Lets divide $3t^4+5t^3-7t^2+2t+2\text{by}{t}^2+3t+1$,
$3t^2-4t+2t^2+3t+1)\overline{3t^4+5t^3-7t^2+2t+2}\underset{–}{{}_{-}3t^4\pm 9t^3\pm 3t^2}\underset{–}{-4t^3-10t^2+2t\mp 4t^3\mp 12t^2\mp 4t}\underset{–}{2t^2+6t+2{}^{}{}_{-}2t^2\pm 6t\pm 2}00$
Thus, the given statement is true.

We know that, by divison algorithm, 
Dividend = Divisor x Quotient + Remainder.
$\therefore P\left(x\right)=(x+2)\times (2x-1)+3=x(2x-1)+2(2x-1)+3=2x^2-x+4x-2+3P\left(x\right)=2x^2+3x+1$
$\therefore \text{The required polynomial is}2x^2+3x+1.$

Let $p\left(x\right)=x^3+5x^2-9x-45$
      $q\left(x\right)=x^3+8x^2+15x$
$p\left(x\right)=x^3+5x^2-9x-45$
$p\left(x\right)=x^2(x+5)-9(x+5)$
$p\left(x\right)=(x+5)(x^2-9)$
$p\left(x\right)=(x+5)(x+3)(x-3)$
$\Rightarrow$ Zeroes are -5,-3 and 3.
$q\left(x\right)=x^3+8x^2+15x$
$q\left(x\right)=x(x^2+8x+15)$
$q\left(x\right)=x(x+5)(x+3)$
$\Rightarrow$ Zeroes are 0, -5 and -3
Therefore, common zeroes are -5 and -3.

Given that,
Sum of zeroes = $\frac{-8}{5}$

Product of zeroes = $\frac{7}{5}$

Required quadratic polynomial is,
$f\left(x\right)=\left[\right(x^2-\left(sumofroots\right)x+\left(productofroots\right)]$
Substituting the given values we get,
$f\left(x\right)=[x^2-\frac{(-8)}{5}x+\frac{7}{5}]$
$f\left(x\right)=[x^2+\frac{8}{5}x+\frac{7}{5}]$ 
multiplying by 5 we get
$f\left(x\right)=5x^2+8x+7$
$\therefore$ Required polynomial is $5x^2+8x+7$.

We know that,
Dividend = Divisor x Quotient +Remainder
$x^3-3x^2+x+2=Divisor\times (x-2)+(-2x+4)$
$\Rightarrow Divisor\times (x-2)=x^3-3x^2+x+2+2x-4$
$\Rightarrow Divisor=\frac{(x^3-3x^2+3x-2)}{x-2}$ 
                                 
$\therefore Divisor=x^2-x+1$

If $\alpha ,\beta$ and $\gamma$ are the zeroes of the cubic polynomial $a{x}^3+b{x}^2+cx+d$, then

sum of its zeroes
= $\alpha +\beta +\gamma$
= $-\frac{\left(coefficientof{x}^2\right)}{\left(coefficientof{x}^3\right)}=-\frac{b}{a}$

$\text{To find the zeroes, equate the given}\text{polynomial to zero.}$
$x^2-2=0$
$\text{Using the identity,}{a}^2-b^2=(a+b)(a-b)$
$\Rightarrow (x+\sqrt{2})(x-\sqrt{2})=0$ 
$\Rightarrow x+\sqrt{2}=0\text{and}x-\sqrt{2}=0$
$\Rightarrow x=-\sqrt{2}\text{and}x=\sqrt{2}$
$\text{Hence, the zeroes are}\sqrt{2},-\sqrt{2}$. 

Let $\alpha \&\beta$ be the roots.
Sum of zeroes,  $\alpha +\beta =\frac{6+\sqrt{3}}{3}+\frac{6-\sqrt{3}}{3}$                                       $=\frac{12}{3}=4$
Product of zeroes, $\alpha \beta =\frac{6+\sqrt{3}}{3}\times \frac{6-\sqrt{3}}{3}=\frac{6^2-(\sqrt{3}{)}^2}{9}\alpha \beta =\frac{36-3}{9}=\frac{33}{9}=\frac{11}{3}$
Required polynomial $f\left(x\right)=\left[\right(x^2-(\alpha +\beta )x+\left(\alpha \beta \right)]=[(x^2-4x+\frac{11}{3}]$
Multiplying by 3
$\Rightarrow f\left(x\right)=3x^2-12x+11$
$\therefore$ the required polynomial is $3x^2-12x+11$.

A polynomial is a mathematical expression of one or more algebraic terms each of which consists of a constant multiplied with one or more variables raised to a non-negative integral power . For example, $(a+bx+c{x}^2)$ where a, b and c are constants, and x is the variable.

Degree is the highest power of the variable.
$4x+2$ is a polynomial of degree 1.

$\frac{1}{(x+1)}=(x+1{)}^{-1}$, the degree of this expression would not be non-negative. Thus, it is not a polynomial.

$4x+2=0$ is an equation and hence not a polynomial.

$x+y$ is a polynomial in two variables, x and y.