Answer: Option D. -> 2
:
D
The zeroes of a polynomial are the points which cuts the $x-\text{axis}$ on the graph. As $f\left(x\right)$ cuts $x-\text{axis}$ at 2 points, the number of zeroes are 2.

Answer: Option A. -> 4x + 2
:
A
A polynomial isa mathematical expression of one or more algebraic terms each of which consists of a constant multiplied withone 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 apolynomialof degree 1.
$\frac{1}{(x+1)}=(x+1{)}^{-1}$, the degree of thisexpression would notbe 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 twovariables, x and y.

Answer: Option C. -> 3√y+y2
:
C
$\text{All the variables of a polynomial}\text{should have whole numbers as}\text{exponents.}$
$\text{For}x+x^2,\text{the exponents of the}\text{variable x are 1 and 2. All the}\text{exponents are whole numbers.}\text{Therefore, it is a polynomial.}$
$\text{In the expression}\sqrt[3]{3y}+y^2,\text{exponent}\text{of}\sqrt[3]{3y}\text{is}\frac{1}{3},\text{but}\frac{1}{3}\text{is not a whole}\text{number.}\text{Hence,}\sqrt[3]{3y}+y^2\text{is not a polynomial.}$
$\text{For}x+\sqrt{3},\text{the exponent of the}\text{variable is 1which is a whole number.}\text{Therefore, it is a polynomial.}$
$\text{For x,the exponentof the variable is}\text{1 which is a whole number.}\text{Therefore, it is a polynomial.}$

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

Answer: Option A. -> −17
:
A
$\text{Given},(49x^2-1)+(1+7x{)}^2=((7x{)}^2-1^2)+(1+7x{)}^2=(7x-1\left)\right(7x+1)+(1+7x\left)\right(1+7x)=(1+7x\left)\right(7x-1+1+7x\left)\right[\text{Taking}(1+7x)\text{common}]=(1+7x\left)\right(14x)$
$\text{To find the zeroes of a polynomial}\text{expression,we must equate it to zero}.(1+7x)\left(14x\right)=0\Rightarrow (1+7x)=0;\left(14x\right)=0\therefore x=\frac{-1}{7},0\text{Zeroes are}\frac{-1}{7}\&0.$

Answer: Option D. -> ±9    
:
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]
$\Rightarrow 81=p^2$
$\Rightarrow p=9\text{or}-9$

Answer: Option A. -> True
:
A
$P\left(x\right)=3x^3+2x^2-x+4$
If $P\left(x\right)$ is divided by $(x-a)$, then the remainder is $P\left(a\right)$
$2x+3=0\Rightarrow x=\frac{-3}{2}$
So, the remainder is the value of polynomial at $\frac{-3}{2}$
Substituting $x=\frac{-3}{2}$ in $p\left(x\right)$
$p\left(\frac{-3}{2}\right)=3(\frac{-3}{2}{)}^3+2(\frac{-3}{2}{)}^2-\left(\frac{-3}{2}\right)+4$
$p\left(\frac{-3}{2}\right)=3\left(\frac{-27}{8}\right)+2\left(\frac{9}{4}\right)-\left(\frac{-3}{2}\right)+4$
$p\left(\frac{-3}{2}\right)=\frac{-81}{8}+\frac{9}{2}+\frac{3}{2}+4$
$p\left(\frac{-3}{2}\right)=\frac{-81+36+12+32}{8}$
$p\left(\frac{-3}{2}\right)=\frac{-81+80}{8}$
$p\left(\frac{-3}{2}\right)=\frac{-1}{8}$
$\therefore$ The remainder is $\frac{-1}{8}.$
Thus, the given statement is true.

Answer: Option A. -> both positive
:
A
In a quadratic equation $a{x}^2+bx+c$,both zeroes are positivewhen a is positive, bis negativeand c is positive.
In this case, $x^2-56x+108$ can be factorized as $(x-2)(x-54)$.
$⟹$the roots are 2 and54; both are positive.

Answer: Option C. -> parabola
:
C
Quadratic polynomial is of degree 2 and the graph of a quadratic polynomial is parabola.

Answer: Option D. -> 10
:
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 remainderof$\frac{4x^4-3x^3+2x^2-x}{x+1}$ is 10.