$\text{All the variables of a polynomial}\text{should have whole numbers as}\text{exponents.}$
$\text{For}x+x^2,\text{the exponents of the}\text{variable x are 1 and 2. All the}\text{exponents are whole numbers.}\text{Therefore, it is a polynomial.}$
$\text{In the expression}\sqrt[3]{3y}+y^2,\text{exponent}\text{of}\sqrt[3]{3y}\text{is}\frac{1}{3},\text{but}\frac{1}{3}\text{is not a whole}\text{number.}\text{Hence,}\sqrt[3]{3y}+y^2\text{is not a polynomial.}$
$\text{For}x+\sqrt{3},\text{the exponent of the}\text{variable is 1 which is a whole number.}\text{Therefore, it is a polynomial.}$
$\text{For x, the exponent of the variable is}\text{1 which is a whole number.}\text{Therefore, it is a polynomial.}$

To find the zeros of a polynomial expression, we must equate it to zero.
$\therefore 5x-75=0$
$\Rightarrow x=\frac{75}{5}$
$\Rightarrow x=15$

$\text{Let}P\left(x\right)=(x+y)(x+y{)}^2-(x+y\left)\right(x^2-xy+y^2)$
$=(x+y)(x^2+2xy+y^2)-(x+y)(x^2-xy+y^2)$
$=(x+y)(x^2+2xy+y^2-x^2+xy-y^2)$
[Taking $(x+y)$ as common]
$=(x+y)\left(3xy\right)$
Hence, from the given options 3xy is a factor of the given polynomial.

Given $f\left(x\right)$ = $3x^2-7x+5$ 
To find the remainder when it is divided by $(x-1)$, we use remainder theorem.

Remainder of $\frac{f\left(x\right)}{(x-a)}$ is $f\left(a\right)$.

$f\left(1\right)=3-7+5=1$
$\Rightarrow$ The remainder when $3x^2-7x+5$ is divided by $(x-1)$ is 1.

$(z-5{)}^3=z^3-5^3-3\times z^2\times 5+3\times z\times 5^2$
$\text{[Using}(a-b{)}^3=a^3-b^3-3a^{2}b+3a{b}^2$]
$\Rightarrow (z-5{)}^3$ = $z^3-125-15z^2+75z$
Hence, the coefficient of $z$ is 75.

The exponents of the variables of a polynomial should be a whole number.
In $y^3+\frac{4}{y}=y^3+4y^{-1}$, the exponent of $y$ in  $\left(4y^{-1}\right)$ is -1 which is not a whole number.
In $\sqrt{2}x-1$, the exponent of  $x$ is a whole number.
In $\frac{x+1}{x-1}=(x+1)\times (x-1{)}^{-1}$, the exponent of $x$ in $(x+1{)}^{-1}$ is not a whole number.
In $z^3+z$, the exponents of $z$ in both the terms is a whole number exponent.
Hence,  $z^3+z$ and $\sqrt{2}x-1$ are polynomials .

$\text{Given},(49x^2-1)+(1+7x{)}^2=((7x{)}^2-1^2)+(1+7x{)}^2=(7x-1\left)\right(7x+1)+(1+7x\left)\right(1+7x)=(1+7x\left)\right(7x-1+1+7x\left)\right[\text{Taking}(1+7x)\text{common}]=(1+7x\left)\right(14x)$
$\text{To find the zeroes of a polynomial}\text{expression, we must equate it to zero}.(1+7x)\left(14x\right)=0\Rightarrow (1+7x)=0;\left(14x\right)=0\therefore x=\frac{-1}{7},0\text{Zeroes are}\frac{-1}{7}\&0.$

In a quadratic equation $a{x}^2+bx+c$,both zeroes are positive when a is positive, b is negative and c is positive.
In this case, $x^2-56x+108$ can be factorized as $(x-2)(x-54)$.
$⟹$the roots are 2 and  54; both are positive.