The zero of $p(x) = x^{2} + 6x + 9$ is:

The zero of the linear polynomial $p(x) = 3x - 6$ is $\ldots$

Divide $3 x^{4}+5 x^{3}-7 x^{2}+2 x+2$ by $x^{2}+3 x+1$.

Given that the zeroes of the cubic polynomial $x^{3}-6x^{2}+3x+10$ are of the form $a, a+b, a+2b$ for some real numbers $a$ and $b$,find the values of $a$ and $b$ as well as the zeroes of the given polynomial.

The zeros of the quadratic polynomial $p(x) = x^{2} - 3x + 2$ are $\alpha$ and $\beta$. Then,$\frac{1}{\alpha} + \frac{1}{\beta} = \ldots$

If $p(x) = x^{3} - 3x^{2} - x + 3$ is divided by $x - 1$,then the remainder is .............