Zero degree polynomial is considered as constant.
Example: $p(x)= k$
Therefore, $k$ is constant.
The constant term in an expression or equation has a fixed value and does not contain variables.

In ancient times, algebra is used to find linear and quadratic equations.

al-Khwarizmi was a persian scientist.

The constant term in an expression or equation has a fixed value and does not contain variables.
So, $-7$ is the constant term in the expression, $4x^2+5x^6-7x^2-7+2x^2-7x^6$

Diophantus of Alexandria used algebraic equations and notations in presenting problems and solutions in Arithmetica.

Given polynomial is $x^3+2x^2-1+x^5-5x(x^2)$

A constant term is a value that do not have variable.
So, in this quadratic equation, −3−3 is a  constant term.

Clearly, the expression $4x^2-3x+7$  is a polynomial in one variable x because there is only one variable x in the expression having the highest power 2, which is whole number.

Clearly, the expression $y^2+\sqrt 2$  is a polynomial in one variable y because there is only one variable y in the expression having the highest power 2, which is whole number.

The expression $3\sqrt t+t\sqrt2$ contain the term $3\sqrt t$, here exponent of $t$ is $\dfrac12$, which is not a whole number.
Therefore, the given expression is not a polynomial in one variable.

The expression $y+\dfrac2y$ contain the term $\dfrac2y$, here

the exponent of $y$ is $-1$, which is not a whole number.
Therefore, the given expression is not a polynomial in one variable.

Clearly, the given expression $x^{10}+y^3+t^{50}$ contains three variables $x,y\space and\space t$
Hence, the given expression is not a polynomial in one variable instead in three variables.

Reciprocals will be $\frac { { { x }^{ 2 } }+1 }{ x+3 } $,$\frac { { x }^{ 2 }+3 }{ { x }^{ 2 }-9 } $
Their sum will be
$\frac { { { x }^{ 2 } }+1 }{ x+3 } +\frac { { x }^{ 2 }+3 }{ { x }^{ 2 }-9 } $
 $=\frac { \left( x-3 \right) \left( { x }^{ 2 }+1 \right) +{ x }^{ 2 }+3 }{ { x }^{ 2 }-9 } $
$=\frac { { x }^{ 3 }+x-3{ x }^{ 2 }-3+{ x }^{ 2 }+3 }{ { x }^{ 2 }-9 } $
 $=\frac { { x }^{ 3 }-2{ x }^{ 2 }+x }{ { x }^{ 2 }-9 } $

$x^{2}-3x+1$

Given $x+\frac{1}{x}=3$ multiply by x both sides

Given $x=2, y=3$

$\therefore x^{2}+1=3x$

Given $x+\frac{a}{x}=b$ Multiply by x both sides

$x<-1$
 ${ x }^{ 2 }<{ \left( -1 \right)  }^{ 2 }\quad squaring\quad both\quad side$
 ${ x }^{ 2 }>1$

$Using\quad { a }^{ 3 }+\quad { b }^{ 3 }+\quad { c }^{ 3 }-3abc=\left( a+b+c \right) \left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca \right) $

Given $a-\frac{1}{3}=\frac{1}{a}$

Option d) contains the variables $x, y$ and $z$

Here, the constant in the given equation is $7$ as it contains no variable.

The variables are $x$ and $y$. Number of variables $=2$

A constant is a symbol having a fixed numerical value.

The number of variables in $(A) $ is zero.

variables in $(1)$ are $ x$ and $ y$ where as variables in $(2$) are $ x, y, z$.

Both the terms of the given expression have either $x$ or $y$ as the variable. 

A variable is a symbol or letter, such as $x$ or $y$, that represents a value. In algebraic equations, the value of one variable is often dependent on the value of another. Hence it's a symbol that takes various numerical values.

The constant in the polynomial $x + 5$ is $5$.

Both the terms in the given expression contain atleast one variable, $x$ and $xy$..

The variables used are $x$ and $y$.

For a particular polynomial, its constant cannot change otherwise polynomial will change.

The terms $y^{3}, y^{2}$ and $y$ are not constants. 

The value of the polynomial changes as the variable changes. 

The value of the polynomial changes as the variable changes. 

al-Khwarizmi is the father of Algebra.

An important development in algebra in the $16^{th}$ century was the introduction to unknown symbols.

The word Algebra literally means the re-union of broken parts.

The constant term in an expression or equation has a fixed value and does not contain variables.
So, $-6$ is the constant term in $q(y)=y^3-3y^2-6+y$

Egyptians used the symbol heap for the unknown in algebra.

The constant term in an expression or equation has a fixed value and does not contain variables.
So, $-12$ is the constant term in the expression, $23x^3+12x^2-6x-12$.

Constant term has zero degree of polynomial.
Because the constant term in an expression or equation has a fixed value and does not contain variables.
Example: $p(x)=k$
Where $k$ is a constant.

Given equation is $0.4x^{7}-75y^{2}-075$

General equation is $ax^3+bx^2+cx+d$ 

Polynomial has only 1 variable $x$ in the equation.

There are two variables in the given equation, i.e. $x$ and $y$ while $1$ is a constant.

Polynomial has only 1 variable $y$ in the equation.

The given polynomial has three variables i.e.  $x, y$ and $z$ .

$\cfrac { { x^{ 3 }+2x+1 } }{ 5 } -\cfrac { 7 }{ 2 } x^ 2-x^ 6$
$=\cfrac { { x^{ 3 }+2x+1 } }{ 5 } -\cfrac { 7 }{ 2 } x^ 2-x^ 6$
$=-x^ 6+\cfrac { x^{ 3 } }{ 5 } -\cfrac { 7x^ 2 }{ 2 } +\cfrac { 2x }{ 5 } +\cfrac { 1 }{ 5 }$
So, the constant term= $\cfrac { 1 }{ 5 }$

A variable which can take real numbers only as its value is called as a real variable.

Given equation is $z^3+2z^2+5z+1$

$x^2+3x+5$
$=(2)^2+3\times 2+5$
$=4+6+5$
$=15$

$x^2+3x+5$
$=(-1)^2+3\times (-1)+5$
$=1-3+5$
$=6-3$
$=3$

$z^3+2z^2+5z+1$
$=(-1)^3+2\times (-1)^2+5\times (-1)+1$
$=-1+2-5+1$
$=-3$ 

$z^3+2z^2+5z+1$
$=(0)+2\times (0)^2+5\times (0)+1$
$=1$

Given, $\dfrac{2+3}{x}=\dfrac{2+x}{3}$

Since the temperature of a day depends on the weather. For example, in summers, the temperature will be higher in Celsius whereas in winters, the temperature is low in Celsius. Therefore, the temperature always varies.

Since the length, breadth and height of any area (Room, park, hall, etc) is measured and decided before its construction and then it gets constructed. Therefore, the length of any area will always remain the same.

Since plants have the unique ability to grow indefinitely throughout their life due to the presence of ‘meristems’ in their body. Meristems in the roots and shoots of plants are responsible for ‘primary growth of the plant’. These increase the height of the plant. Therefore, the height of any growing plant varies.

Since there are $31$ days in January in every year, therefore the number of days in the month of January doesn't vary.

We will solve the given expression $(3x-5)^2+(3x+5)^2=(18x+10)(x-2)$ as shown below:

$x-2$ is factor 

As the point lies on the given line, it should satisfy the equation of the line, if we substitute $ x = 2 $ and $ y = -3 $ in it.

So, $k({ 2) }^{ 2 }3{ (-3) }^{ 2 }+2(2)-32=0$
$ => 4k -27 + 4 - 3 -2 = 0 $
$ 4k = 28 $
$ k = 7 $

For $ n = 1 $, we have $ n^2 - n + 1 = 1^2 -1 + 1 = 1 $ which is an odd number


For $ n =2 $, we have $ n^2 - n + 1 = 2^2 -2 + 1 = 3 $ which is an odd number

For $ n = 3 $, we have $ n^2 - n + 1 = 3^2 -3 + 1 = 7 $ which is an odd number

For $ n = 4 $, we have $ n^2 - n + 1 = 4^2 -4 + 1 = 13 $ which is an odd number

For $ n = 5 $, we have $ n^2 - n + 1 = 5^2 -5 + 1 = 19 $ which is an odd number

Hence, $ n^2 - n + 1 = 1^2 -1 + 1 = 1 $ is an odd number for all $ n \ge 1 $

The variables are $ x$,$ y$ and $z $