Question

Define factor theorem and remainder theorem​

Answer 1

Answer:

In algebra, the factor theorem is a theorem linking factors and zeros of a polynomial. The factor theorem states that a polynomial has a factor if and only if.

In algebra, the polynomial remainder theorem or little Bézout's theorem is an application of Euclidean division of polynomials. It states that the remainder of the division of a polynomial by a linear polynomial is equal to .

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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.

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