Find a polynomial function with integer coefficients that has the given zeros.?

Question

X = 2, 3i, -3i

I don't know what I'm susposed to do.

Answer 1

Ok, it has these three roots:

x = 2, 3i, -3i.

A polynomial root always has the following form:

(x + a) = 0

With a being the negative of the root.

So, we write all those:

(x - 2) = 0
(x - 3i) = 0
(x + 3i) = 0

Well, we multiply all roots to give us the polynomial, we'll multiply them two at a time, we'll use the two 3i's first using the FOIL method:

(x - 3i)(x + 3i)

x^2 + 3ix - 3ix - 9i^2 ; i = sqrt(-1) ; i^2 = -1

= x^2 + 9

Now, we take the equation up there and multiply it by the final root also using the FOIL method:

(x^2 + 9)(x - 2)

x^3 - 2x^2 + 9x - 18

So, we have the final polynomial, which we'll call f(x):

f(x) = x^3 - 2x^2 + 9x - 18

Answer 2

Zeros mean roots. this means that your polynomial function could be 0 once you replace in x = 0, -5 or a million+i (i did not be conscious of i replaced into an integer). So then y = x(x+5)(x-a million-i) may well be a answer. besides the incontrovertible fact that it HAS to have INTEGER coefficients. This function might have complicated coefficients. So subsequently we would desire to introduce yet another root that makes use of the conjugate of a million-i So then y = x(x+5)(x-a million-i)(x-a million+i).

Answer 3

If r is a root of a polynomial P(x) then x-r is a factor.
So P(x) = (x-2)(x+3i)(x-3i)=
(x-2)(x²+9) = x³ - 2x² + 9x - 18.

Answer 4

You would multiply these back together in three parentheses. You need to multiply (x - 2)(x - 3i)(x + 3i) together.
I see that the last two of these are the sum of two squares, so I simplify this to (x^2 + 9)(x - 2)
Multiplying this out I get x^3 - 2x^2 + 9x - 18 = 0 as the polynomial you need.

Answer 5

One thing we know about the poluynomial that we are looking for is that is has roots at x=2, 3I, and -3i
to be a root that means that x-root=0
so one polynomial that has roots at x=2, 31, and -3i is
f(x)=(x-2)(x-3i)(x+3i)
so lets expand it out
f(x)=(x-2)(x^2-3ix+ix-9i^2)=(x-2)(x^2+9) since i^2=-1
now expand some more
f(x)=x^3+9x-2x^2-18=x^3-2x^2+9x-18 is one answer
to get other answers multiply what we are calling f(x) by any factor of x, or any polynomial with integer coefficients

Answer 6

P(x) = (x-2)(x -3i)(x+3i)
expand it

P(x) = x^3 -2x^2 +9x -18