Question

1.) State and prove factor Theorem. 2.) State and prove remainder theorem

Answer 1

FACTOR THEOREM
f f(x) is a polynomial of degree n 1 and ‘ a ‘ is any real number then

(x -a) is a factor of f(x), if f(a) = 0.
and its converse ” if (x-a) is a factor of a polynomial f(x) then f(a) = 0 “
Factor Theorem Proof:

Given that f(x) is a polynomial of degree n 1 by reminder theorem.

f(x) = ( x-a) . q(x) + f(a) . . . . . . . . . . . equation ‘A ‘

1 . Suppose f(a) = 0

then equation ‘A’ f(x) = ( x-a) . q(x) + 0 = ( x-a) . q(x)

Which shows that ( x-a) is a factor of f(x). Hence proved

2 . Conversely suppose that (x-a) is a factor of f(x).

This implies that f(x) = ( x-a) . q(x) for some polynomial q(x).

∴ f(a) = ( a-a) . q(a) = 0.

Hence f(a) = 0 when (x-a) is a factor of f(x).

The factor theorem simply say that If a polynomial f(x) is divided by p(x) leaves remainder zero then p(x) is factor of f(x)

REMAINDER THEOREM
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) degree of p(x) and p(x) 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

(¡) ans. Given in first image send by me.

(¡¡) ans. Given in the second image send my me.

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