USING REMAINDER THEOREM FIND THE REMAINDERS

Use the remainder theorem to find f(k).

Problem 1 :

k = 2; f(x) = x² - 2x + 5

A) -5     B) -3    C) -13     D) 5

Solution :

 f(2) = (2)² - 2(2) + 5

f(2) = 4 - 4 + 5

f(2) = 5

Remainder = 5

So, option (D) is correct.

Problem 2 :

k = -3; f(x) = x² + 2x + 2

A) 1     B) -13     C) 5     D) -17

Solution : 

 f(-3) = (-3)² + 2(-3) + 2

f(-3) = 9 - 6 + 2

f(-3) = 5

Remainder = 5

So, option (D) is correct.

Problem 3 :

k = -2; f(x) = 3x³ - 7x² - 3x + 3

A) 14     B) -55     C) -43     D) -5

Solution :

f(-2) = 3(-2)³ - 7(-2)² - 3(-2) + 3

f(-2) = -24 - 28 + 6 + 3

f(-2) = -52 + 9

f(-2) = -43

Remainder = -43

So, option (C) is correct.

Problem 4 :

k = 4; f(x) = x³ - 2x² + 5x - 2

A) 54     B) 50     C) -78     D) -76

Solution: 

 f(4) = (4)³ - 2(4)² + 5(4) - 2

f(4) = 64 - 32 + 20 - 2

f(4) = 84 - 34

f(4) = 50

Remainder = 50

So, option (B) is correct.

Problem 5 :

k = 2; f(x) = 9x⁴ + 10x³ + 6x² - 6x + 16

A) 360     B) 500     C) 252     D) 36

Solution :

f(2) = 9(2)⁴ + 10(2)³ + 6(2)² - 6(2) + 16

f(2) = 144 + 80 + 24 - 12 + 16

f(2) = 252

Remainder = 252

So, option (C) is correct.

Problem 6 :

k = 5; f(x) = x³ - 3x² - 4x - 5

A) 35     B) 25    C) -225     D) -220

Solution :

f(5) = (5)³ - 3(5)² - 4(5) - 5

f(5) = 125 - 75 - 20 - 5

f(5) = 25

Remainder = 25

So, option (B) is correct.

Problem 7 :

Using the remainder theorem find the remainders obtained when x³ + (kx + 8)x + k is divided by x + 1 and x - 2. Hence, find k if the sum of the two remainders is 1.

Solution :

p(x) = x³ + (kx + 8)x + k

x + 1 = 0

x = -1

p(-1) = (-1)³ + (k(-1) + 8)(-1) + k

p(-1) = -1 + k - 8 + k

p(-1) = -9 + 2k  ----(1)

x - 2 = 0

x = 2

p(2) = 2³ + (k(2) + 8)(2) + k

p(2) = 8 + 4k + 16 + k

p(2) = 24 + 5k -----(2)

(1) + (2)

-9 + 2k + 24 + 5k = 1

7k + 15 = 1

7k = 1-15

7k = -14

k = -2