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Question: If (x-c) is a factor of order m of the polynomial f(x) of degree n(1 < m < n), then x=c is the root ...

If (x-c) is a factor of order m of the polynomial f(x) of degree n(1 < m < n), then x=c is the root of the polynomial(where fr(x){{f}^{r}}\left( x \right) represents rth{{r}^{th}} derivative of f(x) w.r.t. x).

(a) fm(x){{f}^{m}}\left( x \right)
(b) fm1(x){{f}^{m-1}}\left( x \right)
(c) fn(x){{f}^{n}}\left( x \right)
(d) none of these

Explanation

Solution

First, before proceeding for this, we must suppose a polynomial g(x) with degree (n-m) that is multiplied with the factor (x-c) with degree m to get the another polynomial which is defined in the question as f(x) as f(x)=(xc)mg(x)f\left( x \right)={{\left( x-c \right)}^{m}}g\left( x \right) . Then, x=c is a common root for all equations that comes from the derivative of the above function. Then, if we differentiate the above expression to the last degree to get (x=c) as a root, we get the last differentiation as fm1(x)=0{{f}^{m-1}}\left( x \right)=0 which gives the final result.

Complete step-by-step answer:
In this question, we are supposed to find if x=c is the root of the polynomial when (x-c) is a factor of order m of the polynomial f(x) of degree n(1So, before proceeding for this, we must suppose a polynomial g(x) with degree (n-m) that is multiplied with the factor (x-c) with degree m to get the other polynomial which is defined in the question as f(x).
f(x)=(xc)mg(x)f\left( x \right)={{\left( x-c \right)}^{m}}g\left( x \right)
Then, x=c is a common root for all equation that comes from the derivative of the above function as:
f(x)=0,f(x)=0,f(x)=0....f\left( x \right)=0,{f}'\left( x \right)=0,{f}''\left( x \right)=0.... so on till the last term before the degree gets totally eliminated as fm1(x)=0{{f}^{m-1}}\left( x \right)=0 .
So, fr(x){{f}^{r}}\left( x \right) represents the rth{{r}^{th}} derivative of the polynomial function f(x).
Moreover, if we differentiate the above expression to the last degree to get (x=c) as a root, we get the last differentiation as fm1(x)=0{{f}^{m-1}}\left( x \right)=0 .
So, the (x=c) is the root of the polynomial fm1(x){{f}^{m-1}}\left( x \right) .

So, the correct answer is “Option b”.

Note: Now, to solve these type of the questions we need to know some of the basics of the differentiation as let us suppose the polynomial as f(x)=xnf\left( x \right)={{x}^{n}} . Then, if we need to eliminate the entire power of the polynomial with power n, we need to perform the differentiation (n-1) times. So, the factor (x=c ) only occurs when fn1(x)=0{{f}^{n-1}}\left( x \right)=0 is calculated.