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)
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)
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.
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.