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Remainder Theorem of Polynomials – Definition, Proof & Examples

Remainder Theorem

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Polynomials are special types of algebraic expressions which are used to express numbers in almost every field of mathematics and are considered very important in certain branches of math, such as calculus. You can perform the four mathematical operations – addition, subtraction, multiplication, and division on polynomials.

If you just want to check the divisibility of a polynomial by another polynomial, you use the remainder theorem of polynomials. This helps in saving time that is otherwise spent using the long division method.

Let’s understand what the remainder theorem of polynomials is and how it is used to solve problems.

What is Remainder Theorem of Polynomials?

The remainder theorem states that when a polynomial $p \left(x \right)$ (whose degree is greater than or equal to $1$) is divided by a linear polynomial $q \left(x \right)$ whose zero is $x = a$, the remainder is given by $r = p \left(a \right)$. 

The remainder theorem enables us to calculate the remainder of the division of any polynomial by a linear polynomial, without actually carrying out the steps of the long division.

Note: The degree of the remainder polynomial is always $1$ less than the degree of the divisor polynomial. Using this fact, when any polynomial is divided by a linear polynomial (whose degree is $1$), the remainder must be a constant (whose degree is $0$).

Steps to Use Remainder Theorem

The steps to find the remainder of a polynomial $p \left(x \right)$ when it is divided by a linear polynomial $(x – a)$ are

Step 1: Find the zero of the linear polynomial by setting it to zero. i.e., $x – a = 0 => x = a$

Step 2: Substitute $x = a$ in the polynomial $p \left(x \right)$

Step 3: Simplify $p \left(x = a \right)$

Step 4: The result obtained in Step 3 is the remainder of $p \left(x \right)$, when it is divided by a linear polynomial $\left(x – a \right)$

Proof of Remainder Theorem

Let us assume that $q(x)$ and $r$ are the quotient and the remainder respectively when a polynomial $p(x)$ is divided by a linear polynomial $(x – a)$. By division algorithm, we have $\text{Dividend }= \text{Divisor } \times \text{ Quotient } + \text{ Remainder}$.

Using this, $p(x) = (x – a) \times q(x) + r$.

Substitute $x = a$

$p(a) = (a – a) \times q(a) + r$

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

$p(a) = r$

i.e. the remainder = $p(a)$.

Examples

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

Equating $x – 2$ to $0$, we get $x – 2 = 0 => x = 2$.

Now, substituting $x = 2$ in $p(x) = x^3 – 3x^2 + 5x + 1$

$p(a) = p(2) = 2^3 – 3 \times 2^2 + 5 \times 2 + 1 = 8 – 12 + 10 + 1 = 7$.

Therefore, when $(x^3 – 3x^2 + 5x + 1)$ is divided by $(x – 2)$, the remainder is $7$.

Ex 2: Find the remainder when $p(x) = x^2 – 7x + 11$ is divided by $(x + 2)$.

$x + 2 = 0 => x = -2$.

Therefore, $p(-2) = (-2)^2 – 7 \times (-2) + 11 = 4 + 14 + 11 = 29$.

Therefore, when $(x^2 – 7x + 11)$ is divided by $(x + 2)$, the remainder is $29$.

Ex 3: The polynomial $4x^2 – kx + 7$ leaves a remainder of  $–2$ when divided by $x – 3$. Find the value of $k$.

$x – 3 = 0 => x = 3$

By a remainder theorem remainder when $p(x)$ is divided by $(x – a)$ is $p(a)$.

Therefore, the remainder when $4x^2 – kx + 7$ is divided by $x – 3$ is  $4 \times 3^2 – k \times 3 + 7 = 43 – 3k$

$=>43 – 3k = -2 => -3k = -2 – 43 => -3k = -45 => k = \frac{-45}{-3} => k = 15$

Ex 4: If two polynomials $2x^3 + kx^2 + 4x – 12$ and $x^3 + x^2 – 2x + k$ leave the same remainder when divided by $(x – 3)$, find the value of $k$ and the remainder.

The two polynomials are $p(x) = 2x^3 + kx^2 + 4x – 12$ and $q(x) = x^3 + x^2 – 2x + k$

$x – 3 = 0 => x = 3$

Remainder when $p(x)$ is divided by $x – 3$ is $p(3) = 2 \times 3^3 + k \times 3^2 + 4 \times 3 – 12 = 54 + 9k + 12 – 12 = 54 + 9k$.

Remainder when $q(x)$ is divided by $x – 3$ is $q(3) = 3^3 + 3^2 – 2 \times 3 + k = 27 + 9 – 6 + k = 30 + k$.

Therefore, $54 + 9k = 30 + k => 54 + 9k – 30 – k = 0 => 8k + 24 = 0 => 8k = -24 => k = \frac{-24}{8} => k = -3$.

Thus, remainder = $54 + 9 \times (-3) = 54 – 27 = 27$.

Ex 5: Find the value of $m$, if $x = \frac{1}{2}$ is one of the zeroes of the polynomial $p(x) = 4x^4 − 4x^3 − mx^2 + 12x − 3$

Let $p(x) = 4x^4 − 4x^3 − mx^2 + 12x − 3$

$x = \frac{1}{2}$ is one of the zeroes of $p(x)$ => $x – \frac{1}{2}$ is a factor of $p(x)$ => Remainder when $p(x)$ is divided by $\left(x – \frac{1}{2} \right)$ is $0$ => $p \left(\frac{1}{2} \right) = 0$.

$p \left(\frac{1}{2} \right) = 4 \times \left(\frac{1}{2} \right)^4 − 4 \times \left(\frac {1}{2} \right)^3 − m \times \left(\frac{1}{2} \right)^2 + 12 \times \frac{1}{2} − 3 = \frac{1}{4} – \frac{1}{2} – \frac{m}{4} + 6 – 3 = \frac {5}{2} – \frac{m}{4}$

Therefore, $\frac {5}{2} – \frac{m}{4} = 0 => \frac{m}{4} = \frac {5}{2} => m = 10$.

Ex 6: What will be the remainder when $2x^{3} + 3x^{2} – 3x – 2$ is divided by $(2x – 3)$?

Let $p(x) = 2x^{3} + 3x^{2} – 3x – 2$

$2x – 3 = 0 => 2x = 3 => x = \frac{3}{2}$

Remainder when $p(x)$ is divided by $(2x – 3)$ is $p \left(\frac{3}{2} \right) = 2 \times \left(\frac{3}{2} \right)^{3} + 3 \times \left(\frac{3}{2} \right)^{2} – 3 \times \frac{3}{2} – 2 = \frac{27}{4} + \frac{27}{4} – \frac{9}{2} – 2 = \frac{18}{2} – 9 = 0$

Therefore, remainder when when $2x^{3} + 3x^{2} – 3x – 2$ is divided by $(2x – 3)$ is $0$.

Difference Between the Remainder Theorem and Factor Theorem

Following are the differences between the remainder theorem and the factor theorem:

remainder theorem

Points to Remember about the Remainder Theorem

remainder theorem

Practice Problems

  1. Using the remainder theorem find the remainder when
    • $3x^3 – 2x^2 + 5x + 1$ is divided by $x – 1$
    • $2x^2 – 5$ is divided by $x + \frac{2}{3}$
    • $5x^4 + x^2 – x + 7$ is divided by $x – 0.8$
    • $2x^3 – x^2 + 4$ is divided by $x + \frac{1}{2}$
    • $3x^4 – 2x^3 + 7x^{2} – 2x + 3$ is divided by $x + 3$
  2. If $(x – 8)$ is one of the factors of $mx^3 – 24x^2 + 192x – 512$, find the value of $m$.
  3. Find the value of $m$ if $(x – 1)$ divides the polynomial $mx^3 – 2x^2 + 25x – 26$ without leaving the remainder.
  4. Suppose $r$ and $\text{R}$ are the remainders when the polynomials $x^{3} + 2x^{2} – 5kx – 7$ and $x^3 + kx^2 – 12x + 6$ are divided by $(x + 1)$ and $(x – 2)$, respectively. Find the value of $k$ if $2r + \text{R} = 6$.

FAQs

What is the remainder theorem?

The remainder theorem states that when a polynomial $p \left(x \right)$ (whose degree is greater than or equal to $1$) is divided by a linear polynomial $q \left(x \right)$ whose zero is $x = a$, the remainder is given by $r = p \left(a \right)$.

Are the remainder theorem and factor theorem the same?

No, they are different. The factor theorem just ensures whether the given linear polynomial is a factor of the given polynomial, whereas the remainder theorem finds the remainder when a polynomial is divided by another polynomial that is linear.

What if the remainder is zero?

If the remainder is zero, then the remaining quotient and the divisor are the factors of the given expression.

What is the use of the remainder theorem?

The remainder theorem is used to find the remainder when a polynomial $p(x)$ is divided by $(ax + b)$. The remainder theorem is further extended to prove the factor theorem where we can determine whether $(ax + b)$ is a factor of $p(x)$ or not. If the remainder is $0$, then $(ax + b)$ is a factor of a polynomial $p(x)$, otherwise, it is not.

Conclusion

The remainder theorem enables us to calculate the remainder of the division of any polynomial by a linear polynomial, without actually carrying out the steps of the long division.  It states that when a polynomial $p \left(x \right)$ (whose degree is greater than or equal to $1$) is divided by a linear polynomial $q \left(x \right)$ whose zero is $x = a$, the remainder is given by $r = p \left(a \right)$.

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