# Thread:

1. ## Finding polynomials

Hello, I cant figure this problem out for the life of me. Could someone help me?

Find the polynomial P(x) of degree 4 with integer coefficients, and zeroes $3-2i$ and 3 with 3, a zero of multiplicity of 2.

Thanks in advance...

2. Originally Posted by Godzilla
Hello, I cant figure this problem out for the life of me. Could someone help me?

Find the polynomial P(x) of degree 4 with integer coefficients, and zeroes $3-2i$ and 3 with 3, a zero of multiplicity of 2.
Zeros are, therefore, 3-2i, 3+2i, 3, and 3.

$(x - 3 + 2i)(x - 3 - 2i)(x - 3)(x - 3)$

or

$x^4-12x^3+58x^2-132x+117$

3. can someone show me how to factor that? I am confused because of the (3-2i) and (3+2i)? Thanks!!

4. This polynomial:

Originally Posted by Haversine
$x^4-12x^3+58x^2-132x+117$
is actually a result of this factorization:

Originally Posted by Haversine
$(x - 3 + 2i)(x - 3 - 2i)(x - 3)(x - 3)$
That is, if I know the roots are A, B, C, and D, I just set up a series of polynomial factors (x - A)(x - B)(x - C)(x - D).

There's nothing special about the (x - 3 + 2i) factor, for instance. All it's saying is that I want a term that equals zero when "3 - 2i" is substituted for x.

The other imaginary term pops in there because the complex conjugate will always be a root.

5. The whole point is that $x^4-12x^3+58x^2-132x+117$ is from the product (x-3+2i)(x-3- 2i)(x-3)(x-3). For this problem, at least, it is not a matter of "factoring" but of "multiplying"!

As far as (x-(3- 2i))(x-(3+2i))= (x-3+2i)(x-3-2i) is concerned, think of it as a "product of sum and difference= difference of squares": $((x-3)- 2i)((x-3)+ 2i)= (x-3)^2- (2i)^2= x^3- 6x+ 9- 4= x^2- 6x+ 5$
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