How would i write a polynomial equation from the roots 3 and 2+√7?
Note that $\displaystyle x=3$ is a zero, which implies $\displaystyle x-3$ is a factor of the polynomial
Note that $\displaystyle x=2+\sqrt{7}$ is a zero, which implies $\displaystyle x-\left(2+\sqrt{7}\right)$ is a factor of the polynomial. We also need the conjugate of our previous zero $\displaystyle x=2-\sqrt{7}$ to keep the coefficients clear of square roots. That implies that $\displaystyle x-\left(2-\sqrt{7}\right)$ is a factor of the polynomial.
Thus, your polynomial as a product of factors is $\displaystyle \left(x-3\right)\left(x-\left[2+\sqrt{7}\right]\right)\left(x-\left[2-\sqrt{7}\right]\right)$
Foil this out and see what you get.
Does this make sense?
A simpler approach:
$\displaystyle \left( x - \left[ 2 + \sqrt{7} \right] \right)$ $\displaystyle \left(x - \left[ 2 - \sqrt{7} \right] \right) = \left( [x - 2] - \sqrt{7} \right)$ $\displaystyle \left( [x-2] + \sqrt{7} \right) $
which has the form of the factorisation of the difference of two squares:
$\displaystyle (A - B)(A + B) = A^2 - B^2$
(and yes I am attempting the record of how many times I can use the word of in one sentence).
Identifying $\displaystyle A = (x - 2)$ and $\displaystyle B = \sqrt{7}$ you therefore have:
$\displaystyle (x - 2)^2 - \left( \sqrt{7} \right)^2 = x^2 - 4x + 4 - 7 = x^2 - 4x - 3$.