Question

Write the necessary and sufficient condition for the polynomial \[p(x) = {a_0}{x^n} + {a_1}{x^{n - 1}} + {a_2}{x^{n - 2}} + ... + {a_n}\] to be a polynomial of order \[(n - 2)\]

Answer 1

Hint: We use the concept of order of a differential equation and eliminate the coefficients of higher orders than the required orders.
* Order of a polynomial is the highest power of the variable. Example: In a polynomial \[p(x) = {x^4} + 3{x^2} - 7\] has order 4, as the highest power of x is 4.
* Coefficient of any variable is the value (constant term) that is multiplied along with that variable. Example: In a polynomial \[p(x) = {x^4} + 3{x^2} - 7\], coefficient of \[{x^4} = 1\], coefficient of \[{x^2} = - 1\].

Complete step-by-step answer:
We have a polynomial \[p(x) = {a_0}{x^n} + {a_1}{x^{n - 1}} + {a_2}{x^{n - 2}} + ... + {a_n}\]............… (1)
From the definition of order of a polynomial, the highest power of the variable of equation (1) is ‘n’.
Since we want a polynomial of order \[(n - 2)\], the highest power of the variable should be \[(n - 2)\].
We will substitute the value of coefficients of variables having power greater than \[(n - 2)\] as 0.
Since, \[n > n - 1 > n - 2\]
So, we will substitute the coefficients of variables having power \[n,(n - 1)\] equal to 0.
From equation (1),
Coefficient of \[{x^n} = {a_0}\], coefficient of \[{x^{n - 1}} = {a_1}\]
Substitute coefficients as 0.
Substitute \[{a_0} = 0,{a_1} = 0\] in equation (1)
\[ \Rightarrow p(x) = 0 \times {x^n} + 0 \times {x^{n - 1}} + {a_2}{x^{n - 2}} + ... + {a_n}\]
Multiplication of any value with zero gives the answer zero.
\[ \Rightarrow p(x) = {a_2}{x^{n - 2}} + ... + {a_n}\]

\[\therefore \]Necessary and sufficient condition for a polynomial \[p(x) = {a_0}{x^n} + {a_1}{x^{n - 1}} + {a_2}{x^{n - 2}} + ... + {a_n}\] to be of order \[(n - 2)\] is \[{a_0} = 0,{a_1} = 0\].

Note: Students might make mistake of writing the necessary and sufficient condition for the given polynomial to be of order \[(n - 2)\] by substituting the values of n as 2 or greater than 2, this is wrong process as it will give the power in negative terms as well which will further give variable in denominator. On taking LCM we will again get the same power of the variable.