FINDING REMAINDER USING REMAINDER THEOREM

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

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.