Identify which of the following expressions are polynomials? If so,write their degree and kind.
i)$\dfrac{2}{5}{x^4} - \sqrt 3 {x^2} + 5x - 1$
ii)$2{x^2}y - \dfrac{3}{{xy}} + 5{y^3} + \sqrt 3 $
Answer
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Hint: A polynomial of degree n in single variable x is an expression of the form :
$\eqalign{
& {P_n}(x) = {a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + {a_{n - 2}}{x^{n - 2}} + ... + {a_1}{x^1} + {a_0}{x^0} \cr
& \,\,\,where\,\,x\,\,is\,\,the\,\,\operatorname{variable} \,\,and\,{a_i}'s\,\,are\,\,coefficients(real\,nos.\,or\,imaginary\,nos.)\,\,and\,\,{a_n} \ne \,\,0 \cr} $
where exponents of x in all terms are non-negative integers.
For the polynomial of more than one variable, e,g. For two variables,the terms are like $a_n$$x^r$$y^s$ where $a_n$ is constants and r,s are non-negative integers.
We know that,degree of a polynomial is the highest sum of powers of variables in each term.
Complete step-by-step answer:
For option i): $\dfrac{2}{5}{x^4} - \sqrt 3 {x^2} + 5x - 1$
(First part)
Step1:Here all the coefficients are real nos.These are $\dfrac{2}{5}, - \sqrt 3 ,5, - 1$.
Step2:Exponents of x in each term are also non-negative integers.
Step3:It follows the general form of a polynomial.
Hence, i)$\dfrac{2}{5}{x^4} - \sqrt 3 {x^2} + 5x - 1$ is a polynomial.
(Second part)
Option i) is a polynomial in x only with degree 4 as highest power of x is 4 as per different terms.
Again, $\dfrac{2}{5}{x^4} - \sqrt 3 {x^2} + 5x - 1$ has four terms.Therefore, this is a quadrinomial. So, $\dfrac{2}{5}{x^4} - \sqrt 3 {x^2} + 5x - 1$ is a quadrinomial of degree 4.
For option ii): $2{x^2}y - \dfrac{3}{{xy}} + 5{y^3} + \sqrt 3 $
(First part)
Step1:Here all the coefficients are real nos.These are 2,-3,5, $\sqrt 3$ .
Step2:Exponents of x or y in second term are negative integers.
Step3:It does not follow the general form of a polynomial.
Hence,ii) $2{x^2}y - \dfrac{3}{{xy}} + 5{y^3} + \sqrt 3 $ is not a polynomial.
Note: Rene Descartes invented polynomials.Every polynomial should follow the general expression.If power of any variable in an expression be fraction,then this will not be a polynomial also.
$\eqalign{
& {P_n}(x) = {a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + {a_{n - 2}}{x^{n - 2}} + ... + {a_1}{x^1} + {a_0}{x^0} \cr
& \,\,\,where\,\,x\,\,is\,\,the\,\,\operatorname{variable} \,\,and\,{a_i}'s\,\,are\,\,coefficients(real\,nos.\,or\,imaginary\,nos.)\,\,and\,\,{a_n} \ne \,\,0 \cr} $
where exponents of x in all terms are non-negative integers.
For the polynomial of more than one variable, e,g. For two variables,the terms are like $a_n$$x^r$$y^s$ where $a_n$ is constants and r,s are non-negative integers.
We know that,degree of a polynomial is the highest sum of powers of variables in each term.
Complete step-by-step answer:
For option i): $\dfrac{2}{5}{x^4} - \sqrt 3 {x^2} + 5x - 1$
(First part)
Step1:Here all the coefficients are real nos.These are $\dfrac{2}{5}, - \sqrt 3 ,5, - 1$.
Step2:Exponents of x in each term are also non-negative integers.
Step3:It follows the general form of a polynomial.
Hence, i)$\dfrac{2}{5}{x^4} - \sqrt 3 {x^2} + 5x - 1$ is a polynomial.
(Second part)
Option i) is a polynomial in x only with degree 4 as highest power of x is 4 as per different terms.
Again, $\dfrac{2}{5}{x^4} - \sqrt 3 {x^2} + 5x - 1$ has four terms.Therefore, this is a quadrinomial. So, $\dfrac{2}{5}{x^4} - \sqrt 3 {x^2} + 5x - 1$ is a quadrinomial of degree 4.
For option ii): $2{x^2}y - \dfrac{3}{{xy}} + 5{y^3} + \sqrt 3 $
(First part)
Step1:Here all the coefficients are real nos.These are 2,-3,5, $\sqrt 3$ .
Step2:Exponents of x or y in second term are negative integers.
Step3:It does not follow the general form of a polynomial.
Hence,ii) $2{x^2}y - \dfrac{3}{{xy}} + 5{y^3} + \sqrt 3 $ is not a polynomial.
Note: Rene Descartes invented polynomials.Every polynomial should follow the general expression.If power of any variable in an expression be fraction,then this will not be a polynomial also.
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