An Introduction to Quotient Remainder Theorem
The quotient remainder theorem states that when a linear polynomial \[q\left( x \right)\] with a zero \[x\text{ }=\text{ }a\] divides a polynomial \[p\left( x \right)\] (whose degree is greater than or equal to 1), the remainder is given by \[r\text{ }=\text{ }p\text{ }\left( a \right)\]. Without actually doing the long division steps, the quotient remainder theorem allows us to calculate the remainder of the division of any polynomial by a linear polynomial.
Table of Contents
An Introduction to Quotient Remainder Theorem
History of Etienne Bezout
Statement of the Quotient Remainder Theorem
Proof of the Quotient Remainder Theorem
Applications of the Quotient Remainder Theorem
Solved Examples
History of Etienne Bezout
Ettiene Bezout
Name: Ettiene Bezout
Born: 31 March 1930
Died: 27 September 1983
Contribution: He is credited with finding the remainder theorem in Polynomials
Statement of the Quotient Remainder Theorem
Given any integer $A$, and a positive integer $B$, there exist unique integers $Q$ and $R$ such that
$A=BQ+R$ where $0\le R\le B$
Proof of the Quotient Remainder Theorem
Let us assume that $q(x)$ and $r(x)$ are the quotient and the remainder, respectively. When a polynomial $p(x)$ is divided by a linear polynomial $(x-a)$,
By the division algorithm, $Dividend=Divisor\times Quotient+\operatorname{Re}mainder$
Using this, $p(x)=(x-a)q(x)+r(x)$
Substitute $x=a$ and $r(x)=r$
$p(a)=(a-a)q(a)+r$
$p(a)=0\times q(a)+r$
$p(a)=r$
Hence proved.
Applications of the Theorem
This theorem helps us to get the quotient and the remainder without actually performing the long division method.
This theorem also has some real-life applications.
Limitations of the theorem:
The quotient remainder theorem doesn’t work properly with the polynomials having repeated roots.
The remainder theorem does not work when the divisor is not linear
It does not provide us with the quotient.
Solved Examples
1. Find the value of k if $\mathbf{p}\left( \mathbf{x} \right)\text{ }=\text{ }\left( \mathbf{3x}-\text{ }\text{ }\mathbf{2} \right)\left( \mathbf{x}-\text{ }\text{ }\mathbf{k} \right)\text{ }-\text{ }\mathbf{8}$ is divided by $\left( \mathbf{x}-\text{ }\text{ }\mathbf{2} \right)$ leaving the remainder $4$.
Ans: Given,
$p(x)=(3x-2)(x-k)-8$
Also, it is given that the remainder is $4$ when $p(x)$ is divided by $(x-2)$.
So, $a\text{ }=\text{ }2$ and $r\text{ }=\text{ }4$
Using the Remainder theorem,
$p(a)=r$
$p(2)=4$
$[3(2)-2](2-k)-8=4$
$(6-2)(2-k)=4+8$
$4(2-k)=12$
$k=2-3=-1$
The value of \[k\] is \[-1\].
2. If $\left( \mathbf{x}\text{ }-\text{ }\mathbf{8} \right)$ is one of the factors of $\mathbf{m}{{\mathbf{x}}^{\mathbf{3}}}~\text{}\mathbf{24}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{192x}\text{ }\text{ }\mathbf{512}$, find the value of m.
Ans: Let the given polynomial be $p(x)=m{{x}^{3}}-24{{x}^{2}}+192x-512$
Given that $\left( x\text{ }-\text{ }8 \right)$ is one of the factors of p(x):
So, \[r=0\] and \[a=8\].
By the Remainder theorem,
$p(a)=r$
$p(8)=0$
$m{{(8)}^{3}}-24{{(8)}^{2}}+192(8)-512=0$
$512m-1536+1536-512=0$
$512m=512$
$m=512/512=1$
The value of \[m\] is \[1\].
3. Find the remainder using the Remainder theorem for the following expression.
$({{\mathbf{x}}^{\mathbf{4}}}~-\text{}\mathbf{5}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }{{\mathbf{x}}^{\mathbf{2}}}~\text{ }-\mathbf{2x}\text{ }+\text{ }\mathbf{6})\text{ }\div \text{ }\left( \mathbf{x}\text{ }+\text{ }\mathbf{4} \right)$.
Ans: Here, $p\left( x \right)\text{ }=({{\mathbf{x}}^{\mathbf{4}}}~-\text{}\mathbf{5}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }{{\mathbf{x}}^{\mathbf{2}}}~\text{ }-\mathbf{2x}\text{ }+\text{ }\mathbf{6})$
And $x\text{ }+\text{ }4\text{ }=\text{ }x\text{ }-\text{ }\left( -4 \right)$
Comparing with $x\text{ }-\text{ }a$, we have $a\text{ }=\text{ }-4$
By the Remainder theorem,
$r=p(a)$
$r=\text{p}(-4)$
$r={{(-4)}^{4}}-5{{(-4)}^{3}}+{{(-4)}^{2}}-2(-4)+6$
$r=256+5(64)+16+8+6$
$r=256+320+16+8+6$
$r=606$
Hence, the remainder is $606$.
Summary
A remainder Theorem is an approach to the 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.
Important Formulas:
Dividend = (Divisor × Quotient) + Remainder.
When $P(x)$ is divided by $(x-a)$, remainder = $P(a)$
List of Related Articles
FAQs on Quotient-Remainder Theorem
1. What is the Quotient-Remainder Theorem for polynomials as per the CBSE syllabus?
The Quotient-Remainder Theorem, often referred to as the Division Algorithm for Polynomials, states that if p(x) is any polynomial and g(x) is a non-zero polynomial, then there exist unique polynomials q(x) (the quotient) and r(x) (the remainder) such that: p(x) = g(x) ⋅ q(x) + r(x). A key condition is that the degree of the remainder, r(x), must be less than the degree of the divisor, g(x), or the remainder r(x) is zero.
2. Can you provide a simple example of applying the Quotient-Remainder Theorem?
Certainly. Let's find the remainder when the polynomial p(x) = x³ + 2x² - 5x + 8 is divided by g(x) = x - 2. According to the theorem, the remainder is p(2). We find the zero of the divisor (x - 2 = 0, so x = 2) and substitute this value into the dividend:
- p(2) = (2)³ + 2(2)² - 5(2) + 8
- p(2) = 8 + 2(4) - 10 + 8
- p(2) = 8 + 8 - 10 + 8
- p(2) = 14
Thus, the remainder is 14. The complete division would be x³ + 2x² - 5x + 8 = (x - 2)(x² + 4x + 3) + 14.
3. How does the Remainder Theorem provide a shortcut for finding the remainder?
The Remainder Theorem provides a major shortcut by allowing you to find the remainder without performing the entire long division process. Instead of dividing the whole polynomial, you simply need to:
- Find the zero of the linear divisor (the value of 'x' that makes it zero).
- Substitute this value into the dividend polynomial.
- Evaluate the result to get the remainder.
4. What is the proof for the Remainder Theorem?
The proof is based on the Division Algorithm. Let p(x) be any polynomial of degree ≥ 1, and let 'a' be any real number. When p(x) is divided by the linear polynomial (x - a), let the quotient be q(x) and the remainder be r(x).
- By the Division Algorithm, we have: p(x) = (x - a) ⋅ q(x) + r(x).
- Since the degree of the divisor (x - a) is 1, the degree of the remainder r(x) must be less than 1, meaning its degree is 0. A polynomial of degree 0 is a constant. Let's call this constant 'r'.
- The equation becomes: p(x) = (x - a) ⋅ q(x) + r.
- This relation holds for all values of x. By substituting x = a, we get:
- p(a) = (a - a) ⋅ q(a) + r
- p(a) = (0) ⋅ q(a) + r
- p(a) = r
This proves that the remainder 'r' is precisely the value of the polynomial p(x) at x = a.
5. Why does substituting the zero of the divisor into the dividend give the remainder?
This works because of the fundamental structure of the division equation: Dividend = (Divisor × Quotient) + Remainder. When the divisor is a linear polynomial like (x - a), the equation is p(x) = (x - a)q(x) + r. By substituting x = a (the zero of the divisor), the term (x - a) becomes (a - a), which is 0. This makes the entire '(Divisor × Quotient)' part of the equation zero, regardless of the quotient's value. You are left with just the remainder: p(a) = 0 + r, which simplifies to p(a) = r.
6. How are the Remainder Theorem and the Factor Theorem related?
The Factor Theorem is a special case of the Remainder Theorem. The Remainder Theorem states that when a polynomial p(x) is divided by (x - a), the remainder is p(a). The Factor Theorem takes this one step further by adding a condition on the remainder:
- It states that if the remainder p(a) is zero, then (x - a) must be a factor of the polynomial p(x).
- Conversely, if (x - a) is a factor of p(x), then dividing by it will leave no remainder, meaning p(a) must be zero.
In essence, the Remainder Theorem finds the remainder, while the Factor Theorem uses a remainder of zero to identify factors.
7. What is the main difference between the quotient and the remainder in polynomial division?
In the division of a polynomial p(x) by g(x), the quotient and remainder are two distinct results:
- The Quotient (q(x)) is the primary result of the division, representing how many times the divisor g(x) 'fits into' the dividend p(x).
- The Remainder (r(x)) is what is 'left over' after the division is complete. Its defining property is that its degree is always strictly less than the degree of the divisor g(x). If the remainder is 0, the division is exact.
8. What are the conditions for using the Remainder Theorem as a shortcut?
The Remainder Theorem, when used as a shortcut, has one crucial condition: the divisor must be a linear polynomial, meaning it is of the form (x - a) or (ax + b). The theorem allows you to find the remainder by substituting the zero of this linear divisor. While the general Division Algorithm applies to divisors of any degree, the simple shortcut of evaluating p(a) only works for linear divisors. For divisors with a degree of 2 or higher (e.g., quadratic), you must use polynomial long division to find the remainder.
