Remainder Theorem: Statement & Applications

The Remainder Theorem is a fundamental concept in algebra that provides a convenient way to find the remainder when a polynomial, denoted by $p(x)$, is divided by a linear divisor of the form $(x – a)$. Instead of performing long division, the theorem allows us to determine the remainder simply by evaluating the polynomial at the value ‘a’. This greatly simplifies the process and saves time, especially with higher-degree polynomials.

Formulae

The Remainder Theorem can be summarized with the following:

• When a polynomial $p(x)$ is divided by $(x – a)$, the remainder is $p(a)$.
• This can be written as: $p(x) = (x – a) \cdot q(x) + r$, where $q(x)$ is the quotient and $r$ is the remainder. According to the Remainder Theorem, $r = p(a)$.

Examples

Let’s illustrate the Remainder Theorem with a couple of examples.

Example-1: Find the remainder when $p(x) = x^3 – 4x^2 + 5x – 2$ is divided by $(x – 1)$.

Solution: According to the Remainder Theorem, the remainder is $p(1)$.

Substitute $x = 1$ into $p(x)$:

$p(1) = (1)^3 – 4(1)^2 + 5(1) – 2 = 1 – 4 + 5 – 2 = 0$.

Therefore, the remainder is 0. This also implies that $(x-1)$ is a factor of $p(x)$.

Example-2: Find the remainder when $p(x) = 2x^4 + 3x^3 – x^2 + 5x – 8$ is divided by $(x + 2)$.

Solution: We are dividing by $(x+2)$, which is equivalent to $(x – (-2))$. Therefore, $a = -2$. The remainder is $p(-2)$.

Substitute $x = -2$ into $p(x)$:

$p(-2) = 2(-2)^4 + 3(-2)^3 – (-2)^2 + 5(-2) – 8 = 2(16) + 3(-8) – 4 – 10 – 8 = 32 – 24 – 4 – 10 – 8 = -14$.

Therefore, the remainder is -14.

Theorem with Proof

Theorem: If a polynomial $p(x)$ is divided by $(x – a)$, then the remainder is $p(a)$.

Proof:

When a polynomial $p(x)$ is divided by $(x-a)$, we can express it in the following form using the division algorithm:

$p(x) = (x – a) \cdot q(x) + r$

where:

• $q(x)$ is the quotient.
• $r$ is the remainder (which must be a constant since we are dividing by a linear expression).

Now, let’s substitute $x = a$ into the equation:

$p(a) = (a – a) \cdot q(a) + r$

Since $(a – a) = 0$, the equation simplifies to:

$p(a) = 0 \cdot q(a) + r$

$p(a) = r$

Therefore, the remainder, $r$, is equal to $p(a)$, which proves the Remainder Theorem.

Common mistakes by students

• Incorrectly identifying ‘a’: Students often struggle to correctly identify the value of ‘a’ from the divisor $(x – a)$. For example, if the divisor is $(x + 3)$, then ‘a’ is -3, not +3.
• Substitution Errors: Making arithmetic errors when substituting the value of ‘a’ into the polynomial $p(x)$ is a common mistake. Double-checking calculations, especially when dealing with negative numbers or exponents, is crucial.
• Not Recognizing the Special Case of Factors: Failing to understand that if the remainder is 0, then $(x – a)$ is a factor of the polynomial.

Real Life Application

The Remainder Theorem, while seemingly abstract, has practical applications in several areas:

• Engineering: In circuit design and signal processing, polynomials are used to model system behavior. The Remainder Theorem can help analyze system stability and predict the output based on a given input.
• Computer Science: The theorem is used in error detection and correction codes, ensuring data integrity during transmission.
• Cryptography: Polynomials, and therefore the Remainder Theorem, are utilized in certain cryptographic algorithms.

Fun Fact

The Remainder Theorem is a specific case of the polynomial remainder theorem, which applies even when the divisor isn’t necessarily linear! This broader concept shows the interconnectedness of mathematical ideas.

Recommended YouTube Videos for Deeper Understanding

Q.1 If the polynomial $P(x) = x^3 – 2x^2 + kx – 1$ is divided by $(x-1)$, the remainder is 2. What is the value of $k$?

Check Solution

Ans: C

By the Remainder Theorem, $P(1) = 2$. Thus, $(1)^3 – 2(1)^2 + k(1) – 1 = 2$. This simplifies to $1 – 2 + k – 1 = 2$, so $k – 2 = 2$, and $k = 4$.

Q.2 When the polynomial $f(x) = 3x^3 + ax^2 – 14x + 8$ is divided by $(x+2)$, the remainder is zero. Find the value of $a$.

Check Solution

Ans: A

Since the remainder is zero, $f(-2) = 0$. Thus, $3(-2)^3 + a(-2)^2 – 14(-2) + 8 = 0$. This simplifies to $-24 + 4a + 28 + 8 = 0$, which gives $4a + 12 = 0$. Hence $4a = -12$ and $a = -3$. None of the options are correct. The option closest to -3 is -1. However, since the provided answer options are limited, choose the option closest to the correct answer (which is a minus sign here)

Q.3 What is the remainder when $x^4 + 3x^3 – 2x^2 + x – 1$ is divided by $(x-1)$?

Check Solution

Ans: C

Using the Remainder Theorem, we find $P(1) = (1)^4 + 3(1)^3 – 2(1)^2 + 1 – 1 = 1 + 3 – 2 + 1 – 1 = 2$.

Q.4 Determine the remainder when the polynomial $2x^3 – 5x^2 + 8x – 3$ is divided by $(2x-1)$.

Check Solution

Ans: B

Set $2x-1=0$ or $x = 1/2$. Thus, $P(1/2) = 2(1/2)^3 – 5(1/2)^2 + 8(1/2) – 3 = 2(1/8) – 5(1/4) + 4 – 3 = 1/4 – 5/4 + 1 = -4/4 + 1 = -1 + 1 = 0$.

Q.5 If $x-3$ is a factor of $x^3 – 5x^2 + kx – 6$, then find the value of $k$.

Check Solution

Ans: D

Since $x-3$ is a factor, the remainder when dividing by $x-3$ is 0. So, $P(3) = 0$. Therefore, $(3)^3 – 5(3)^2 + k(3) – 6 = 0$. This gives $27 – 45 + 3k – 6 = 0$, or $3k – 24 = 0$. Thus, $3k = 24$, and $k = 8$.

Next Topic: Factor Theorem: Statement & Applications