The Remainder Theorem is a formula for calculating the remainder when dividing a polynomial by a linear polynomial. The amount that is left after dividing a particular number of things into an equal number of things in each group is known as the Reminder. For example; if we divide 16 by 5 we get the quotient 3 and remainder 1. It can also be written as 16=(5×3)+1 We can observe here the remainder 1 is less than the divisor 5. 

In simple terms, a reminder is when something remains after division. The question is can a polynomial be divided by a linear polynomial. When the divisor is a linear polynomial, there is no need to apply the long division method to find the remainder.

The remainder theorem helps to find the remainder when a polynomial is divided by another polynomial without using the long division method. In this article, we will read more about Remainder Theorem with a solved example,

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Polynomials

An algebraic expression that consists of variables with exponents as whole numbers, coefficients, and constants combined using basic mathematical operations like addition, subtraction, and multiplication is called a polynomial.

For example, \({x^2} + x + 1,{x^2} + 2,y – 3x + 2,\) etc.

Zeros of a Polynomial

The zeros of the polynomial \(p(x)\) are all the \(x\)- values that make the polynomial equal to zero.

Consider the polynomial \(f(x) = {x^3} – 6{x^2} + 11x – 6\)

The value of this polynomial at \(x = 1\) is \(f(1) = 1 – 6 + 11 – 6 = 0\)

So, we can say that \(1\) is the root or zero of the polynomial \(f(x)\).

Remainder Theorem Definition

We use the remainder theorem in the division of polynomials to find the remainder when a polynomial is divided by a linear polynomial. Doesn’t it sound incredible that we can find remainder without doing the long division? Let us learn about the remainder theorem in detail.

Theorem: Let \(p(x)\) be any polynomial of degree greater than one or equal to one, and \(a\) be any real number. If \(p(x)\) is divided by \((x – a)\) then the remainder is equal to \(p(a)\).

Proof: Let \(q(x)\) be the quotient and \(r(x)\) be the remainder when \(p(x)\) is divided by \((x – a)\).

Then,

\(p(x) = (x – a)q(x) + r(x)………\left( {\rm{i}} \right)\)

Where \(r(x)\)\( = 0\) or degree \(r(x) < \) degree \((x – a)\). But \((x – a)\) is a polynomial of degree one, and a polynomial of degree less than one is a constant. Therefore, either \(r(x) = 0\) or \(r(x) = \) constant.

Let \(r(x)\,\, = \,\,r\) Then,

\(p(x) = (x – a)q(x) + r\,………..\left( {\rm{ii}} \right)\)

Putting \(x\, = a\) in \(\left( {\rm{i}} \right)\) we get

\(p(a) = (a – a)q(a) + r\)

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

\( \Rightarrow p(a) = 0\, + \,r\)

\( \Rightarrow p(a) = \,r\)

This shows that the remainder is \(p(a)\) when \(p(x)\) is divided by \((x – a)\).

Let us see how we can find the difference in finding the remainder using the long division method and by remainder theorem.

Let us take an example.

Divide \(x + 3{x^2} – 1\,\,by\,\,(1 + x)\)

First, let us see by the long division method.

(i) Organize the terms of the dividend and divisor in descending order of their degrees.
Here, the divisor is \((1 + x)\) and dividend is \(x + 3{x^2} – 1\) . So, arranging the terms of the divisor and the dividend in descending order, we have, \((x + 1)\) and \(3{x^2} + x – 1\) respectively.

(ii) Divide the first term of the dividend by the first term of the divisor to get the first term of the quotient.

So, \(\frac{{3{x^2}}}{x} = \,3x\)

(iii) Multiply each term of the divisor by the first term of the quotient. Then, subtract the result from the dividend.

(iv) Consider the remainder \( – 2x – 1\) as a new dividend and divide the first term of the dividend by the first term of the divisor to get the first term of the quotient.
So, \(\frac{{ – 2x}}{x} =  – 2\)

(v) Multiply each term of the divisor by the first term of the quotient. So, \(( – 2)(x + 1) =  – 2x – 2\) Then, subtract the result from the dividend.

(vi) Finally, we get the remainder as \( + 1\)

After doing, such a long and tedious process, we got the remainder. Now let us see how we can find the remainder using the remainder theorem.

The dividend is \(p(x) = x + 3{x^2} – 1\)

The divisor is \(q(x) = 1 + x\)

Now let us equate the divisor \(1 + x\) to \(0.\)

i.e., \(1 + x\,\, = \,0\)

\( \Rightarrow x\,\, = \, – 1\)

Let us now compute \(p( – 1)\)  i.e., the value of \(p(x)\) when \((x)\) is replaced by \( – 1\)

We have,

\(p( – 1) =  – 1 + 3{( – 1)^2} – ( – 1) =  3\)

Thus, we find that the remainder when \(p(x)\) is divided by \((1 + x)\) is equal to \(p( – 1)\)  i.e., the value of \(p(x)\) at \(x =  – 1\).

Hence, the remainder is \( 3\)

See how quickly and easily we found the remainder.

Remainder theorem finds excellent applications in cases where the quotient does not have any significant value but the remainder matters.

Factor Theorem

A polynomial \(q(x)\) is said to divide a polynomial \(p(x)\) completely if the remainder is zero. If \(q(x)\) divides \(p(x)\) we say that \(p(x)\) is divisible by \(q(x)\) or, \(q(x)\) is a factor of \(p(x)\).
Therefore, \(p(x)\) be a polynomial of degree greater than or equal to \(1\) and \(a\) be a real number such that \(p(a) = 0\) then \((x – a)\) is a factor of \(p(x)\).

For example: Use factor theorem when the dividend is \({x^3} – 3{x^2} + 4x – 12\) and the divisor is \((x – 3)\)

Let \(p(x) = {x^3} – 3{x^2} + 4x – 12\) be the given polynomial.

By factor theorem, \((x – a)\) is a factor of polynomial \(p(x)\) if \(p(a)\)\( = 0\)

Therefore, to prove that \((x – 3)\) is a factor of \(p(x)\) it is sufficient to show that \(p(3)\)\( = 0\)

Now, \(p(x) = {x^3} – 3{x^2} + 4x – 12\)

\( \Rightarrow p(3) = {3^3} – 3 \times {3^2} + 4 \times 3 – 12 = 27 – 27 + 12 – 12 = 0\)

Hence, \((x – 3)\) is a factor of \(p(x) = {x^3} – 3{x^2} + 4x – 12\)

Solved Examples

Q.1. Find the remainder when the polynomial \(3{x^2} + 4x – 12\) is divided by \((x – 1)\) using the remainder theorem.
Ans: Let \(p(x) = 3{x^2} + 4x – 12\) be the given polynomial.
We know, if \(p(x)\) be any polynomial of degree greater than one or equal to one, and \(a\) be any real number and if \(p(x)\) is divided by \((x – a)\), then the remainder is equal to \(p(a)\).
Let us now compute \(p(1)\) i.e., the value of \(p(x)\) when \((x)\) is replaced by \( + 1\).
Therefore, to prove that \((x – 1)\) is a factor of \(p(x)\) it is sufficient to show that \(p(1) = 0\).
Now, \(p(x) = 3{x^2} + 4x – 12\)
\( \Rightarrow p(1) = 3 \times {1^2} + 4 \times 1 – 12 = 3 + 4 – 12 =  – 5\)
Hence, the remainder when the polynomial \(3{x^2} + 4x – 12\) is devided by \((x – 1)\) is \( – 5\)

Q.2. Find the remainder, when \({x^3} + 3{x^2} + 5x + 6\) is divided by \(x – 2\).
Ans: The zero of \(x – 2\) is \( + 2\)
\(p(x) = {x^3} + 3{x^2} + 5x + 6\)
Then, \(p(2) = {(2)^3} + 3 + {2^2} + 5 \times (2) + 6 = 8 + 12 + 10 + 6 = 36\)
So, by the remainder theorem, the remainder is \(36\).

Q.3. Find the remainder using the long division method of the following \(({x^3} + 3{x^2} + 4x + 10) \div (x + 1)\)
Ans: For the long division method, we will follow the steps given below.
(i) Divide the first term of the dividend by the first term of the divisor to get the first term of the quotient.
(ii) Perform multiplication with each term of the divisor by the first term of the quotient. Then, subtract the result from the dividend.
(iii) Take Into consideration the remainder (if any) as a new dividend and go ahead as before.
(iv) Repeat this process until we get a remainder either \(0\) or a polynomial of lower degree than the divisor.

Hence, the remainder is \(2\).

Q.4. Find the value of \(a\) if \((x – a)\) is a factor of \({x^3} – {a^2}x + x + 2\).
Ans: Let \(p(x) = {x^3} – {a^2}x + x + 2\)
If \((x – a)\) is a factor of \({x^3} – {a^2}x + x + 2\) then the remainder will be zero.
\( \Rightarrow p(a) = 0\)
Now, \(p(a) = {a^3} – {a^2} \times a + a + 2 = 0\)
\( \Rightarrow a + 2 = 0\)
\(\Rightarrow a =  – 2\)
Therefore, \((x – a)\) is a factor of the polynomial, if \(a = \, – 2\)

Q.5. Determine the value of \(a\) for which the polynomial \(2{x^4} – a{x^3} + 4{x^2} + 2x + 1\) is divisible by \(1 – 2x\).
Ans: Let \(p(x) = 2{x^4} – a{x^3} + 4{x^2} + 2x + 1\).
If the polynomial \(p(x)\) is divisible by \((1 – 2x)\), then \((1 – 2x)\) is a factor of \(p(x)\).
Therefore, \(p\left( {\frac{1}{2}} \right) = 0\)
\( \Rightarrow p\left( {\frac{1}{2}} \right) = 2{\left( {\frac{1}{2}} \right)^4} – a{\left( {\frac{1}{2}} \right)^3} + 4{\left( {\frac{1}{2}} \right)^2} + 2\left( {\frac{1}{2}} \right) + 1 = 0\)
\( \Rightarrow \frac{2}{{16}} – \frac{a}{8} + \frac{4}{4} + \frac{2}{2} + 1 = 0\)
\( \Rightarrow \frac{1}{8} – \frac{a}{8} + 1 + 1 + 1 = 0\)
\( \Rightarrow \frac{{25}}{8} = \frac{a}{8}\)
\( \Rightarrow a = 25\)
Hence, the given polynomial is divisible by \(1 – 2x\) if \(a = 25\)

Summary

This article explains the remainder theorem, proof of the remainder theorem. We have learnt that we can avoid the long division method of the polynomial to find the remainder when the divisor is a linear polynomial. Also, we learned some example problems using the remainder theorem, which will help you understand the concept better.

FAQs

Q.1. What is the remainder theorem formula?
Ans: There is no particular formula for the remainder theorem. The theorem states that let \(p(x)\) be any polynomial of degree one or greater than one, and \(a\) be any real number. If \(p(x)\) is divided by \((x – a)\) then the remainder is equal to \(p(a)\).

Q.2. How do you use the remainder theorem?
Ans: The remainder theorem states that if \(p(x)\) be any polynomial of degree one or  greater than one, and \(a\) be any real number, then if \(p(x)\) is divided by \((x – a)\) then the remainder is equal to \(p(a)\).
For example, the dividend is \(p(x) = 3{x^2} – 1\) and the divisor \(q(x) = 1 + x\)
Now we equate the divisor \(1 + x\) to \(0.\)
\( \Rightarrow x\, =  – 1\)
Let us now compute \(p( – 1)\) i.e., the value of \(p(x)\) when \(x\) is replaced by \( – 1\)
We have, \(p( – 1) = 3{( – 1)^2} – ( – 1) = 3 + 1 = 4\)
Thus, we find that the remainder when \(p(x)\)  is divided by \((1 + x)\) is equal to \(p( – 1)\)  i.e., the value of \(p(x)\) at \(x =  – 1\)
Hence, the remainder is \(4\).

Q.3. Can you use the remainder theorem if the remainder is zero?
Ans: If the remainder is zero, then the divisor is known as the factor, and the factor theorem tells us that, let \(p(x)\) be a polynomial of degree one or greater than one and \(a\) be a real number such that \(p(a) = 0,\) then \((x – a)\) is a factor of \(p(x)\).

Q.4. Is the factor theorem and remainder theorem the same?
Ans: Two theorems are similar, but these are different.
The remainder theorem states that for any polynomial \(p(x)\) if you divide it by the linear polynomial \((x – a)\) the remainder is equal to the value of \(p(a)\).
The factor theorem states that if \(a\) is a zero of a polynomial \(p(x)\) then \((x – a)\) is a factor of \(p(x)\) and vice versa.

Q.5. Where is the remainder theorem used?
Ans: We use the remainder theorem in the division of polynomials, especially to identify the remainder when performing the division of polynomials using the linear polynomial as the divisor.

We hope this detailed article on the remainder theorem proves to be helpful. If you have any questions, feel to use the comments section and we will update you on the same.