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April 18, 2015

April 18, 2015

Posted by **Zach** on Sunday, November 15, 2009 at 12:16am.

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^4-9x^3+7x^2-91x-348,zero 5-2i

I normally would use synthetic division with the root 5-2i 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 5-2i 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 5-2i, 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^4-9x^3+7x^2-91x-348) 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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