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Last updated on October 16, 2025

Remainder Theorem

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The remainder theorem is a basic concept in algebra that saves a lot of time by speeding up the long division process of polynomials. It also helps in testing the values of zeros of a polynomial and the factorization of polynomials. In this article, we will be looking into the details of the remainder theorem.

Remainder Theorem for US Students
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What is the Remainder Theorem?

The remainder theorem is a helpful rule in algebra that lets us find the remainder when a polynomial is divided by an expression like x - c.

We don’t have to do the full division process here. Instead of dividing, we can substitute the value c into the polynomial. The result that we get will be the remainder.

It’s an effective method that saves time and helps you check if a number is a root of the polynomial. 
 

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Remainder Theorem Statement and Proof

The remainder theorem states that when a polynomial p(x) is divided by x - a, the remainder will be equal to the value of the polynomial at x = a. That is, r = p(a). 

Follow these steps to find the remainder:

Step 1: Identify the value of a from the divisor x - a.

Step 2: Substitute x = a in p(x)

Step 3: The result gives the remainder. 
 

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Remainder Theorem for Polynomials

The remainder obtained when a polynomial is divided by a linear polynomial can be determined directly using substitution. Evaluate the polynomial at the zero of the divisor.
 

For better understanding, consider the following example, of two polynomials

Dividend, \(p(x): 4x^3 + 5x^2 - 3x + 1\) 

Divisor: \(x - 2 \)

Step 1: Find the zero of the divisor 

\(x - 2 = 0  \)

\(x = 0  \)

Step 2: Apply the remainder theorem 
The Remainder Theorem states, 
Remainder = p(zero of divisor) \(= p(2) \)

Substitute \(x = 2 \) into \(p(x) \):

\(p(2) = 4(2)^3 + 5(2)^2 - 3(2) + 1\)

Step 3: Simplify further

\(4(8) + 5(4) - 3(2) + 1 \)

\(32 + 20 - 6 + 1 = 47 \) 
\(\)
So, the remainder is 47.
 

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Difference Between the Remainder Theorem and Factor Theorem

Now that we know what the remainder theorem is, let us learn the difference between the factor theorem and the remainder theorem.

Although the factor theorem is similar to the remainder theorem, they both serve a slightly different purpose. Let’s see how they differ in the table below:
 

 

Remainder Theorem

Factor Theorem
 

Purpose: Tells you the remainder when a polynomial is divided by x - c.

Purpose: Helps you find factors (or roots) of a polynomial.

How it works: Place c into the polynomial f(x), and the result is the remainder.

How it works: If f(c) = 0, then x - c is a factor of the polynomial.

Focus: Focuses on quickly finding the remainder. 

Focus: Focuses on identifying the factors of a polynomial. 

 

 

What is The Remainder Theorem Standard and Proof? 

 

The remainder theorem states that if a polynomial f(x) is divided by x - c, then the remainder is f(c). This means that we don’t have to do long division; just place c into the polynomial, and the answer you get is the remainder of the division.

When p(x) is divided by (x - a): \(remainder = p(a) \)

or,

when p(x) is divided by (ax + b): \(remainder = p (-\frac {b}{a})\)

Furthermore, for different types of linear polynomials, we can extend the remainder theorem as given below:
 

  • If p(x) is divided by x - a, the remainder is \(p(a) \; (\because \, x - a = 0 \Rightarrow x = a) \)
     
  • When p(x) is divided by ax + b , the remainder is \(p\left(-\frac{b}{a}\right) \; (\because \, ax + b = 0 \Rightarrow x = -\frac{b}{a}) \)
     
  • The remainder when p(x) is divided by ax - b is \(p\left(\frac{b}{a}\right) \; (\because \, ax - b = 0 \Rightarrow x = \frac{b}{a}) \)
     
  • When p(x) is divided by bx - a is \(p\left(\frac{a}{b}\right) \; (\because \, bx - a = 0 \Rightarrow x = \frac{a}{b}) \)


Proof of Remainder Theorem

Assuming q(x) as the quotient and r as the remainder. Let us divide a polynomial p(x) by a linear polynomial (x - a).

According to the division algorithm,

\(Dividend = (divisor × quotient) + remainder\)

so, 

\(p(x)=(x−a)⋅q(x)+r\)

Substituting x = a, we get

\(p(a) = (a - a) ⋅ q(x) + r\)

\(p(a) = 0 ⋅ q(a) + r\)

\(p(a) = r \)

Hence, proved, \(remainder = p(a) \)

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Tips and Tricks to Master the Remainder Theorem

The tips and tricks given below will help students get a good command on the topic by providing efficient methods to work with the topic.
 

  •  It is key to identify the zero of the divisor, before any calculations, always find the zero first.
     
  • The direct substitution method is quicker and more efficient when compared to the long division method.
     
  • Always check for negative divisors if the divisor is of the form \(ax + b\). Convert to zero by solving \(ax + b = 0 \)\(x = - \frac{b}{a}\)
     
  • A quick way to verify if a number is a factor of the polynomials is: if p(a) = 0, then x - a is a factor.
     
  • Practice with higher degree polynomials to improve speed and efficiency. 
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Common Mistakes and How to Avoid Them in Remainder Theorem

Students may find the concept of remainder theorem simple. However, there are some common misconceptions and mistakes that occur.

Mistake 1

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Using the divisor instead of zero.

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Always solve x - a = 0 (or ax + b = 0) first to find the correct value to substitute. 

Mistake 2

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Plugging in the wrong value or miscalculating powers and coefficients.

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Carefully substitute and simplify step by step, especially with negative signs.

Mistake 3

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Directly substituting without accounting for negative coefficients in \(ax + b\)

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Solve \(ax + b = 0\)\(x = - \frac {b}{a}\) before substitution.

Mistake 4

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Thinking that the remainder is always less

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The remainder can be any number, it does not have to be less than the divisor.

Mistake 5

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Confusing remainder with quotient

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Remember that p(a) = r. Do not involve the quotient in the final calculation of the remainder.

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Real-Life Applications of the Remainder Theorem

The remainder theorem is an algebraic tool often used in real world situations such as:
 

  • Checking divisibility – Businesses can use it to determine if quantities of items or payments can be evenly distributed without performing lengthy calculations.
     
  • Coding and cryptography – The theorem aids in algorithms that detect errors in data transmission by evaluating polynomial-based codes.
     
  • Engineering calculations – Engineers can quickly check values in polynomial models of circuits, beams, or fluid flows without full computation.
     
  • Computer graphics – It is used in rendering curves and animations modeled by polynomials, helping identify specific points efficiently.
     
  • Calendar calculations – Mathematicians and programmers use it to calculate recurring events or determine weekdays for given dates using polynomial formulas.
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Solved Examples on Remainder Theorem

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Problem 1

Find the remainder when p(x) = 2x^3 −5x^2 +3x − 4 is divided by x - 2.

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Remainder = 2.

Explanation

Zero of the divisor \(x-2 = 0\) 
\(x = 2 \)
Substitute x = 2 into p(x)
\(p(2) = 2(2)^3 - 5(2)^3 + 3(2) - 4 \)
\(16 - 20 + 6 - 4 = 2\)

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Problem 2

Find the remainder when 𝑝(𝑥) = 𝑥^4 + 3𝑥^3 − 𝑥 + 1 p(x) = x^4 + 3x^3 − x + 1 is divided by 𝑥 + 1 x + 1.

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Remainder = 4

Explanation

Zero of the divisor \(x + 1 = 0, x = -1\) 

\(p(-1) = (-1)^4 + 3(-1)^3 - (-1) + 1 = 1 - 3 + 1 + 1 = 0\)

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Problem 3

Find the remainder when 𝑝(𝑥) = 3𝑥^3 − 4𝑥^2 + 5𝑥 − 6 is divided by 2𝑥 − 3.

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Remainder = \(- \frac{1}{8}\)

Explanation

Zero of the divisor \(2x - 3 = 0\)\(x = \frac{3}{2}\)

\(p \frac {3}{2} = 3 (\frac {3}{2})^3 - 4 (\frac {3}{2}) + 5 (\frac {3}{2}) - 6\)

\(= \frac {81}{8} - 9 + \frac {15}{2} - 6 \)

\(= \frac {81}{8} - \frac{72}{8} + \frac{60}{8} - \frac {48}{8} = \frac{11}{8}\)

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Problem 4

Find the remainder when p(x) = x^3 - 6x^2 + 11x - 6 is divided by x -1.

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Remainder = 0

Explanation

Zero of the divisor \(x - 1 = 0\) 
\(​​​​​​​x = 1 \)
 \(p(1) = 1 - 6 + 11 - 6 = 0\)

Hence, x - 1 is a factor of p(x).

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Problem 5

Find the remainder when 𝑝(𝑥) = 2𝑥^4 + 3𝑥^3 − 𝑥 + 5 is divided by x + 2.

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Remainder = 13

Explanation

Zero of divisor \(x + 2 = 0 \)

\(x = -2\)

\(p(-2) = 2(-2)^4 + 3(-2)^3 - (-2) + 5\)

\(2(16) + 3(-8) + 2 + 5 = 32 - 24 + 2 + 5 = 15\)

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1.How do you use the Remainder Theorem?

To find the remainder, identify the zero of the divisor \(x - a = 0\)
\(x = a \) ,then substitute it into p(x). 

 

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2.What are the applications of the remainder theorem formula?

Checking divisibility of polynomials, verifying factors, coding and cryptography, engineering calculations, and calendar or date-related calculations.

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3.What is the need for teaching my child the remainder theorem?

It builds algebraic thinking, simplifies polynomial calculations, and prepares students for higher-level math.

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4.How can I help my child practice the remainder theorem?

Encourage substitution exercises, factor-checking problems, and simple polynomial remainder calculations.

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5.Will the Remainder Theorem help in exams?

Absolutely. It saves time in solving polynomial division problems and factorization questions. 

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6.Who invented the Remainder Theorem?

The Remainder Theorem has origins in early algebra, with contributions from mathematicians in the 17th and 18th centuries; it evolved as part of polynomial theory.

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Jaskaran Singh Saluja

About the Author

Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.

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Fun Fact

: He loves to play the quiz with kids through algebra to make kids love it.

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