Given $\alpha ,\beta$ and $\gamma$ are the zeroes of the polynomial $f\left(x\right)=a{x}^3+b{x}^2+cx+d$
$\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }$
$=\frac{\beta \gamma +\alpha \gamma +\alpha \beta }{\left(\alpha \beta \gamma \right)}$ $=\frac{\left(\frac{c}{a}\right)}{(-\frac{d}{a})}$ $=-\frac{c}{d}$

Given $p\left(x\right)=4x^4-3x^3+2x^2-x$
To find the remainder of  $\frac{4x^4-3x^3+2x^2-x}{x+1}$,
we will use remainder theorem.
Remainder of $\frac{p\left(x\right)}{x-a}$ is $p\left(a\right)$.
$\Rightarrow$ Remainder of  $\frac{4x^4-3x^3+2x^2-x}{x+1}$ is $p(-1)$.

$[\because x+1=x-(-1\left)\right]$

$p(-1)=4(-1{)}^4-3(-1{)}^3+2(-1{)}^2-(-1)$
$\Rightarrow p(-1)=4+3+2+1=10$

Hence, the remainder of $\frac{4x^4-3x^3+2x^2-x}{x+1}$ is 10.

$\text{If k is a zero of the polynomial p(x),}\text{then,}p\left(k\right)=0.$
 
$\text{Here, 1 is a zero of the polynomial}\text{p(x).}$
$\Rightarrow p\left(1\right)=0$
$\text{Also,}p\left(2\right)-p\left(1\right)=p\left(2\right)-0=p\left(2\right)$

Answer: Option D. -> D)

We know that, the values at which a quadratic polynomial cuts the x-axis are its zeroes.
Option A: Polynomial cuts the x-axis at : -3, -2 ; sum of zeroes: -5
Option B: Polynomial cuts the x-axis at : 2, 3; sum of zeroes: 5
Option C: Polynomial cuts the x-axis at : 2, 3; sum of zeroes: 5
Option D: Polynomial cuts the x-axis at : -2, 2; sum of zeros: 0 
Hence, the answer is D.

Given polynomial $x^2-px+36$
On comparing with the standard form of a quadratic polynomial $a{x}^2+bx+c$, we get
a = 1, b = -p, c = 36
Here, $\alpha$ and $\beta$ are the zeroes of  the polynomial.
$\Rightarrow \alpha +\beta =\frac{-b}{a}=p$
and $\alpha \beta =\frac{c}{a}=36$
Now, ${\alpha }^2$+${\beta }^2={(\alpha +\beta )}^2$ - 2$\alpha \beta$
$\Rightarrow$ $9=p^2-2\times 36$  [$\because {\alpha }^2+{\beta }^2$ = 9]
$\Rightarrow 81=p^2$
$\Rightarrow p=9\text{or}-9$