Find the zeros of the cubic polynomial $p(x)=4 x^{3}+10 x^{2}+6 x$ and also verify the relationship between the zeros and the coefficients.

(N/A) To find the zeros of $p(x),$ set $p(x)=0$.
$4 x^{3}+10 x^{2}+6 x=0$
$2 x(2 x^{2}+5 x+3)=0$
$2 x(2 x^{2}+2 x+3 x+3)=0$
$2 x(2 x(x+1)+3(x+1))=0$
$2 x(2 x+3)(x+1)=0$
Thus,the zeros are $x=0, x=-\frac{3}{2}, x=-1$.
For $p(x)=4 x^{3}+10 x^{2}+6 x+0$,we have $a=4, b=10, c=6, d=0$.
Sum of zeros: $0 + (-\frac{3}{2}) + (-1) = -\frac{5}{2} = -\frac{10}{4} = -\frac{b}{a}$.
Sum of product of zeros taken two at a time: $(0)(-\frac{3}{2}) + (-\frac{3}{2})(-1) + (-1)(0) = 0 + \frac{3}{2} + 0 = \frac{3}{2} = \frac{6}{4} = \frac{c}{a}$.
Product of zeros: $(0)(-\frac{3}{2})(-1) = 0 = -\frac{0}{4} = -\frac{d}{a}$.