Remainder theorem questions and solutions are provided here to help the students learn how to find the remainder when a polynomial is divided by another polynomial without performing division. Let’s practise questions on the Remainder theorem, and verify the answers with the solutions provided. Also, get additional practice problems on the Remainder theorem in this article.
What is the Remainder theorem?
Suppose p(x) is a polynomial of degree at least 1 and a is any real number. When p(x) is divided by (x − a), the remainder will be p(a).
Thus, the Remainder theorem for polynomials can be written as:
p(x) = (x – a) q(x) + r
Here,
p(x) = Dividend
(x – a) = Divisor
q(x) = Quotient
r = Remainder
Substituting x = a in p(x), we have;
p(a) = (a – a) q(a) + r
= 0.q(a) + r
= r
Hence proved.
This can be compared with the basic division theorem, i.e., Dividend = Divisor × Quotient + Remainder.
Also, read: Remainder theorem
Remainder Theorem Questions and Answers
1. Using the Remainder Theorem, find the remainder when x6 – 5x4 + 3x2 + 10 is divided by x – 2.
Solution:
Let p(x) = x6 – 5x4 + 3x2 + 10
Now, divide p(x) by x – 2.
Here, remainder = r(x) = 6
Substitute x = 2 in p(x).
p(2) = (2)6 – 5(2)4 + 3(2)2 + 10
= 64 – 80 + 12 + 10
= 86 – 80
= 6
Thus, r = p(a) = p(2) = 6
2. The polynomial 4x2 – kx + 7 leaves a remainder of –2 when divided by x – 3. Find the value of k.
Solution:
Let p(x) = 4x2 – kx + 7
From the given,
-2 is the remainder when p(x) is divided by x – 3.
So, a = 3 and r = -2
By the Remainder theorem,
p(a) = r
p(3) = -2
4(3)2 – k(3) + 7 = -2
36 – 3k + 7 = -2
43 – 3k = -2
⇒ 3k = 43 + 2
⇒ 3k = 45
⇒ k = 15
Therefore, the value of k is 15.
3. If two polynomials 2x3 + kx2 + 4x – 12 and x3 + x2 – 2x + k leave the same remainder when divided by (x – 3), find the value of k and the remainder.
Solution:
Let p(x) = 2x3 + kx2 + 4x – 12
f(x) = x3 + x2 – 2x + k
Given that p(x) and f(x) leave the same remainder when divided by (x – 3).
So, a = 3
r = p(a) = f(a)
Thus, p(3) = f(3)
2(3)3 + k(3)2 + 4(3) – 12 = (3)3 + (3)2 – 2(3) + k
54 + 9k + 12 – 12 = 27 + 9 – 6 + k
54 + 9k = 30 + k
9k – k = 30 – 54
8k = -24
k = -24/8
k = -3
Substituting k = -3 in p(x), we get;
p(x) = 2x3 + (-3)x2 + 4x – 12
= 2x3 – 3x2 + 4x – 12
Now,
p(3) = 2(3)3 – 3(3)2 + 4(3) – 12
= 54 – 27 + 12 – 12
= 27
Therefore, k = -3 and the remainder = 27.
4. Find the value of m, if x = 1/2 is one of the zeroes of the polynomial p(x) = 4x4 − 4x3 − mx2 + 12x − 3.
Solution:
Given,
p(x) = 4x4 − 4x3 − mx2 + 12x − 3
x = 1/2 is the zero of p(x).
So, x – (1/2) is the factor of p(x).
That means, x – (1/2) divides the polynomial p(x) exactly and the remainder is 0.
By the Remainder theorem,
Remainder (r) = p(a)
p(1/2) = 0
4(1/2)4 − 4(1/2)3 − m(1/2)2 + 12(1/2) − 3 = 0
(4/16) – (4/8) – (m/4) + 6 – 3 = 0
(1/4) – (1/2) – (m/4) + 3 = 0
⇒ m/4 = 3 – (1/4)
⇒ m/4 = (12 – 1)/4
⇒ m = 11
Hence, the value of m is 11.
5. Given p(x) = x3 + 2x2 + 5x + 4, a = −1. Determine p(a) using the Remainder Theorem. If p(a) = 0, factor p(x) = (x – a) q(x).
Solution:
Given,
p(x) = x3 + 2x2 + 5x + 4
a = −1
By the Remainder theorem,
r = p(a)
= p(-1)
= (-1)3 + 2(-1)2 + 5(-1) + 4
= -1 + 2 – 5 + 4
= 0
Here, p(a) = 0.
Thus, x – (-1) = x + 1 is one of the factors of p(x).
Now, divide p(x) by x + 1.
Thus, q(x) = x2 + x + 4
Therefore, p(x) = (x + 1)(x2 + x + 4)
This is of the form p(x) = (x – a) q(x).
6. Find the value of k if p(x) = (3x – 2)(x – k) – 8 is divided by (x – 2) leaving the remainder 4.
Solution:
Given,
p(x) = (3x – 2)(x – k) – 8
Also, it is given that the remainder is 4 when p(x) is divided by (x – 2).
So, a = 2 and r = 4
Using the Remainder theorem,
p(a) = r
p(2) = 4
[3(2) – 2](2 – k) – 8 = 4
(6 – 2)(2 – k) = 4 + 8
4(2 – k) = 12
2 – k = 12/4
2 – k = 3
⇒ k = 2 – 3 = -1
Therefore, the value of k is -1.
7. What will be the remainder when 2x3 + 3x2 – 3x – 2 is divided by (2x – 3)?
Solution:
Let the given polynomial be:
p(x) = 2x3 + 3x2 – 3x – 2
Thus, remainder = 7.
Alternative method:
From (2x – 3), we can write the value of a as 3/2.
And p(a) = p(3/2) = 2(3/2)3 + 3(3/2)2 – 3(3/2) – 2
= 2(27/8) + 3(9/4) – (9/2) – 2
= (27/4) + (27/4) – (9/2) – 2
= (27 + 27 – 18 – 8)/4
= 28/4
= 7
Hence, by the remainder theorem, the remainder is 7.
8. If (x – 8) is one of the factors of mx3 – 24x2 + 192x – 512, find the value of m.
Solution:
Let the given polynomial be: p(x) = mx3 – 24x2 + 192x – 512
Given that (x – 8) is one of the factors of p(x).
That means x – 8 divides p(x) exactly.
Consequently, the remainder is 0, i.e., r = 0.
And let a = 8.
By the Remainder theorem,
p(a) = r
p(8) = 0
m(8)3 – 24(8)2 + 192(8) – 512 = 0
512m – 1536 + 1536 – 512 = 0
512m = 512
m = 512/512 = 1
Hence, the value of m is 1.
9. Determine the remainder when 4x3 – 3x2 + 2x – 4 is divided by x + (½) without actually performing the division.
Solution:
Let p(x) = 4x3 – 3x2 + 2x – 4
We have to find the remainder when p(x) is divided by x + (½), i.e., x – (-½).
So, let a = -½
By the Remainder theorem,
r = p(a)
= p(-½)
= 4(-½)3 – 3(-½)2 + 2(-½) – 4
= 4(-⅛) – 3(¼) – 1 – 4
= (-½) – (¾) – 5
= (-2 – 3 – 20)/4
= -25/4
Therefore, the remainder is -25/4.
10. Find the remainder using the Remainder theorem for the following expression.
(x4 – 5x3 + x2 – 2x + 6) ÷ (x + 4)
Solution:
(x4 – 5x3 + x2 – 2x + 6) ÷ (x + 4)
Here, p(x) = x4 – 5x3 + x2 – 2x + 6
And x + 4 = x – (-4)
Comparing with x – a, we have a = -4.
By the Remainder theorem,
r = p(a)
= p(-4)
= (-4)4 – 5(-4)3 + (-4)2 – 2(-4) + 6
= 256 + 5(64) + 16 + 8 + 6
= 256 + 320 + 16 + 8 + 6
= 606
Hence, the remainder is 606.
Practice Questions on Remainder Theorem
- Find the value of m if (x – 1) divides the polynomial mx3 – 2x2 + 25x – 26 without leaving the remainder.
- What is the remainder when p(x) = x3 – 2x2 + 6x + 7 is divided by x – 3?
- Determine the remainder for p(x) = x4 – 2x2 + 4 and a = 3/2, using the Remainder theorem.
- Suppose r and R are the remainders when the polynomials x3 + 2x2 –5kx–7 and x3 + kx2 – 12x + 6 are divided by x + 1 and x – 2, respectively. Find the value of k if 2r + R = 6.
- What will be the remainder when x2022 + 2022 is divided by (x – 1)?
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