PROBLEMS ON REMAINDER THEOREM

Problem 1 :

Without performing division, find the remainder when

x³ + 2x² - 7x + 5 is divided by x - 1

Solution :

let p(x) = x³ + 2x² - 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

x⁴ - 2x² + 3x - 1 is divided by x + 2

Solution:

Let p(x) = x⁴ - 2x² + 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 x³ - 2x + a is divided by x - 2, the remainder is 7.

Solution:

let p(x) = x³ - 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 2x³ + x² + ax - 5 is divided by x + 1, the remainder is -8.

Solution :

Let p(x) = 2x³ + x² + 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 x³ + 2x² + 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) = x³ + 2x² + 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 :

2xⁿ + ax² - 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) = 2xⁿ + 2x² + 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 z² - 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.