For an expression to be a polynomial, the exponents of $x$ should be whole numbers. The above expression can be written as  $x^g+x^{-a}$ . For this to be a polynomial, $-a$ should be a whole number. This is possible if $a$ is a negative integer or zero.

Since a monomial can contain only one term, only one of $a$, $b$ and $c$ can be non-zero to leave a single term.

A zero degree polynomial is essentially a constant. i.e., any non-zero real number.
So, if $a{x}^n$ is a zero degree polynomial, then
$a{x}^n$ = $a{x}^0$ = a
So, n has to be zero, and $a$ should not be zero.

Let $p\left(x\right)=(x^3+2x^2+2x+1)$ be the given polynomial
By observation, we can see that by putting
$x=-1,p\left(x\right)=0$
$\therefore x=-1$ is the root of $p\left(x\right)$
$\Rightarrow (x+1)$ is a factor of $p\left(x\right)$ 
To find the other factor, we perform long division of $p\left(x\right)$ by $(x+1)$
 

$\therefore$ The two factors of $p\left(x\right)$ are $(x+1)$ and $(x^2+x+1)$

$\text{Let y be the width.}$
$\text{Given: Area}=25a^2-35a+69$
$\text{We know that, area of a rectangle}$$\text{= Length}\times \text{breadth}$
$\text{Hence,}y\times (5a-3)=25a^2-35a+69$
$y=(25a^2-35a+69)÷(5a-3)$
By long division method, 
$5a-3\stackrel{5a-23}{)\overline{25a^2-35a+69}}25a^2-12a-+\overline{}-23a+69-23a+69+-\overline{}0$
$\therefore y=5a-23\text{Width of the rectangle =}5a-23$

Suppose $f\left(x\right)$ is the polynomial.
Now since it cuts x-axis at 2 and -4, the roots of $f\left(x\right)$ are 2 and -4.
Sum of roots = $-a=2-4=-2$
$\Rightarrow a=2$
Product of roots = $-c=2\times -4=-8$
$\Rightarrow c=8$
Hence, $(a+c)=10$

Using the algebraic Identity $(a^2-b^2)=(a-b)(a+b)$

$\Rightarrow (a-b)$ and $(a+b)$ are factors of $a^2-b^2$.

Consider,
$({101}^2-{99}^2)$

$=(101+99)(101-99)$

$=\left(200\right)\left(2\right)$

We find that 2 and 200 both are the factors of $({101}^2-{99}^2)$​.

Since a and b are the zeroes of the polynomial f(x), both f(a) = 0 and f(b) = 0, i.e. a and b will be the roots of the equation f(x) = 0.
Thus, f(a) = f(b) = 0. Hence, the given statement is true.