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Answers
3. Divisor
4. x=0
5. Many
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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).