Remainder Theorem (original) (raw)

Last Updated : 22 Apr, 2026

The Remainder Theorem is an algebraic concept that allows us to quickly determine the remainder when a polynomial is divided by a linear expression of the form (x − a), without performing a long division.

**Statement:

Instead of actually dividing the polynomial, we can just substitute x = ai nto the polynomial to get the remainder.

**Example: Divide 2x3 + 3x2 + 4x + 5 by x + 2

Given,

• Dividend = p(x) = 2x3 + 3x2 + 4x + 5
• Divisor = s(x) = (x + 2)

**Using long division:

• Quotient = q(x) =2x2 - x + 6
• Remainder = r(x) = -7

Now, let’s simplify factorization using the Remainder Theorem.

First, find the zero of the divisor:

• x + 2 = 0
• x = -2

According to Remainder Theorem, substituting x = -2 in p(x),

Now, Remainder for p(-2) = -7

Thus, Remainder Theorem is verified.

Alternate Method

We know that any polynomial p(x) can be written as :

p(x) = (x−a)*q(x)+r, where q(x) is the quotient and r is the remainder.

If we substitute x = a in the above equation, we get

p(a) = r

Thus, the remainder is equal to the value of the polynomial at x=a

Similarly, for the divisor (x+2),

p(x) = (x+2)*q(x)+r

Putting x = −2 , we get

p(−2) = r

Hence, the remainder is p(−2)
Hence, proved.

**Applications

• **Simplifying polynomial division: The theorem eliminates the need for long division in certain cases.
• **Factoring polynomials: Using the theorem with the Factor Theorem allows for easier factoring.
• **Solving polynomial equations: It helps in checking whether a given number is a root of the polynomial.
• **Signal processing: In engineering, polynomials are used to model signals, and the Remainder Theorem aids in simplifying these models.

**Related Articles

• Euler's Theorem
• Types of Polynomials
• Multiplying Polynomials
• Dividing Polynomial

Solved Examples

**Example 1: Find the remainder when p(x) = x4 - x3 + x2 - 2x + 1 is divided by g(x) = x - 2.

Given,

p(x) = x4 - x3 + x2 - 2x + 1

g(x) = x-2

Using Remainder theorem,

p(2) = (2)4 - (2)3 + (2)2 - 2(2) + 1 = 9

Thus, the remainder when p(x) is divided by g(x) then we get remainder as, 9

**Example 2: Find the root of the polynomial x2 - 5x + 4

We know that if for any p(x) we get p(a) = 0, then x-a is the factor of p(x) or a is the root of equation.

Given,

f(x) = x2 - 5x + 4

By hit and try method.

f(4) = 42 - 5(4) + 4

f(4) = 20 - 20 = 0

Since f(4) = 0, x = 4 is a root of the equation, and (x − 4) is a factor of x2 − 5x + 4.

**Example 3: Find the remainder when t3 - 2t + 4t + 5 is divided by t - 1.

Given,

p(t) = t3 - 2t2 + 4t + 5

g(t) = t - 1

Using Remainder theorem,

g(1) = (1)3 - 2(1)2 + 4 + 5 = 8

By the Remainder Theorem, 8 is the remainder when t3 - 2t2 + 4t + 5 is divided by t - 1

**Example 4: Find the remainder when x² - x + 2 is divided by x - 2.

Given,
p(x) = x² − x + 2

Divisor: x − 2
So, x = 2

Using Remainder Theorem,
Remainder = p(2)

p(2) = (2)² − (2) + 2
= 4 − 2 + 2
= 4

**Example 5: Find the root of the polynomial 3x2 - 7x + 2

We know that if for any p(x) we get p(a) = 0, then x-a is the factor of p(x) or a is the root of equation.

Given,

f(x) = 3x2 - 7x + 2

By hit and try method.

f(2) = 3(2)2 - 7(2) + 2

f(2) = 12 -14 + 2= 0

Since f(2) = 0, x = 2 is a root of the equation, and (x − 2) is a factor 3x2 - 7x + 2.