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Remainder Theorem Worksheet with Answers | Remainder Theorem Questions PDF
Problem 1:
Find the remainder when f(x) = x3 + 3x2 + 3x + 1 is divided by (x + 1), using the remainder theorem?
Solution:
Given that,
The function is f(x) = x3 + 3x2 + 3x + 1.
Here, the divisor is (x + 1).
First, equate the divisor to zero. i.e.,
x + 1 = 0
Now, solve the value of x.
The value of x is -1.
Next, to find the remainder. So 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
f(-1) = -1 + 3 – 3 + 1
f(-1) = 0
Hence, the remainder is 0.
Problem 2 :
Using Remainder Theorem, find the remainder when f(x) = x3-3x + 1 is divided by (2 – 3x)?
Solution:
As given in the question,
The function is f(x) = x3-3x + 1.
Here, the divisor is (2 – 3x).
Now, we can equate the divisor to zero.
Then, it will be 2 – 3x = 0
Next, solve the value of x.
-3x = -2
x = 2/3
Now, we need to find the value of a remainder.
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=Â (8 – 27)/27
f(2/3) = -19/27
Therefore, the 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:
Let the given function is f(x) = 2x4 + 3x3 + 2kx2 + 3x + 6
Next, the divisor is (x + 2).
Now, equate the divisor to zero.
x + 2 = 0
After solving, the value of x is,
x = -2
Next, to find the remainder value. So substitute -2 for x into the function f(x).
f(-2) = 2(-2)4 + 3(-2)3 + 2k(-2)2 + 3(-2) + 6
f(-2) = 2(16) + 3(-8) + 2k(4) – 6 + 6
f(-2) = 32 – 24 + 8k – 6 + 6
f(-2) = 8 + 8k
Then the remainder is (8 + 8k).
Next, if f(x) is exactly divisible by (x + 2), then the remainder must be zero.
So, 8 + 8k = 0
Solving of k is,
8k = -8
k = -1
Hence, f(x) is exactly divisible by (x+2) when k = –1.
Problem 4: If the polynomial expression is (x) = x4−2x3+3x2−ax+b such that when divided by x−1 and x+1, the remainder is respectively 5 and 19. Find the remainder value when f(x) is divided by (x−2).
Solution:
Given that,
The polynomial expression is f(x) = x4−2x3+3x2−ax+b.
If f(x) is divided by x-1 and x+1 then the remainders are 5 and 19 respectively.
Therefore f(1)=5 and f(−1)=19.
First, substitute the value of x as 1 in the given polynomial equation.
Then it is p(1) = (1)4−2(1)3+3(1)2− a(1)+b = 5
Now, substitute the x value as −1, then it will be
p(-1) = (-1)4−2(−1)3+3(−1)2− a(−1)+b=19
Then the equations is
1−2+3−a+b=5
and 1+2+3+a+b=19
⇒2−a+b=5 and 6+a+b=19
⇒−a+b=3 and a+b=13
Adding these two equations, we get
(−a+b)+(a+b)=3+13
⇒2b=16 ⇒b=8
Now, Putting b=8 and −a+b=3, we get
−a+8=3 ⇒-a=−5 ⇒a=5
Putting the values of a and b in
f(x)=x4−2x3+3x2−5x+8
The remainder when f(x) is divided by (x-2) is equal to f(2).
So, Remainder = f(2) = (2)4−2(2)3+3(2)2−5(2)+8
= 16−16+12−10+8 =10
Hence, by dividing f(x) by x-2 we get the remainder is 10.
Problem 5:  If two polynomials 2x3 + ax2 + 4x – 12 and x3 + x2 –2x +a leave the same remainder when divided by (x – 3), find the value of a, and what is the remainder value?
Solution:
In the given question,
The two polynomial functions are 2x3 + ax2 + 4x – 12 and x3 + x2 –2x +a.
The divisor is (x – 3).
First, equate the divisor to zero.
x – 3 = 0
Now, Solve the value of x.
x = 3
Let p(x) = 2x3 + ax2 + 4x – 12 and q(x) = x3 + x2 –2x +a.
First, take p(x) = 2x3 + ax2 + 4x – 12
Subsitute x value in the function of p(x).
p(3) = 2(3)3 + a(3)2 + 4(3) – 12 = 2(27) + a(9) + 12 – 12
p(3) = 54 + 9a ——-(1)
Next take the function q(x) = x3 + x2 –2x+ a.
Subsitute x value in the q(x) function.
q(3) = 33 + 32 –2(3) + a
=27 + 9 – 6 + a
q(x) = 30 + a ——(2)
Now equate, 1 and 2 equations.
(1) = (2) —-> 54 + 9a = 30 + a
9a – a = 30 – 54
8a = -24
a = -3
Therefore, the value of the a is -3.
Problem 6: What is the remainder when x2018 + 2018 is divided by (x – 1)?
Solution:
Given that,
The function p(x) is x2018 + 2018 and the divisor is (x – 1).
Now, equate the divisor to zero.
Then it is x – 1 = 0
Solve the value of x. i.e.,
x = 1
Substitute the value 1 for x in p(x).
p(1) = (1)2018 + 2018
p(1) = 1 + 2018
p(1) = 2019
Hence, the remainder of the given polynomial function is 2019.
Problem 7: For what value of k is the polynomial p(x) = 2x3 – kx2 + 3x + 10 is exactly divisible by (x – 2).
Solution:
Given that,
The divisor is (x – 2). Now, Equate the divisor to zero.
x – 2 = 0
For solving the value of x is, x = 2
Substitute 2 for x in p(2). Then we get,
p(2) = 2(2)3 – k(2)2 + 3(2) + 10
p(2) = 2(8) – k(4) + 6 + 10
p(2) = 16 – 4k + 16
p(2) = 32 – 4k
So, the remainder is (32 – 4k).
If the p(x) is exactly divisible by (x – 2), then the remainder must be zero. Then,
32 – 4k = 0
Solve for k value, 32 – 4k = 0
-4k = -32
k = 8
Therefore, p(x) is exactly divisible by (x-2) when the value of k is 8.
Problem 8: By remainder theorem, find the remainder when, p(x) is divided by g(x) where,
(i) p(x) = x3 – 2x2 – 4x – 1 and g(x) = x + 1
(ii) p(x) = 4x3– 12x2 + 14x – 3 and g(x) = 2x – 1
(iii) p(x) = x3 – 3x2 + 4x + 50 and g(x) = x – 3
Solution:
Given the p(x) and g(x) values.
Now, we need to find the value of the remainder.
(i) p(x) = x3 – 2x2 – 4x – 1 and g(x) = x + 1
First, equate the divisor to zero. i.e.,
x + 1 = 0
x = -1
p(x) = x3 – 2x2 – 4x – 1
p(-1) = (-1)3 – 2(-1)2 – 4(-1) – 1
p(-1) = -1 – 2(1) + 4 – 1 = -1 – 2 + 4 – 1
So, the p(-1) is 0.
Hence, the remainder of the given polynomial is 0.
(ii) p(x) = 4x3 – 12x2 + 14x – 3 and g(x) = 2x – 1
Equate the divisor to zero. Then it will be,
2x – 1 = 0
x = 1/2
Next substitute the value of x in the p(x).
p(1/2) = 4(1/2)3 – 12(1/2)2 + 14(1/2) – 3
p(1/2) = (4/8) – (12/4) + 7 – 3 = (1/2) – 3 + 7 – 3
p(1/2) = (1/2) + 1 = 3/2.
Therefore, the given polynomial remainder is 3/2.
(iii) p(x) = x3 – 3x2 + 4x + 50 and g(x) = x – 3
Now, equate the divisor to zero. That is,
x – 3 = 0
x = 3
p(x) = x3 – 3x2 + 4x + 50
After substituting the value of x.
p(3) = 33 – 3(3)2 + 4(3) + 50
= 27 – 27 + 12 + 50 = 62.
Thus, the polynomial remainder is 62.
Problem 9:Â Divide 3x3 + x2 + 2x + 5 by x + 1 and find the remainder?
Solution:
As given in the question,
The Dividend is p(x) = 3x3 + x2 + 2x + 5
The Divisor = g(x) = (x + 1)
The divisor is (x + 1), i.e. it is a factor of the given polynomial p(x).
Now, equate the divisor to zero.
So, x + 1 = 0
x = -1
Now, substituting x = -1 in p(x), then it will be
p(-1) = 3(-1)3 + (-1)2 +2(-1) +5
= 3(-1) + 1 – 2 + 5
= -3 + 4 = 1
So, the Remainder Value of p(x) at x = -1. Hence proved the remainder theorem.
Alternatively,
We know that, p(x) = (x – a)·q(x) + r
Now, observe what happens when we have x equal to a
p(a) = (a – a)·q(a) + r
Substituting the values, in the above formula. We get,
p(-1) = [-1 – (-1)]·q(-1) + (1)
p(-1) = 0.q(-1) +1
p(-1) = 1
p(-1) = remainder
Hence, it is proved.
Problem 10: (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, Find the values of the a and b?
Solution:
In the given question,
The polynomial function p(x) is x3 + ax2 + bx – 9
(x-3) is a factor of the polynomial function.
Now, we need to find the values of a and b.
First, equate the divisor to zero. i. e.,
x-3 =0 , x=3
So, the p(3) is 0.
Now, substitute x value in the polynomial function. Then it will be
p(3) = (3)3 + a(3)2 + b(3) – 9 = 0
After simplification, the function is,
27 + 9a + 3b – 9 = 0
3a + b = -6 … Equation (i)
Next, when p(x) is divided by (x – 2) then the remainder is 1 will be given.
Now, equate (x-2) is zero.
Then x-2 =0
x=2.
Substitute x value in the polynomial function.
p(2) = 1
p(2) = (2)3 + a(2)2 + b(2) – 9 = 1
p(2) = 4a + 2b
2a + b = 1 …. Equation (ii)
By solving equations (i) and Equation (ii), we get the value of a and b.
So, the value of a is -7 and the value of b is 14.