Question

State and prove remainder theorem and factor theorem​

Answer 1

Answer:

The remainder theorem states the following: If you divide a polynomial f(x) by (x - h), then the remainder is f(h). The theorem states that our remainder equals f(h). Therefore, we do not need to use long division, but just need to evaluate the polynomial when x = h to find the remainder

Answer 2

$\huge\boxed{\fcolorbox{lime}{yellow}{⭐ANSWER⤵࿐}}$

$\sf\large\underline\green{✴Introduction}$

In mathematics, factor theorem is used when factoring the polynomials completely. It is a theorem that links factors and zeros of the polynomial.

According to factor theorem, if 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.

Also, we can say, if (x-a) is a factor of polynomial f(x), then f(a) = 0. This proves the converse of the theorem. Let us see the proof of this theorem along with examples.

$\sf\large\underline\red{⭐Factor Theorem :-}$

Factor theorem is commonly used for factoring a polynomial and finding the roots of the polynomial. It is a special case of a polynomial remainder theorem.

As discussed in the introduction, a polynomial f(x) has a factor (x-a), if and only if, f(a) = 0. It is one of the methods to do the factorisation of a polynomial.

$\sf\large\underline\blue{✴Proof:-}$

Here we will prove the factor theorem, according to which we can factorise the polynomial.

Consider a polynomial f(x) which is divided by (x-c), then f(c)=0.

Using remainder theorem,

➡f(x)= (x-c)q(x)+f(c)

Where f(x) is the target polynomial and q(x) is the quotient polynomial.

➡Since, f(c) = 0, hence,

➡f(x)= (x-c)q(x)+f(c)

➡f(x) = (x-c)q(x)+0

➡f(x) = (x-c)q(x)

Therefore, (x-c) is a factor of the polynomial f(x).

$\sf\large\underline\red{⭐RemainderTheorem :-}$

Remainder Theorem is an approach of Euclidean division of polynomials. According to this theorem, if we divide a polynomial P(x) by a factor ( x – a); that isn’t essentially an element of the polynomial; you will find a smaller polynomial along with a remainder. This remainder that has been obtained is actually a value of P(x) at x = a, specifically P(a). So basically, x -a is the divisor of P(x) if and only if P(a) = 0. It is applied to factorize polynomials of each degree in an elegant manner.

For example: if f(a) = a3-12a2-42 is divided by (a-3) then the quotient will be a2-9a-27 and the remainder is -123.

if we put, a-3 = 0

then a = 3

Hence, f(a) = f(3) = -123

Thus, it satisfies the remainder theorem.

$\sf\large\underline\blue{✴Proof :-}$

Theorem functions on an actual case that a polynomial is comprehensively dividable, at least one time by its factor in order to get a smaller polynomial and ‘a’ remainder of zero. This acts as one of the simplest ways to determine whether the value ‘a’ is a root of the polynomial P(x).

That is when we divide p(x) by x-a we obtain

p(x) = (x-a)·q(x) + r(x),

as we know that Dividend = (Divisor × Quotient) + Remainder

But if r(x) is simply the constant r (remember when we divide by (x-a) the remainder is a constant)…. so we obtain the following solution, i.e

p(x) = (x-a)·q(x) + r

Observe what happens when we have x equal to a:

p(a) = (a-a)·q(a) + r

p(a) = (0)·q(a) + r

p(a) = r

Hence, proved.

$\sf\large\underline\red{⭐Hope It's Help You !!✔ }$