What is the complete polynomial with these certain zeroes?

Question

Okay, so they actually give you the zeroes in this problem. It says, not exactly:

The zeroes of a given polynomial of the 4th degree are 3 and (3-i). The 3 has a multiplicity of 2. What polynomial has the zeroes of those stated above?

Answer 1

The FOIL part is actually quite messy. I'd suggest noting the following :

(x + a + bi)(x + a - bi) = x^2 + 2ax + (a^2 + b^2)

For your problem, a = -3 and b = 1. Therefore, the quadratic part with the imaginary roots is x^2 - 6x + 10. Combine that with the quadratic with the real roots, and you get:

(x^2 - 6x + 10)(x^2 - 6x + 9)

If my math is right, that expands to:

x^4 - 12x^3 + 55x^2 - 114x + 90

OK, let's check with x = 3. The positive terms come to 81 + 495 + 90 = 666. The negative terms come to -324 + -342 = -666. OK, the math checks, but these two 666 really creep me out...

Answer 2

(x-3)^2 * (x-(3+i)) * (x - (3-i))
To do this one you need the (x-3)^2 so that the 3 can have a multiplicity of 2, and you need (3+i) to go with the (3-i) because the roots of real coefficient polynomials always occur in conjugate pairs.
You can multiply this out if you want, or if your teacher requires. When you multiply it out, be careful, the product of two conjugates is the difference of two squares. Since you have i's in there it will come out to be the -sum- of two squares. If you do it correctly you'll get a 4th degree polynomial with all real coefficients.

Answer 3

Since they give you the roots, write them as factors:

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

Remember multiplicity of two means that you use that factor twice.

Now multiply the factors together to get the polynomial.

The first two give:
(x^2-6x+9)

(x^2-6x+9)(x-(3-i))(x-(3+i))
(x^2-6x+9)(x-3+i)(x-3-i)

FOIL from here...