Question 1191999
<pre>
Instead of doing your problem for you, I'm going to do 
one exactly like it instead, step by step.  Here's the one I'm 
going to do. Use it as a model for yours.</pre>
Form a fifth-degree polynomial function with real coefficients 
such that 6i, 1-4i, and -4 are zeros and f(0)=7344
<pre>

If a polynomial with real coefficients has a complex imaginary zero,
its conjugate is also a zero.

Since 6i is a zero, so is -6i.
Since 1-4i is a zero, so is 1+4i.

x-6i, x-(-6i), x-(1-4i), x-(1+4i), x-(-4) are all factors. Simplifying them:

x-6i, x+6i, x-1+4i, x-1-4i, x+4

So f(x) must be some constant "a" times the product of those factors:

{{{"f(x)"}}}{{{""=""}}}{{{a(x-6i^"")(x+6i^"")(x-1+4i^"")(x-1-4i^"")(x+4^"")}}}

Multiply that out:

It makes it a little easier to group the real parts of conjugates:

{{{"f(x)"}}}{{{""=""}}}{{{a(x-6i^"")(x+6i^"")((x-1)+4i^"")((x-1)-4i^"")

(x+4^"")}}}
 
{{{"f(x)"}}}{{{""=""}}}{{{a(x^2-36i^2)((x-1)^2-16i^2)

(x+4^"")}}}

Since i<sup>2</sup> = -1

{{{"f(x)"}}}{{{""=""}}}{{{a(x^2+36)((x-1)^2+16)

(x+4^"")}}}

{{{"f(x)"}}}{{{""=""}}}{{{a(x^2+36)((x^2-2x+1)^""+16)

(x+4^"")}}}

{{{"f(x)"}}}{{{""=""}}}{{{a(x^2+36)(x^2-2x+17)

(x+4^"")}}}

{{{"f(x)"}}}{{{""=""}}}{{{a(x^4-2x^3+17x^2+36x^2-72x+612)(x+4^"")}}}

{{{"f(x)"}}}{{{""=""}}}{{{a(x^4-2x^3+53x^2-72x+612)(x+4^"")}}}

{{{"f(x)"}}}{{{""=""}}}{{{a(x^5-2x^4+53x^3-72x^2+612x+4x^4-8x^3+212x^2-288x+2448)}}}

{{{"f(x)"}}}{{{""=""}}}{{{a(x^5+2x^4+45x^3+140x^2+324x+2448)}}}

Since f(0)=7344,

{{{"f(0)"}}}{{{""=""}}}{{{a(0^5+2(0)^4+45(0)^3+140(0)^2+324(0)+2448)}}}

{{{7344}}}{{{""=""}}}{{{2448a}}}

{{{3}}}{{{""=""}}}{{{a}}}

So

{{{"f(x)"}}}{{{""=""}}}{{{3(x^5+2x^4+45x^3+140x^2+324x+2448)}}}

{{{"f(x)"}}}{{{""=""}}}{{{3x^5+6x^4+135x^3+420x^2+972x+7344)}}}

Edwin</pre>