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Last updated on October 16, 2025
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.
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.
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.
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.
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:
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, \(remainder = p(a) \)
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.
Students may find the concept of remainder theorem simple. However, there are some common misconceptions and mistakes that occur.
The remainder theorem is an algebraic tool often used in real world situations such as:
Find the remainder when p(x) = 2x^3 −5x^2 +3x − 4 is divided by x - 2.
Remainder = 2.
Zero of the divisor \(x-2 = 0\)
\(x = 2 \)
Substitute x = 2 into p(x)
\(p(2) = 2(2)^3 - 5(2)^3 + 3(2) - 4 \)
\(16 - 20 + 6 - 4 = 2\)
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.
Remainder = 4
Zero of the divisor \(x + 1 = 0, x = -1\)
\(p(-1) = (-1)^4 + 3(-1)^3 - (-1) + 1 = 1 - 3 + 1 + 1 = 0\)
Find the remainder when 𝑝(𝑥) = 3𝑥^3 − 4𝑥^2 + 5𝑥 − 6 is divided by 2𝑥 − 3.
Remainder = \(- \frac{1}{8}\)
Zero of the divisor \(2x - 3 = 0\), \(x = \frac{3}{2}\)
\(p \frac {3}{2} = 3 (\frac {3}{2})^3 - 4 (\frac {3}{2}) + 5 (\frac {3}{2}) - 6\)
\(= \frac {81}{8} - 9 + \frac {15}{2} - 6 \)
\(= \frac {81}{8} - \frac{72}{8} + \frac{60}{8} - \frac {48}{8} = \frac{11}{8}\)
Find the remainder when p(x) = x^3 - 6x^2 + 11x - 6 is divided by x -1.
Remainder = 0
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).
Find the remainder when 𝑝(𝑥) = 2𝑥^4 + 3𝑥^3 − 𝑥 + 5 is divided by x + 2.
Remainder = 13
Zero of divisor \(x + 2 = 0 \)
\(x = -2\)
\(p(-2) = 2(-2)^4 + 3(-2)^3 - (-2) + 5\)
\(2(16) + 3(-8) + 2 + 5 = 32 - 24 + 2 + 5 = 15\)
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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