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$2+i$ is given to be one of the roots of the polynomial.

I am doing this as a practice for exam prep.

Since $2+i$, is a root, then $(z-2-i)$ is a factor?

So I have:

$(z-2-i)(z^3-Az^2-Bz+C) = z^4-2z^3-z^2+2z+10$

Then, I group the $z$ terms of the same degrees after I multiply out?

Am I on the right track? Because it seems like this will take quite a while in exam conditions? Thanks for any direction!

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    TO your first question, you are correct. Now, instead of doing that multiplication, I would suggest carrying out polynomial longdivision.2012-06-01
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    Or note that if $2+i$ is a root, then $2-i$ is also a root.2012-06-01
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    Imaginary factors always come in pairs!2012-06-01
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    @ThomasAndrews But only if the polynomial has real coefficients right?2012-06-01
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    Right, @mathstudent, but in this case that is true.2012-06-01
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    Out of curiosity, why that choice of signs in $z^3-Az^2-Bz+C$?2012-06-01
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    @JonasMeyer - Yes it was missing, Thanks. I used those signs to match the original equation? I guess its not necessary though?2012-06-01
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    @mathstudent: It's not right or wrong, so whatever helps you is best. If instead I used the general form $z^3+Az^2+Bz+C$, we'd end up with the same polynomial, but our "$A$" and "$B$" would have opposite signs.2012-06-01

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