# Polynomial Remainder Theorem Important Formulas, Definitions, & Examples:

#### Overview:

 Topic Included: Formulas, Definitions & Exmaples. Main Topic: Quantitative Aptitude. Quantitative Aptitude Sub-topic: Number System Aptitude Notes & Questions. Questions for practice: 10 Questions & Answers with Solutions.

#### What is Polynomial:

A polynomial is an algebraic expression consisting of variables, coefficients, constants, and positive integer exponents, that are connected with operators such as addition, subtraction, multiplication but not by division.

#### Polynomial Degree:

The highest value of the exponent of the variable decides the degree of any polynomial expression. the above equation is a 2-degree polynomial equation because the highest value of the exponent of variable $$x$$ is 2.

$$2$$, Here the degree is 0 because It is a constant. It is also called zero polynomial.

$$2x - 1$$, Here the degree is 1 because the highest value of the exponent of the variable $$x$$ is 1. It is a linear equation or linear polynomial or binomial.

$$3x^2 + 2x - 1$$, Here the degree is 2 because the highest value of the exponent of the variable $$x$$ is 2. It is a quadratic polynomial or trinomial.

$$4x^3 + 3x^2 + 2x - 1$$, Here the degree is 3 because the highest value of the exponent of the variable $$x$$ is 3. It is a cubic polynomial.

#### What is Polynomial Remainder Theorem:

Polynomial Remainder Theorem is used to find the remainder of any polynomial algebraic expression. According to Polynomial remainder theorem-

$$(k + x)^n = k^n + n C_1 k^{n-1} x^1 + n C_2 k^{n-2} x^2 +.............n C_{n-1} k^1 x^{n-1} + x^n$$

$$(k + x)^n =$$ $$= k^n + n C_1 k^{n-1} x^1 + n C_2 k^{n-2} x^2 +...\\..........n C_{n-1} k^1 x^{n-1} + x^n$$

Dividing by k into both sides, we will get.

$$\frac{(k + x)^n}{k} = \frac{k^n + n C_1 k^{n-1} x^1 + n C_2 k^{n-2} x^2 +.............n C_{n-1} k^1 x^{n-1} + x^n}{k}$$

$$\frac{(k + x)^n}{k} =$$ $$= \frac{k^n + n C_1 k^{n-1} x^1 + n C_2 k^{n-2} x^2 +...\\..........n C_{n-1} k^1 x^{n-1} + x^n}{k}$$

In the above polynomial remainder theorem equation, the last term $$x^n$$ is the only term that is not divisible by $$k$$, but the rest of the terms contains $$'k'$$ which is completely divisible by $$k$$, that's why the Remainder of parent equation will be. $$\frac{(k + x)^n}{k} = \frac{x^n}{k}$$

#### Polynomial Remainder Theorem Examples:

Example(1): Find the remainder of the expression $$\frac{7^{99}}{6}$$ by using the polynomial remainder theorem?

Solution: According to the polynomial remainder theorem, $$\frac{7^{99}}{6} = \frac{(6 + 1)^{99}}{6}$$ $$= \frac{1^{99}}{6} = 1 \ (Remainder)$$

Example(2): Find the remainder of the expression $$\frac{2x^2 - 3x - 1}{x - 3}$$ by using the polynomial remainder theorem?

Solution: According to the polynomial remainder theorem, $$\require{enclose} \begin{array}{r} 2x + 3 \\[-3pt] x - 3 \enclose{longdiv}{2x^2 - 3x - 1} \\[-3pt] \underline{2x^2 - 6x}\phantom{ - 1} \\[-3pt] 3x - 1 \\[-3pt] \underline{3x - 9} \\[-3pt] 0 + 8 \end{array}$$ Here 8 is the remainder and $$(2x + 3)$$ is the quotient.

Example(3): Find the remainder of the expression $$\frac{8^{100}}{6}$$ by using the polynomial remainder theorem?

Solution: According to the polynomial remainder theorem, $$\frac{8^{100}}{6} = \frac{(6 + 2)^{100}}{6} = \frac{2^{100}}{6}$$ $$= \frac{2^{99} \times 2}{6} = \frac{(2^3)^{33} \times 2}{6}$$ $$= \frac{(6 + 2)^{3} \times 2}{6} = \frac{2^3 \times 2}{6}$$ $$= \frac{8 \times 2}{6} = \frac{(6 + 2) \times 2}{6}$$ $$= \frac{2 \times 2}{6} = \frac{4}{6} = 4 \ (Remainder)$$

Example(4): Find the remainder of the expression $$\frac{2^{100}}{7}$$ by using the polynomial remainder theorem?

Solution: According to the polynomial remainder theorem, $$\frac{2^{100}}{7} = \frac{(2)^{99} \times 2}{7} = \frac{(2^3)^{33} \times 2}{7}$$ $$= \frac{(7 + 1)^{33} \times 2}{7} = \frac{1^{33} \times 2}{7}$$ $$= \frac{1 \times 2}{7} = 2 \ (Remainder)$$

Example(5): Find the remainder of the expression $$\frac{2^{10}}{3}$$ by using the polynomial remainder theorem?

Solution: According to the polynomial remainder theorem, $$\frac{2^{10}}{3} = \frac{2^9 \times 2}{3} = \frac{(2^3)^3 \times 2}{3}$$ $$= \frac{(3 + 5)^3 \times 2}{3} = \frac{5^3 \times 2}{3}$$ $$= \frac{(3 + 2)^{3} \times 2}{3} = \frac{2^3 \times 2}{3}$$ $$= \frac{16}{3} = 1 \ (Remainder)$$