Posted by **jordan** on Thursday, September 24, 2009 at 8:48pm.

Find a polynomial of lowest degree with only real coefficients and having the given zeros.

-2+i, -2-i, 3, -3

- algebra 3 -
**Emily**, Thursday, September 24, 2009 at 8:57pm
Zeros are also "x = ..." statements, so x = -2 + i, x = -2 - i, etc...

Since you need REAL coefficients in the polynomial, change the x = -2 + i and x = -2 - i statements to "clean" polynomials, like this:

x = -2 + i

x + 2 = i (square each side to get rid of the i)

(x+2)^2 = i^2

x^2 + 4x + 4 = -1

==> x^2 + 4x + 5 = 0

x = -2 - i

(x+2)^2 = (-i)^2

I'm not going to go further with this because you get the same exact equation as the one before. So now when you multiply all the zeroes together, you get this:

0 = (x^2 + 4x + 5)(x-3)(x+3)

Multiply it out and you get:

x^4 + 4x^3 - 4x^2 - 36x - 45 = 0

Does this make sense?

- algebra 3 -
**Christiaan**, Thursday, September 24, 2009 at 9:05pm
The polynomial would have to have 4 zeros, meaning it would have to be a polynomial of the 4th degree. The general form for a polynomial of the 4th degree with zeros a, b, c and d would be:

f*(x-a)*(x-b)*(x-c)*(x-d)

where f is a random real number (lets take this to be 1 in this case).

So, if we fill in the zeros you were given we get that:

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

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

When we multiply the first two factors and the last two, we get:

(x^2 + 2x + ix + 2x + 4 + 2i -ix - 2i +1) * (x^2 - 9) =

(x^2 + 4x + 5)*(x^2 - 9) =

(x^4 - 9x^2 + 4x^3 - 36x + 5x^2 -45) =

x^4 + 4x^3 - 4x^2 -36x - 45

this is a polynomial of the 4th degree which has the given values as its zeros

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