Posted by Zach on .
I am familiar with this type of problem but can't seem to get the right answer.
Use the given zero to find the remaining zeros of each function
f(x)=x^49x^3+7x^291x348,zero 52i
I normally would use synthetic division with the root 52i bringing it down to x^3. Then synthetic with 5+2i, bring it down to a quadratic solve the quadratic then I'll have zeros, but I can't get past the 52i with the synthetic division. Every time I do it I can't get it to equal zero. I have done it multiple times and it won't work.

Pre Calc 
Reiny,
One property of complex roots is that they always come in conjugate pairs.
So if one root is 52i, there will be another 5+2i
so there are two factors,
(x  5  2i) and (x  5 + 2i)
I multiplied these and go
(x^2  10x + 29)
Now do a long division of
(x^49x^3+7x^291x348) by (x^2  10x + 29)
That should leave you with a quadratic, that can be solved for 2 more roots.
Let me know if it worked for you.