aren't going to be able to have anything smaller than a cubic. Complex roots come in pairs (the root and its conjugate).

zepdrix · Answer

Hmm darn :d

unklerhaukus · Answer

$$f(x)=(x-2)(x+2i)=x^2-2x+2ix-4i$$
$$f(2)=((2)-2)((2)+2i)=0(2+2i)=0$$$$f(-2i)=((-2i)-2)((-2i)+2i)=(-2i-2)0=0$$

zepdrix · Answer

That polynomial doesn't have real coefficients Uncle :'O

anonymous · Answer

unkle that's what I put as my first answer as well. it was wrong. it seemed logical to me. simple roots. and zep, I used the first one as well, that wasn't accept, the second IDK

tamtoan · Answer

use original zepdrix answer : (x -2)(x^2 +4) = 0
                                            x^3 - 2x^2 + 4x -8 = 0    <===

anonymous · Answer

tam, that was the answer that killed me. answers I got, x^3 -2x^2-4x+8 ... x^3-2x^2+2x-4 ... I tried putting them as the zeros themselves ....

anonymous · Answer

either way I'm out of guesses on the hw, and I cannot seem to figure this out to save my life, so I guess I'll accept this one wrong on the test tuesday.

tamtoan · Answer

x = -2i  ==> x^2 = 4i^2 = -4 ==> x^2 +4 = 0
x = 2                                       ==>  x     -2 = 0
so P(x) = 0 = (x - 2)(x +2i) = (x -2)(x^2 +4) = x^3 - 2x^2 + 4x - 8

i don't know where the logic fail so that the answer wasn't right, check with your teacher again ? if you got other explanation from your teach, let us all hear and learn from it. thanks

tamtoan · Answer

your teacher i mean.