A quartic function in which the coefficients are relatively prime is known to have zeros of 1+3i and -2 - √5 What is the coefficient of the cubic term of this function?

Question

A quartic function in which the coefficients are relatively prime is known to have zeros of 1+3i  and  -2 - √5  What is the coefficient of the cubic term of this function?

anonymous · Answer

Is there a quicker way than expanding this?

anonymous · Answer

You don't have to solve! Just tell me if there's a quicker way!

btaylor · Answer

The other zeroes are 1-3i and -2 + √5. I don't know about the -2 +/- √5, but I have a shortcut for the imaginary root.

The two imaginary roots will multiply to form a trinomial. In the trinomial (ax^2 + bx + c), 
-b/a = (1-3i)+(1+3i)
c/a=(1-3i)(1+3i). 
a is almost always 1.

IDK, it may work for the squares. It should. Try it!

anonymous · Answer

Thank you! I never thought of doing it that way! I find the two quadratics and multiply to get the cubic coefficient! :)

btaylor · Answer

yup.