Given that $\displaystyle 3+\sqrt{2}$ and 2-i are the roots of the equation f(x)=0 . If the degree of f is 4 and f(1)=8 . Find the polynomial f(x)

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- Aug 27th 2009, 07:35 AMthereddevilspolynomial
Given that $\displaystyle 3+\sqrt{2}$ and 2-i are the roots of the equation f(x)=0 . If the degree of f is 4 and f(1)=8 . Find the polynomial f(x)

- Aug 27th 2009, 08:46 AMrunning-gag
Hi

I suppose that f(x) is a polynomial with integer coefficients

Therefore if 2-i is a solution then 2+i is a solution

And if $\displaystyle 3 - \sqrt{2}$ is a solution then $\displaystyle 3 + \sqrt{2}$ is a solution

Finally $\displaystyle f(x) = \alpha (x-(2-i))(x-(2+i))(x-(3 - \sqrt{2})(x-(3 + \sqrt{2}))$

Expand and determine $\displaystyle \alpha$ using the value of f(1) - Aug 27th 2009, 09:18 AMthereddevils
- Aug 27th 2009, 11:21 AMQM deFuturo
I can't quote a "rule" for Real roots, but I do know that if any roots of an equation are complex, they *always* come in pairs of conjugates. So if a +bi is a root, a - bi *has* to be a root as well.

There's probably something similar for Real roots, but I do not know any formal properties in this area. - Aug 28th 2009, 05:59 AMstapel
When have you

*ever*(and*only*) gotten solutions of these forms, with radicals or imaginaries (or both)? From the "plus-minus" part of**the Quadratic Formula**!

Working backwards here**from the zeroes**, you know that, if one zero is "(something) plus (a square root)", then another zero must be "(that same thing) minus (that same square root)". Using the same reasoning with the complex root, you obtain the fourth zero.

From these, and the fact that, if x = a is a root, then x - a is a factor, you can find all four factors, and multiply them out to find the original polynomial. (Wink)