Remainder Theorem - Polynomials, Statement, Proof and Examples

Polynomial expression

An expression of the form $f(x)=a_0 x^n+a_1 x^{n-1}+a_2 x^{n-2}+\ldots+a_{n-1} x+a_n$, is called a polynomial expression.

Where $x$ is variable and $a_0, a_1, a_2, \ldots \ldots . ., a_n$ are constant, known as coefficients and $a_0 \neq 0, n$ is non-negative integer,

Degree: The highest power of the variable in the polynomial expression is called the degree of the polynomial. In $a_0 \cdot x^n+a_1 \cdot x^{n-1}+\ldots+a_n$ , the highest power of x is n, so the degree of this polynomial is n.

If coefficients are real numbers then it is called a real polynomial, and when they are complex numbers, the polynomial is called a complex polynomial.

The root of polynomial

If $\mathrm{f}(\mathrm{x})$ is a polynomial, $\mathrm{f}(\mathrm{x})=0$ is called a polynomial equation.
The value of x for which the polynomial equation, $\mathrm{f}(\mathrm{x})=0$ is satisfied is called a root of the polynomial equation.
If $x=a$ is a root of the equation $f(x)=0$, then $f(a)=0$.
Eg, $x=2$ is a root of $x^2-3 x+2=0$, as $x=2$ satisfies this equation.
A polynomial equation of degree n has n roots (real or imaginary).

Equation of higher degree

An equation of the form $a_0 x^n+a_1 x^{n-1}+\ldots+a_{n-1} x+a_n=0$ where $\mathrm{a}_0, \mathrm{a}_1, \ldots, \mathrm{a}_{\mathrm{n}}$ are constant and $\mathrm{a}_0 \neq 0$

is known as the polynomial equation of degree n which has exactly n roots (i.e., number of real roots + number of imaginary roots = n)

Relation between its coefficients and roots

sum of all roots $=\sum \alpha_1=\alpha_1+\alpha_2+\ldots+\alpha_{n-1}+\alpha_n=(-1) \frac{a_1}{a_0}$
sum of products taken two at a time

$
\sum \alpha_1 \alpha_2=\alpha_1 \alpha_2+\alpha_1 \alpha_3+\ldots+\alpha_1 \alpha_{\mathrm{n}}+\alpha_2 \alpha_3+\ldots+\alpha_2 \alpha_{\mathrm{n}}+\ldots+\alpha_{\mathrm{n}-1} \alpha_{\mathrm{n}}=(-1)^2 \frac{\mathrm{a}_2}{\mathrm{a}_0}
$

sum of products taken three at a time

$
\sum \alpha_1 \alpha_2 \alpha_3=(-1)^3 \frac{a_3}{a_0}
$

product of all roots $=\alpha_1 \alpha_2 \ldots \alpha_{\mathrm{n}}=(-1)^{\mathrm{n}} \frac{\mathrm{a}_{\mathrm{n}}}{\mathrm{a}_0}$

For example,

Suppose $\mathrm{n}=3$ and $a x^3+b x^2+c x+d=0$ is polynomial equation with a ≠ 0 and $\alpha$, $\beta$ and $\gamma$ are the roots of the equation then :

$\begin{aligned} & \alpha+\beta+\gamma=-\frac{\mathrm{b}}{\mathrm{a}} \\ & \sum \alpha \beta=\alpha \beta+\beta \gamma+\gamma \alpha=(-1)^2 \frac{\mathrm{c}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{a}} \\ & \alpha \beta \gamma=(-1)^3 \frac{\mathrm{d}}{\mathrm{a}}=-\frac{\mathrm{d}}{\mathrm{a}}\end{aligned}$

Transformation of roots

For the transformation of roots, we can use the same procedure we used in the case of quadratic equations.

Remainder theorem

The remainder theorem states that if a polynomial f(x) is divided by a linear function (x - k), then the remainder is f(k).

In Division,

Dividend = Divisor x Quotient + Remainder

For polynomials also we can use this theorem

$f(x)=d(x) \cdot q(x)+r(x)$

where $f(x)$ is the divisor, $d(x)$ is the divisor, $q(x)$ is the quotient and $r(x)$ is the remainder. And these 4 are polynomials
The degree of remainder $r(x)$ is always less than degree of divisor $d(x)$
Now, if divisor $d(x)$ is a linear polynomial $(x-k)$. Let $q(x)$ be the quotient, remainder $r(x)$ will be a constant value equal to $R$ :

$
f(x)=(x-k) q(x)+R
$

Now if we put $x=k$
i.e. $\quad f(k)=(k-k) q(x)+R=0+R$

$
f(k)=R
$
So, remainder is $\mathrm{f}(\mathrm{k})$, when $\mathrm{f}(\mathrm{x})$ is divided by a linear polynomial $(\mathrm{x}-\mathrm{k})$
Eg. To find remainder when $f(x)=2 x^3-3 x-4$ is divided by $(x-3)$,
Here $k=3$, So remainder will be $f(k)=f(3)=2 .(3)^3-3(3)-4=54-9-4=41$

Steps to Divide a Polynomial by a Non-Zero Polynomial

• First, arrange the polynomials (dividend and divisor) in the decreasing order of its degree
• Divide the first term of the dividend by the first term of the divisor to produce the first term of the quotient
• Multiply the divisor by the first term of the quotient and subtract this product from the dividend, to get the remainder.
• This remainder is the dividend now and divisor will remain same
• Again repeat from the first step, until the degree of the new dividend is less than the degree of the divisor.

Euler Remainder Theorem

Euler's theorem states that if $n$ and $X$ are two co-prime positive integers, then:

$
X^{\varphi(n)} \equiv 1 \quad(\bmod n)
$

where $\varphi(n)$ is Euler's totient function, defined as:

$
\varphi(n)=n\left(1-\frac{1}{a}\right)\left(1-\frac{1}{b}\right)\left(1-\frac{1}{c}\right)
$

for a natural number $n$ expressed in terms of its prime factorization as:

$
n=a^p \cdot b^q \cdot c^r
$

where $a, b, c$ are distinct prime factors of $n$, and $p, q, r$ are positive integers.

Factor Theorem

Now if $f(k)=0$, then this means that the remainder when $f(x)$ is divided by $(x-k)$ is 0.
As the remainder is 0 , so $(x-k)$ is a factor of $f(x)$
So, the factor theorem states that if $\mathrm{f}(\mathrm{k})-0$, then $(\mathrm{x}-\mathrm{k})$ is a factor of $\mathrm{f}(\mathrm{x})$.
Eg, $f(x)=x^3+3 x-4$
Now we can observe by hit and trial that $f(1)=1+3-4=0$, so $(x-1)$ is a factor of $f(x)$.

Differences Between the Remainder Theorem and Factor Theorem

Applications of Remainder Theorem

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.

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.

Recommended Video Based on Remainder Theorem

Solved Examples based on Remainder Theorem

Example 1: If $2+3 i$ is one of the roots of the equation, $2 x^3-9 x^2+k x-13=0, k \in R$ then the real root of this equation:

1) does not exist.

2) exists and is equal to $\frac{1}{2}$

3) exists and is equal to $-\frac{1}{2}$

4) exists and is equal to 1

Solution

As we have learned

The sum of roots of cubic Equation -

$\alpha+\beta+\gamma=\frac{-b}{c}$

Product of roots of the cubic equation -

$\alpha \beta \gamma=\frac{-d}{a}$

- wherein

$a x^3+b x^2+c x+d=0$

is the cubic equation

As complex roots always exist as conjugates,

$\begin{aligned} & \alpha=2+3 i \\ & \beta=2-32 \\ & \gamma=? \\ & \alpha+\beta+\gamma=9 / 2 \\ & \text { and } \alpha \beta \gamma=13 / 2 \\ & (4+9) \gamma=13 / 2 \\ & \gamma=1 / 2\end{aligned}$

Hence, the answer is the option 2.

Example 2: The sum of the real roots of the equation

$\left|\begin{array}{ccc}x & -6 & -1 \\ 2 & -3 x & x-3 \\ -3 & 2 x & x+2\end{array}\right|=0$, is equal to :

1) 6

2) 0

3) 1

4) -4

Solution:

Sum of roots of cubic Equation -

$\alpha+\beta+\gamma=\frac{-b}{a}$

- wherein

$a x^3+b x^2+c x+d=0$ is the cubic equation

$\left|\begin{array}{ccc}x & -6 & -1 \\ 2 & -3 x & x-3 \\ -3 & 2 x & x+2\end{array}\right|=0$

$\begin{aligned} & \Rightarrow x \cdot(-3 x(x+2)-2 x(x-3))-(-6)(2 \cdot(x+2)-(-3)(x-3))+(-1)(2.2 x-(-3)(-3 x)] \\ & \Rightarrow x^3-7 x+6 \\ & \text { Root of equation }(-3,1,2)\end{aligned}$

So,

Sum of real root of equation$=-3+1+2=0$

Example 3: Let $\alpha, \beta$ are two roots of $x^3+p x^2+q x+r=0$ \& satisfies $\alpha \beta=-1$, if $r \neq 0$ then $r^2+p r+q \mid$equals

1) 0

2) 1

3) 2

4) 3

Solution

As we learnt in

Product of roots of cubic equation -

$\alpha \beta \gamma=\frac{-d}{d}$

- wherein

$a x^3+b x^2+c x+d=0$

is the cubic equation

$\begin{aligned} & \alpha \beta \gamma=-r \\ & \therefore \alpha \beta=-1 \Rightarrow \gamma=r \\ & \therefore \text { it will satisfy the equation } \\ & \Rightarrow r^3+p r^2+q r+r=0 \\ & \Rightarrow r^2+p r+q=-1 \\ & \therefore\left|r^2+p r+q\right|=1\end{aligned}$

Hence, the answer is the option 2.

Example 3: If $\alpha, \beta, \gamma$ are roots of $x^3-x^2-1=0$ then $\frac{\alpha+1}{\alpha}+\frac{\beta+1}{\beta}+\frac{\gamma+1}{2}$ equals
1) 2

2) 3

3) 4

4) 5

Solution

As we learnt in

Sum of product of pair of roots in cubic equation -

$\alpha \beta+\beta \gamma+\gamma \alpha=\frac{c}{a}$

- wherein

$a x^3+b x^2+c x+d=0$

is the cubic equation

$\begin{aligned} & \alpha+\beta+\gamma=1 \\ & \alpha \beta+\beta \gamma+\gamma \alpha=0 ; \alpha \beta \gamma=1 \\ & \frac{\alpha+1}{\alpha}+\frac{\beta+1}{\beta}+\frac{\gamma+1}{\gamma}=1+\frac{1}{\alpha}+1+\frac{1}{\beta}+1+\frac{1}{\gamma}=3+\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} \\ & 3+\frac{\alpha \beta+\beta \gamma+\gamma \alpha}{\alpha \beta \gamma}=3+\frac{0}{1}=3\end{aligned}$

Hence, the answer is the option 2.

Example 5: If $\alpha, \beta, \gamma$ are roots of $x\left(1+x^2\right)+x^2(6+x)+2=0$ then $\alpha^{-1}+\beta^{-1}+\gamma^{-1}$ equals

1) -1

2) -0.5

3) 0

4) 0.5

Solution

As we learnt in

Sum of product of pair of roots in cubic equation -

$\alpha \beta+\beta \gamma+\gamma \alpha=\frac{c}{\sigma}$

- wherein

$a x^3+b x^2+c x+d=0$

is the cubic equation

Equation becomes : $2 x^3+6 x^2+x+2=0$

$\because \alpha, \beta, \gamma$, are roots , so

$\begin{aligned} \alpha \beta+\beta \gamma+\gamma \alpha=\frac{1}{2} \text { and } \alpha \beta \gamma & =-1 \\ \therefore \frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=\frac{\beta \gamma+\alpha \gamma+\alpha \beta}{\alpha \beta \gamma} & =\frac{\frac{1}{2}}{-1}=\frac{-1}{2}\end{aligned}$

Hence, the answer is the option 2.