A third degree polynomial equation with rational coefficients has roots -4 and -4i. If the leading coefficient of the equation is 3/2, what is the equation?

Question

tkhunny · Answer

The REALLY important clue is "with rational coefficients".  This means that Complex roots must come in congugate pairs!

-4
-4i
4i <== This is the ne we construed from the hint.

There are the roots.  What are the factors?

anonymous · Answer

(x+4)(x-4i)(x+4i) Is that what you mean?

tkhunny · Answer

Perfect!  Write "=0" and you're almost done.

anonymous · Answer

(x+4)(x-4i)(x+4i)=0

tkhunny · Answer

The problem statement MAY want you to multiply it out to its polynomial form.

anonymous · Answer

When I multiply it out I do not get 3/2 as the leading coefficient, which is what I am supposed to have in the final equation. Instead I get $$x ^{3}+4x ^{2}+16x+64$$

tkhunny · Answer

Where did the "=0" go?

So, multiply it by 3/2.  Done.

anonymous · Answer

Multiply the whole equation by 3/2 ?

anonymous · Answer

-4i is a pure imaginary number, I thought only complex numbers in the form a+bi came in pairs. Should I assume that that is necessary anyway?

tkhunny · Answer

-4i = 0-4i
Complex Conjugate is 0+4i

Yes, just multiply by 3/2.  That will not change roots.