As per the given figure,the graph of $y=p(x)$ has $\ldots \ldots$ real zeros.

Prove that $-6, -\frac{1}{2}$ and $1$ are the zeros of the cubic polynomial $p(x) = 2x^3 + 11x^2 - 7x - 6$. Also,verify the relationship between the zeros and the coefficients.

One of the factors of $p(x) = x^{3} - 3x^{2} + 9x - 27$ is.........

State the degree of the following polynomial: $p(x) = x^{15} - (x^5)^6 + x^3 - 7$.

If the sum of the zeros is $-7$ and the product of the zeros is $12$,then the quadratic polynomial is $\ldots \ldots \ldots$

Are the following statements 'True' or 'False'? Justify your answers.
If the graph of a polynomial intersects the $x$-axis at exactly two points,it need not be a quadratic polynomial.