Question

Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form.
5, -3, and -1 + 3i
how do I do this?

Answer 1

So you have three zeros listed.
\(\Large\rm x=5,\qquad x=-3,\qquad x=-1+3\mathcal i\)
Recall that `complex roots always come in pairs`.
So the `conjugate` of our complex root will also be a root of this polynomial.

Answer 2

So the minimum degree polynomial will have these 4 roots/zeros,\[\Large\rm x=5,\qquad x=-3\]\[\Large\rm x=-1+3\mathcal i,\qquad x=-1-3\mathcal i\]

Answer 3

With the first root, I'll subtract 5 from each side,\[\Large\rm x=5\qquad\to\qquad x-5=0\]I'm gonna put some brackets around it just so it's clear that this will be a `factor` of my polynomial.\[\Large\rm (x-5)=0\]

Answer 4

We'll do the same with the others, and then we have to multiply a bunch of stuff out :\ blah... ummm.. let's deal with the complex roots first I guess...

Answer 5

What do you think miss Chris? :) This process a little confusing?

Answer 6

(x-5)(x+3)(x-(1+3i))(x-(1-3i)
I thought I would have to do something like this
but i dont understand your way

Answer 7

I only did the first factor, (x-5)
Yes, doing the next one x=-3 becomes (x+3)=0
and the others as well, x=-1+3i becomes (x+(1-3i))=0
and yah, you multiply them all together to get your polynomial,\[\Large\rm (x-5)(x+3)\left[x+(1-3\mathcal i)\right]\left[x+(1+3\mathcal i)\right]=0\]K looks good so far :)
Oh oh your 1's should be -1 if you want to do it that way.
x-(-1+3i) and x-(-1-3i).
I think it's a little easier with the addition sign though.

Answer 8

So then uhhh, multiply out the stuff, ya?
With the complex I guess we get something like this:\[\Large\rm x^2+(1-3\mathcal i)x+(1+3\mathcal i)x+(1-3\mathcal i)(1+3\mathcal i)\]Kind of a pain to FOIL all that out, I know :( But can you see how I did that?

Answer 9

ok so the final answer would be f(x)= x^4 - 9x^2 - 50x - 150 ?

Answer 10

Hmmmm that doesn't look quite right :o

Answer 11

You should be getting a third power in there somewhere, and your first and second power coefficients look a little off.

Answer 12

For the complex part, it simplifies further to,\[\Large\rm x^2+2x+10\]

Answer 13

\[\Large\rm (x-5)(x+3)\color{orangered}{\left[x+(1-3\mathcal i)\right]\left[x+(1+3\mathcal i)\right]}=0\]
\[\Large\rm (x-5)(x+3)\color{orangered}{(x^2+2x+10)}=0\]

Answer 14

\[\Large\rm (x^2-2x-15)\color{orangered}{(x^2+2x+10)}=0\]And then expand a little further, ya? :)

Answer 15

AHHHH I input the equation wrong :(
no no no,
you are correct,

x^4-9x^2-50x-150
ya ya ya ya ya, good job!
sorry sorry :)

Answer 16

ohh ok thanks so much :)