Math, asked by munnavathdeepika, 1 month ago

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Answered by kumariaradhana371
2

3. Divisor

4. x=0

5. Many

hope this will help you

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Answered by jhanviv
0

Answer:

3) divisor

4) c) x=0

5) c) depends on whether the equation is in the form ax+by=c

6) Let f(x) be any polynomial of degree greater than or equal to one and let ‘ a‘ be any number. If f(x) is divided by the linear polynomial (x-a) then the remainder is f(a).

Remainder Theorem Proof:

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).

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