Problem 1 :
Without performing division, find the remainder when
x^{3} + 2x^{2 }- 7x + 5 is divided by x - 1
Solution :
let p(x) = x^{3} + 2x^{2} - 7x + 5
x - 1 = 0
x = 1
When p(x) is divided by x - 1, the remainder is p(1).
p(1) = (1)^{3} + 2(1)^{2} - 7(1) + 5
p(1) = 1 + 2 - 7 + 5
p(1) = 1
So, the remainder is 1.
Problem 2 :
Without performing division, find the remainder when
x^{4} - 2x^{2} + 3x - 1 is divided by x + 2
Solution:
Let p(x) = x^{4} - 2x^{2} + 3x - 1
Zero of x + 2 is -2
When p(x) is divided by x + 2, the remainder is p(-2).
p(-2) = (-2)^{4} - 2(-2)^{2} + 3(-2) - 1
p(-2) = 16 - 8 - 6 - 1
p(-2) = 1
So, the remainder is 1.
Problem 3 :
Find a given that:
when x^{3} - 2x + a is divided by x - 2, the remainder is 7.
Solution:
let p(x) = x^{3} - 2x + a
When p(x) is divided by x - 2, the remainder is p(2).
Given that p(2) = 7.
This implies that, (2)^{3} - 2(2) + a = 7
8 - 4 + a = 7
4 + a = 7
a = 7 - 4
a = 3
So, the value of a is 3.
Problem 4 :
Find a given that:
when 2x^{3} + x^{2} + ax - 5 is divided by x + 1, the remainder is -8.
Solution :
Let p(x) = 2x^{3} + x^{2} + ax - 5
When p(x) is divided by x + 1, the remainder is p(-1).
Given that p(-1) = -8.
This implies that 2(-1)^{3} + (-1)^{2} + a(-1) - 5 = -8
-2 + 1 - a - 5 = -8
-6 - a = -8
a = 8 - 6
a = 2
So, the value of a is 2.
Problem 5 :
Find a and b given that when x^{3} + 2x^{2} + ax + b is divided by x - 1 the remainder is 4, and when divided by x + 2 the remainder is 16.
Solution :
Let p(x) = x^{3} + 2x^{2} + ax + b
When p(x) is divided by x - 1, the remainder is p(1).
Given that p(1) = 4.
This implies that (1)^{3} + 2(1)^{2} + a(1) + b = 4
1 + 2 + a + b = 4
3 + a + b = 4
a + b = 1
a = 1 - b ---> (1)
When p(x) is divided by x + 2, the remainder is p(-2).
Given that p(-2) = 16.
This implies that (-2)^{3} + 2(-2)^{2} + a(-2) + b = 16
-8 + 8 - 2a + b = 16
-2a + b = 16
-2a = 16 - b ---> (2)
Substitute a = 1 - b in (2)
-2(1 - b) = 16 - b
-2 + 2b = 16 - b
2b + b = 16 + 2
3b = 18
b = 18/3
b = 6
By applying b = 6 in (1)
a = 1 - 6
a = -5
So, the values of a and b is -5 and 6.
Problem 5 :
2x^{n} + ax^{2} - 6 leaves a remainder of -7 when divided by x - 1, and 129 when divided by x + 3. Find a and n given that n∈Z^{+}.
Solution :
Let p(x) = 2x^{n} + 2x^{2} + ax + b
When p(x) is divided by x - 1, the remainder is p(1).
Given that p(1) = -7.
This implies that 2(1)^{n} + a(1)^{2} - 6 = -7
2 + a - 6 = -7
a = -7 + 6 - 2
a = -3
When p(x) is divided by x + 3, the remainder is p(-3).
Given that p(-3) = 129.
This implies that 2(-3)^{n} + (-3)(-3)^{2} - 6 = 129
2(-3)^{n} - 27 - 6 = 129
2(-3)^{n} = 129 + 33
2(-3)^{n} = 162
(-3)^{n} = 81
(-3)^{n} = (-3)^{4}
n = 4
So, the values of a and n is -3 and 4.
Problem 6 :
When P(z) is divided by z^{2} - 3z + 2 the remainder is 4z - 7. Find the remainder when P(z) is divided by:
a. z - 1 b. z - 2
Solution:
Given, let P(z) = 4z - 7
a.
When P(z) is divided by z - 1, the remainder is P(1).
P(1) = 4(1) - 7
P(1) = 4 - 7
P(1) = -3
b.
When P(z) is divided by z - 2, the remainder is P(2).
P(2) = 4(2) - 7
P(2) = 8 - 7
P(2) = 1
Problem 7 :
When P(z) is divided by z + 1 the remainder is -8 and when divided by z - 3 the remainder is 4. Find the remainder when P(z) is divided by (z - 3) (z + 1).
Solution :
If P(z) be the required polynomial, then Q(z) be quotient and R(z) be the remainder.
We divide the given polynomial by (z - 3) (z + 1), since the
= (z - 3) (z + 1)
Given that P(-1) = -8 and P(3) = 4.
We want to write,
P(z) = (z - 3) (z + 1) Q(z) + R(z)
Where R(z) = az + b
P(-1) = R(-1) = -a + b = -8 ---> (1)
P(3) = R(3) = 3a + b = 4 ---> (2)
Solving (1) and (2),
a = 3
By applying a = 3 in (1),
-3 + b = -8
b = -8 + 3
b = -5
Therefore, the remainder is 3z - 5.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
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