Answer: The remainder theorem applies only if the divisor is monic linear binomial which is x-a. If there is polynomial P(x) and dividing the same by x-a, in such scenario, remainder is P(a). It should be noted that the remainder theorem does not give the quotient, therefore, it cannot be used for questions which are looking for quotient or remainder.

For instance- P(x) = 2x2-x-1 which is divided by x-2. When there is application of synthetic or long division, the solution received is= Q(x)= 2x+3 and R(x)=5. However, with usage of remainder theorem, there can be remainder which is P (2) = 2*22-2-1=8-2-1=5

When there is combination of remainder theorem with factor theorem, it can be used for finding or verifying factors of polynomial. Therefore, x-2 is not considered to be factor of P(x). However, P (1) = 2*12-1-1=0, therefore, x-1 is considered to be factor of P(x).

Instead, there can be other aspect which can be tried that is: p (0) = 2*02-0-1= -1, therefore, x-0 is not considered to be the factor. However, consider that P (0) is below the x-axis and P (2) is above x axis. It means that P(x) must be crossing x-axis somewhere which will be between 0 and 2. It will be leading to try P (1) as the factor.

From the instance, it can be seen and analyzed that polynomial is and should be divided by x-k, the remainder which will be found in a quick manner through evaluation of polynomial function at k, which is, f(k) and the proof of the theorem is shown as follows:

The Division Algorithm has been successful in stating the fact that given the polynomial dividend f(x) and the non-zero polynomial divisor d(x), wherein the degree of d (x) is equal or less to the degree of f(x), in such scenario, there is existence of the unique kind of polynomials q(x) as well as r(x) which are described as follows:

For instance- When the divisor d(x), is considered to be f (x)= d (x) q (x)+ r (x)

If the divisor d(x) is x-k, it takes the entire form of:

F (x) = (x−k) q (x) + r

As the divisor is x-k is considered to be linear, the remainder will become constant r, and when we will be evaluating the same for x=k, then,

{f (k) = (k−k) q (k) + r

= 0* q (k) + r

= r

Therefore, from the entire instance or example, the f(k) is considered to be the remainder obtained by dividing f(x) by x-k.

One of the Examples which will be playing a vital role in utilising the remainder theorem for evaluating polynomial is shown as follows:

Question- Evaluate f (x) = 6x4??−x?3??−15x?2??+2x−7 at x= 2Solution- The synthetic division can be done to divide the polynomial by x-2

{2) {6 -1 -15 2 -7

{12 22 14 32

{6 11 7 16 25

Therefore, from the overall solution, it can be seen that the remainder is 25 and therefore, f (2) = 25

The solution can be checked by evaluating f (2) {f(x) = 6x4- x3 – 15x2 + 2x -7

{F (2) = 6 (2)4 – (2)3 -15 (2)2 + 2 (2) – 7

= 25

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