Question 1 :
What is the remainder when
x2018 + 2018
is divided by (x – 1).
Solution :
Let
p(x) = x2018 + 2018
Here, the divisor is (x - 1).
Equate the divisor to zero.
x - 1 = 0
Solve for x.
x = 1
Substitute 1 for x in p(x).
p(1) = (1)2018 + 2018
p(1) = 1 + 2018
p(1) = 2019
So, the remainder is 2019.
Question 2 :
For what value of k is the polynomial
p(x) = 2x3 - kx2 + 3x + 10
exactly divisible by (x – 2).
Solution :
Here, the divisor is (x - 2).
Equate the divisor to zero.
x - 2 = 0
Solve for x.
x = 2
Substitute 2 for x in p(2).
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 p(x) is exactly divisible by (x - 2), then the remainder must be zero.
Then,
32 - 4k = 0
Solve for k.
32 - 4k = 0
-4k = -32
k = 8
Therefore, p(x) is exactly divisible by (x-2) when k = 8.
Question 3 :
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 also find the remainder.
Solution :
Here, the divisor is (x - 3).
Equate the divisor to zero.
x - 3 = 0
Solve for x.
x = 3
Let
p(x) = 2x3 + ax2 + 4x – 12 and q(x) = x3 + x2 –2x+ a
p(x) = 2x3 + ax2 + 4x – 12
p(3) = 2(3)3 + a(3)2 + 4(3) – 12
= 2(27) + a(9) + 12 - 12
p(3) = 54 + 9a -------(1)
q(x) = x3 + x2 –2x+ a
q(3) = 33 + 32 –2(3) + a
27 + 9 - 6 + a
30 + a ------(2)
(1) = (2) ---->
54 + 9a = 30 + a
9a - a = 30 - 54
8a = -24
a = -3
Question 4 :
Determine whether (x -1) is a factor of the following polynomials:
(i) x3 + 5x2 - 10x + 4
Solution :
x - 1 = 0
x = 1
p(x) = x3 + 5x2 - 10x + 4
p(1) = 13 + 5(1)2 - 10(1) + 4
= 1 + 5 - 10 + 4
= 10 - 10
= 0
(ii) x4 + 5x2 - 5x + 1
Solution :
x - 1 = 0
x = 1
p(x) = x4 + 5x2 - 5x + 1
p(1) = 14 + 5(1)2 - 5(1) + 1
= 1 + 5 - 5 + 1
= 2
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