Pre Calc (For Reiny)
posted by Zach .
Thanks so much your method worked perfectly!
My solution was: the remaining zeros were 5+2i, 3, and 4
I have one question that's bugging me though, why is it that synthetic division did not work for me? I thought that long division and synthetic division were both interchangeable.
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, Friday, November 13, 2009 at 11:21pm
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.
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