α and β are the zeroes of $3x^2-7x-6$. Which polynomial with integer coefficients has −α and −β as its zeroes?

$\alpha +\beta =\frac{7}{3}$ and $\alpha \beta =-\frac{6}{3}=-2$. New sum $=-\alpha -\beta =-\frac{7}{3}$; new product $=(-\alpha )(-\beta )=\alpha \beta =-2$, i.e. the product is UNCHANGED. So the polynomial is $x^2+\frac{7}{3}x-2$, and multiplying through by 3 gives $3x^2+7x-6$. (Equivalently, replace $x$ by −$x$: only the odd-power term changes sign. The zeroes 3 and $-\frac{2}{3}$ become −3 and $\frac{2}{3}.)3x^2-7x+6$ negates the product as well; $3x^2+7x+6$ negates only the product; $3x^2-7x-6$ is the original polynomial, left untouched.