What Is the Quotient Remainder Theorem Definition Formula Proof and Solved Examples
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 in Polynomial Division
1. What is the Quotient Remainder Theorem?
The Quotient Remainder Theorem states that for any polynomial P(x) divided by a non-zero polynomial D(x), there exist unique polynomials Q(x) and R(x) such that P(x) = D(x)·Q(x) + R(x), where the degree of R(x) is less than the degree of D(x).
- P(x) = dividend polynomial
- D(x) = divisor polynomial
- Q(x) = quotient polynomial
- R(x) = remainder polynomial
2. What is the formula for the Quotient Remainder Theorem?
The formula for the Quotient Remainder Theorem is P(x) = D(x)Q(x) + R(x), where deg R(x) < deg D(x).
- If D(x) is linear, such as (x − a), then R(x) is a constant.
- If D(x) has degree n, then R(x) has degree less than n.
3. How do you use the Quotient Remainder Theorem to divide polynomials?
To use the Quotient Remainder Theorem, divide the polynomial using long division or synthetic division and express the result as P(x) = D(x)Q(x) + R(x).
- Step 1: Arrange polynomials in descending powers.
- Step 2: Divide the leading terms.
- Step 3: Multiply and subtract.
- Step 4: Repeat until the degree of remainder is less than the divisor.
4. What is the remainder when a polynomial is divided by (x − a)?
When a polynomial P(x) is divided by (x − a), the remainder is P(a).
- This result follows from the Remainder Theorem.
- If P(a) = 0, then (x − a) is a factor.
5. What is the difference between the Quotient Remainder Theorem and the Remainder Theorem?
The Quotient Remainder Theorem gives the full expression P(x) = D(x)Q(x) + R(x), while the Remainder Theorem specifically finds the remainder when dividing by (x − a).
- Quotient Remainder Theorem applies to any polynomial divisor.
- Remainder Theorem applies only to linear divisors of the form (x − a).
6. Can you give an example of the Quotient Remainder Theorem?
An example of the Quotient Remainder Theorem is dividing P(x) = x³ − 1 by D(x) = x − 1, which gives P(x) = (x − 1)(x² + x + 1) + 0.
- Quotient Q(x) = x² + x + 1
- Remainder R(x) = 0
7. Why is the degree of the remainder less than the degree of the divisor?
The degree of the remainder must be less than the degree of the divisor because otherwise further division would still be possible.
- If deg R(x) ≥ deg D(x), divide again.
- Division stops only when deg R(x) < deg D(x).
8. How is synthetic division related to the Quotient Remainder Theorem?
Synthetic division is a shortcut method of applying the Quotient Remainder Theorem when dividing by (x − a).
- It directly computes Q(x) and the remainder.
- The final number in synthetic division equals P(a).
9. What happens if the remainder is zero in the Quotient Remainder Theorem?
If the remainder R(x) is 0, then the divisor D(x) is an exact factor of P(x).
- This means P(x) = D(x)Q(x).
- The division is exact.
10. How is the Quotient Remainder Theorem used in factorization of polynomials?
The Quotient Remainder Theorem helps factor polynomials by testing possible divisors and checking if the remainder is zero.
- Use the Remainder Theorem to test values.
- If P(a) = 0, then (x − a) is a factor.
- Divide to find the quotient polynomial.
