Question:

1. Check the given numbers as zeroes\
• For \(x = 3\):\
\[p(3)=3(3)^3-5(3)^2-11(3)-3=81-45-33-3=0\]\
Hence \(x=3\) is a zero.\
• For \(x = -1\):\
\[p(-1)=3(-1)^3-5(-1)^2-11(-1)-3=-3-5+11-3=0\]\
Hence \(x=-1\) is a zero.\
• To find the third zero, divide \(p(x)\) by \((x-3)(x+1)\).\
First divide by \(x-3\) (synthetic division):\
\[\begin{array}{c|cccc}
3 & 3 & -5 & -11 & -3 \\
& & 9 & 12 & 3 \\
\hline
& 3 & 4 & 1 & 0 \\
\end{array}\]
Quotient = \(3x^2+4x+1\).\
Factor the quadratic:\
\[3x^2+4x+1 = (3x+1)(x+1)\]\
Hence \(p(x)= (x-3)(x+1)(3x+1)\).\
The remaining zero is obtained from \(3x+1=0\) ⇒ \(x = -\frac{1}{3}\).\
Therefore the three zeroes are \(\alpha=3,\; \beta=-1,\; \gamma=-\frac13\).\

2. Verify Vieta’s relations\
The cubic is of the form \(ax^3+bx^2+cx+d\) with \(a=3,\; b=-5,\; c=-11,\; d=-3\).\
- Sum of zeroes:\
\[\alpha+\beta+\gamma = 3+(-1)+\left(-\frac13\right)=\frac{5}{3}\]\
\[-\frac{b}{a}= -\frac{-5}{3}=\frac{5}{3}\]\
Both are equal, so the first relation holds.\
- Sum of products of zeroes taken two at a time:\
\[\alpha\beta+\beta\gamma+\gamma\alpha = (3)(-1)+(-1)\left(-\frac13\right)+(3)\left(-\frac13\right) = -3+\frac13-1 = -\frac{11}{3}\]\
\[\frac{c}{a}= \frac{-11}{3}= -\frac{11}{3}\]\
Hence the second relation is satisfied.\
- Product of the three zeroes:\
\[\alpha\beta\gamma = 3\times(-1)\times\left(-\frac13\right)=1\]\
\[-\frac{d}{a}= -\frac{-3}{3}=1\]\
The third relation also holds.\

Thus, the numbers \(3, -1, -\frac13\) are indeed the zeroes of \(p(x)\) and they satisfy the standard relationships between zeroes and coefficients for a cubic polynomial.