Question

write statements of remainder theorem and also write the proof of this.​

Answer 1

Answer:

Step-by-step explanation:

Let f(x) be any polynomial with degree greater than or equal to 1.

Further suppose that when f(x) is divided by a linear polynomial p(x) = ( x -a), the quotient is q(x) and the remainder is r(x).

In other words , f(x) and p(x) are two polynomials such that the degree of f(x) \geq degree of p(x) and p(x) \neq 0 then we can find polynomials q(x) and r(x) such that, where r(x) = 0 or degree of r(x) < degree of g(x).

By division algorithm

f(x) = p(x) . q(x) + r(x)

∴ f(x) = (x-a) . q(x) + r(x) [ here p(x) = x – a ]

Since degree of p(x) = (x-a) is 1 and degree of r(x) < degree of (x-a)

∴ Degree of r(x) = 0

This implies that r(x) is a constant , say ‘ k ‘

So, for every real value of x, r(x) = k.

Therefore f(x) = ( x-a) . q(x) + k

If x = a,

then f(a) = (a-a) . q(a) + k = 0 + k = k

Hence the remainder when f(x) is divided by the linear polynomial (x-a) is f(a).

Answer 2

If P(x) is a polynomial function, then the remainder of P(x) when divided by x-k is equal to P(k).

Let p(x) when divided by d(x) the quotient=q(x) and Remainder= R

Then p(x)=d(x)*q(x)+R

here p(x)=(x-k)*q(x)+R

Putting x=k

p(k)=(k-k)*q(x)+R

p(k)=0+R

So R=p(k)