# Problems on Remainder Theorem | Remainder Theorem Practice Questions

When you divide one polynomial by another the process will be very long, so we can avoid this long division method by providing certain rules using the Remainder Theorem. We will learn about how to solve the problems using the 10th Grade Math Concept Remainder Theorem in this article. In the case of divisibility of the polynomial by a linear polynomial, we use a well-known theorem that is called the Remainder Theorem.

## What is Remainder Theorem?

The Remainder Theorem will be the way of addressing Euclideanâ€™s division of polynomials. Another name for the Remainder Theorem is Bezoutâ€™s theorem of approaching polynomials of Euclideanâ€™s division. The remainder theorem definition states that when a polynomial f(x) is divided by the factor (x -a) when the factor is not necessarily an element of the polynomial, then you will find a smaller polynomial along with a remainder.

The resultant obtained is the value of the polynomial f(x) where x = a and this is possible only if f(a) = 0. To factorize the polynomials easily, we can apply the remainder theorem.

### Solving Remainder Theorem Problems and Solutions | Remainder Theorem Question and Answers

Problem 1: Find the remainder when f(x) = x3 + 3x2 + 3x + 1 is divided by (x + 1), using the Remainder Theorem.

Solution :
In the question, given that
The divisor is (x + 1).
The function is f(x) = x3 + 3x2 + 3x + 1.
First, equate the divisor to zero. i.e.,
x + 1 = 0
Solve for x is, x = -1.
Now, we will find the remainder of the given function.
First, substitute -1 for x into the function f(x).
f(-1) = (-1)3 + 3(-1)2 + 3(-1) + 1
f(-1) = -1 + 3(1) – 3 + 1
So, the value is f(-1) = -1 + 3 – 3 + 1
f(-1) = 0
Hence, the remainder of the given function is 0.

Problem 2 : Find the remainder when f(x) = x3 – 3x + 1 is divided by (2-3x), using the Remainder Theorem.

Solution :
As given in the question,
The divisor is (2 – 3x).
The polynomial equation is f(x) = x3 – 3x + 1
Now, Equate the divisor to zero. i.e.,
2 – 3x = 0
Solve for x isÂ -3x = -2
x = 2/3
Next, to find the remainder of the given equation.
So, substitute 2/3 for x into the function f(x).
f(2/3) = (2/3)3 – 3(2/3) + 1
f(2/3) = 8/27 – 2 + 1
f(2/3) = 8/27 – 1
f(2/3) = 8/27 – 27/27
f(2/3) = (8 – 27)/27Â = -19/27
Therefore, the given polynomial function remainder is -19/27.

Problem 3 : For what value of k is the polynomial 2x4 + 3x3 + 2kx2 + 3x + 6 is divisible by (x + 2).

Solution :
In the given question,
The polynomial equation is f(x) = 2x4 + 3x3 + 2kx2 + 3x + 6.
The divisor will be (x + 2).
First, equate the divisor to zero. It will be
x + 2 = 0
Next, solve for x is,Â x = -2
Now to find the remainder value, so substitute -2 for x into the function f(x).
After substituting f(-2) = 2(-2)4 + 3(-2)3 + 2k(-2)2 + 3(-2) + 6
f(-2) = 2(16) + 3(-8) + 2k(4) – 6 + 6
= 32-24+8k-6+6
f(-2) = 8 + 8k
Thus, the remainder is (8 + 8k).
If f(x) is exactly divisible by (x + 2), then the remainder must be zero. Then,
8 + 8k = 0
Solve for k is 8k = -8
k = -1
Hence, the f(x) is exactly divisible by (x+2) when the k = â€“1.

Problem 4: Using the remainder theorem to find the remainder when p(x) = x4 – 9x3 – 5x2 – 3x + 4 is divided by x + 3.

Solution:
Given the polynomial equation p(x) is x4 – 9x3 – 5x2 – 3x + 4
It is divided by x + 3
For the division, we can use the remainder theorem
So, when p(x) is divided by (x-a) then p(a) is the remainder.
From this, p(-3) gives the remainder.
So, the expression is (x4 – 9x3 – 5x2 – 3x + 4) Ã· (x – 3)
Then p(x) = x4 – 9x3 – 5x2 – 3x + 4
p(-3) = (-3)4 – 9(-3)3 – 5(-3)2 – 3(-3) + 4
p(-3) = 81 – 9(-27) – 5(9) – 3(-3) + 4
p(-3) = 81 – 9(-27) – 5(9) – 3(-3) + 4
= 81 – 243 – 45 + 13
= 292
So, the remainder of the function is 292.

Problem 5: The (x-3) is a factor of x to the power 3 + ax to the power 2+ bx – 9 and when it is divided by (x – 2) the remainder is 1. What is the value of a and b?

Solution:
In the Given question,
The p(x) is x3 + ax2 + bx – 9
The (x-3) is a factor of the polynomial equation x3 + ax2 + bx – 9
Equate divisor to zero i.e.,Â p(3) = 0
Substitute the value in the equation, we get
(3)3 + a(3)2 + b(3) – 9 = 0
27 + 9a + 3b – 9 = 0
So, the a + b isÂ  -6 … (i)
Next, when p(x) is divided by (x – 2) then the remainder is 1, and it is given.
So the p(2) is 1.
Now substituting the values. We get,
(2)3 + a(2)2 + b(2) – 9 = 1
4a + 2b = 2
2a + b = 1 …. (ii)
Solving (i) and (ii), we get the values of the a and b.
a = -7 and b = 14
So, the value of a is -7 and the value of the b is 14.

Problem 6: Let p(x) be a polynomial, which when divided by xâˆ’3 and xâˆ’5 leaves remainders 10 and 6, respectively. If the polynomial is divided by (xâˆ’3)(xâˆ’5), then what is the remainder?

Solution:Â
Given that,
Let p(x) is divided by xâˆ’3 and xâˆ’5, leaving the remainder 10 and 6.
From the remainder theorem of polynomial, p(3)=10 and P(5)=6.
If the polynomial is divided by (xâˆ’3)(xâˆ’5), then the remainder must be of the form ax+b (Since the degree of remainder is less than that of the divisor).
p(x)=q(x)(xâˆ’3)(xâˆ’5)+(ax+b)
where q(x) is some polynomial (quotient obtained).
Now, Substituting for x=3 and x=5. We get,
p(3)=q(x)(3âˆ’3)(3âˆ’5)+a(3)+b=0+3a+b=3a+bâŸ¹10=3a+bâˆ’âˆ’(i)
p(5)=q(x)(5âˆ’3)(5âˆ’5)+a(5)+b=0+5a+b=5a+bâŸ¹6=5a+bâˆ’âˆ’(ii).
Next, solving the equation 1 and equation 2. We get,
(i)âˆ’(ii) = 10âˆ’6 = (3âˆ’5)a
4 =âˆ’2a
a=âˆ’2âˆ’âˆ’(a).
(a) in (ii), we get
6 = 5(âˆ’2)+b
6=âˆ’10+b
b=6+10, b=16.
Therefore, the value of a is âˆ’2, and the value of b is16.
Hence, the Remainder is âˆ’2x+16.

Problem 7: Write the remainders in each of the following cases,
(i) Find the remainder value, when f(x) is divided by (x – 2).
(ii) What is the value, when g(x) is divided by (2x – 1).
(iii) When h(x) is divided by (3x+4).

Solution:
Let us find,
By the remainder theorem formula, the remainder when p(x) is divided by (ax + b) is p(-b/a).
So, find the following,
(i) x- 2 = 0
After equating the value of x is 2.
Hence, the remainder of f(x) is divided by (x – 2) is f(2).

(ii) When g(x) is divided by (2x-1)
that is 2x – 1 = 0
After equating the value of x is 1/2.
Hence, the remainder when g(x) is divided by (2x – 1) is g(1/2).

(iii) Find the remainder when h(x) is divided by (3x+4)
i.e., 3x + 4 = 0
Now, equating the value. We get,
x = -4/3
Thus, the remainder when h(x) is divided by (3x + 4) is h(-4/3).

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### FAQ’s on Questions Based on Remainder Theorem

1. What are the Applications of the Remainder Theorem Formula?
The main application of the remainder theorem formula is the factor theorem. While we are proving the factor theorem, we need the remainder theorem. So, the factor theorem says that, if the remainder is obtained by dividing the p(x) by (x – a) is 0, then the (x – a) is a factor of p(x).

2. How To Derive the Remainder Theorem Formula?
Assume that we have to find the remainder r when a polynomial p(x) is divided by (x – a). In this case, assume that the quotient is q(x). Now, we can apply Euclid’s division lemma in this case and then substitute x = a to derive the remainder theorem formula.

3. How To Use Remainder Theorem Formula?
The remainder theorem formula can be used to find the remainder when a polynomial p(x) is divided by (ax + b). By finding the remainder, we will determine whether (ax + b) is a factor of p(x) as well. If the remainder is 0, then (ax + b) is a factor of p(x), otherwise, it is not.

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