Remainder Theorem - Polynomials, Statement, Proof and Examples

Polynomials play an important role in algebra and calculus. A polynomial equation is the equation of degree n which has exactly n roots. The remainder theorem provides insights into polynomial division and equations. Further, a remainder theorem has numerous applications in physics, engineering, astronomy, etc.

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 $f(k)=f(3)=2 .(3)^3-3(3)-4=54-9-4=41$
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 $f(x)$.

Summary

Polynomials of higher degrees are important in various fields of science and mathematics. Understanding the properties and solving the equation can help solve complex problem and taking out insights. By mastering these concepts, one can gain deeper insights into the behavior of polynomial functions and their applications in real-world scenarios.

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Solved Examples

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

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.