शेषफल प्रमेय

शेषफल प्रमेय सूत्र का उपयोग किसी बहुपद को रैखिक बहुपद से विभाजित करने पर शेषफल ज्ञात करने के लिए किया जाता है।

शेषफल प्रमेय

शेषफल प्रमेय में कहा गया है कि "जब एक बहुपद $p(x)$को एक रैखिक बहुपद $(x-a)$ ,से विभाजित किया जाता है, तो शेषफल $p(a)$होता है।"

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शेषफल प्रमेय

The remainder theorem is a fundamental concept in algebra and is used to find the remainder when a polynomial is divided by a linear expression. We know that the remainder by dividing one polynomial by other can be found by long division of polynomials. Instead, we can use the remainder theorem to find the remainder easily. This has many important applications:

• It is used to find the remainder when a polynomial is divided by another linear polynomial.
• It helps in the factorization of polynomials.
• It helps in determining the zeros of a polynomial.

But the remainder theorem has some limitations. This theorem works only when the divisor is linear.

What is the Remainder Theorem?

The remainder theorem states that when a polynomial $p(x)$ is divided by a linear polynomial $(x-a)$), then the remainder is equal to $p(a)$. The remainder theorem enables us to calculate the remainder of the division of any polynomial by a linear polynomial, without actually carrying out the steps of the long division.

Note that the degree of the remainder polynomial is always $1$ less than the degree of the divisor polynomial. Using this fact, when any polynomial is divided by a linear polynomial (whose degree is $1$), the remainder must be a constant (whose degree is $0$).

Remainder Theorem Statement and Proof

According to the remainder theorem, when a polynomial $p(x)$ (whose degree is greater than or equal to $1$) is divided by a linear polynomial $x-a$, the remainder is given by

$r=p(a)$. i.e., to find the remainder, follow the steps below:

• Find the zero of the linear polynomial by setting it to zero. i.e., $x-a=0\Rightarrow x=a$
• Then just substitute it in the given polynomial. The result would give the remainder.

Here is the remainder theorem formula depending on the type of divisor (linear polynomial).

When $p(x)$is divided by$(x-a)$

Remainder = $p(a)$

or

When $p(x)$ is divided by $(ax+b)$

Remainder = $p\left({\frac {-b}{a}}\right)$

Similarly, we can extend the remainder theorem for different types of linear polynomials as follows:

• The remainder when $p(x)$ is divided by $(x-a)$is $p(a)$
• The remainder when $p(x)$ is divided by $(ax+b)$is $p\left({\frac {-b}{a}}\right)$
• The remainder when $p(x)$ is divided by $(ax-b)$ is $p\left({\frac {b}{a}}\right)$
• The remainder when $p(x)$ is divided by $(bx-a)$ is $p\left({\frac {a}{b}}\right)$

Proof of Remainder Theorem

Let us assume that $q(x)$ and '$r$' are the quotient and the remainder respectively when a polynomial $p(x)$ is divided by a linear polynomial $(x-a)$. By division algorithm, Dividend = (Divisor × Quotient) + Remainder.

Using this, $p(x)=(x-a)\times q(x)+r$.

Substitute $x=a$

$p(a)=(a-a)\times q(a)+r$

$p(a)=(0)\times q(a)+r$

$p(a)=r$

i.e. the remainder = $p(a)$

Hence, proved.

Example

Find the remainder when the polynomial $p(x)=3x^{3}+x^{2}+2x+5$ is divided by $x+1$.

	$3x^{2}-2x+4$
$x+1$	$3x^{3}$	$+$	$x^{2}$	$+$	$2x$	$+$	$5$
	$3x^{3}$	$+$	$3x^{2}$
		$-$	$2x^{2}$	$+$	$2x$	$+$	$5$
		$-$	$2x^{2}$	$-$	$2x$
				$+$	$4x$	$+$	$5$
				$+$	$4x$	$+$	$4$
							1

Here, quotient = $3x^{2}-2x+4$

Remainder = $1$

Verification :

Given, the divisor is $x+1$, i.e. it is a factor of the given polynomial $p(x)$

Let $x+1=0$

$x=-1$

Substituting $x=-1$ in $p(x)$,

$p(x)=3x^{3}+x^{2}+2x+5$

$p(-1-)=3(-1)^{3}+(-1)^{2}+2(-1)+5$

$p(-1-)=3(-1)+1-2+5$

$p(-1-)=-3+1-2+5$

$p(-1-)=1$

Remainder  = Value of $p(x)$ at $x=-1$.

Hence proved the remainder theorem.

Important Notes on Remainder Theorem:

• The remainder theorem says "when a polynomial p(x) is divided by a linear polynomial whose zero is x = k, the remainder is given by p(k)".
• The basic formula to check the division is: Dividend = (Divisor × Quotient) + Remainder.
• The remainder theorem does not work when the divisor is not linear.
• Also, it does not help to find the quotient.

जब एक बहुपद को एक रैखिक बहुपद द्वारा विभाजित किया जाता है तो शेषफल ज्ञात करने के लिए शेषफल प्रमेय सूत्र का उपयोग किया जाता है।

शेषफल प्रमेय

शेषफल प्रमेय में कहा गया है कि "जब एक बहुपद $p(x)$ को एक रैखिक बहुपद $(x-a)$ से विभाजित किया जाता है, तो शेषफल $p(a)$ होता है"।

उदाहरण

बहुपद $p(x)=3x^{3}+x^{2}+2x+5$ को $x+1$ से विभाजित करने पर शेषफल ज्ञात कीजिए

	$3x^{2}-2x+4$
$x+1$	$3x^{3}$	$+$	$x^{2}$	$+$	$2x$	$+$	$5$
	$3x^{3}$	$+$	$3x^{2}$
		$-$	$2x^{2}$	$+$	$2x$	$+$	$5$
		$-$	$2x^{2}$	$-$	$2x$
				$+$	$4x$	$+$	$5$
				$+$	$4x$	$+$	$4$
							1

यहाँ, भागफल = $3x^{2}-2x+4$

शेषफल = $1$

सत्यापन:

दिया गया है कि भाजक $x+1$ है, अर्थात यह दिए गए बहुपद $p(x)$ का एक गुणनखंड है।

मान लीजिए $x+1=0$

$x=-1$

$x=-1$ को $p(x)$ में प्रतिस्थापित करने पर ,

$p(x)=3x^{3}+x^{2}+2x+5$

$p(-1-)=3(-1)^{3}+(-1)^{2}+2(-1)+5$

$p(-1-)=3(-1)+1-2+5$

$p(-1-)=-3+1-2+5$

$p(-1-)=1$

शेषफल  = $x=-1$ पर $p(x)$ का मान ।

अतः शेषफल प्रमेय सिद्ध हुआ।

शेषफल प्रमेय पर महत्वपूर्ण टिप्पणियाँ:

• शेष प्रमेय कहता है "जब एक बहुपद p(x) को एक रैखिक बहुपद से विभाजित किया जाता है जिसका शून्य x = k है, तो शेष p(k) द्वारा दिया जाता है"।
• विभाजन की जाँच करने का मूल सूत्र है: लाभांश = (भाजक × भागफल) + शेषफल।
• जब भाजक रैखिक नहीं होता है तो शेषफल प्रमेय काम नहीं करता है।
• साथ ही, यह भागफल ज्ञात करने में सहायता नहीं करता है।