# REMAINDER THEOREM AND FACTOR THEOREM WORKSHEET

Remainder Theorem and Factor Theorem Worksheet :

Worksheet given in this section will be much useful for the students who would like to practice solving problems on remainder theorem and factor theorem.

Before look at the worksheet, if you would like to know about remainder theorem and factor theorem in detail,

## Remainder Theorem and Factor Theorem Worksheet - Problems

Problem 1 :

Using Remainder Theorem, find the remainder when

f(x)  =  x3 + 3x2 + 3x + 1

is divided by (x + 1).

Problem 2 :

Using Remainder Theorem, find the remainder when

f(x)  =  x3 - 3x + 1

is divided by (2 - 3x).

Problem 3 :

For what value of k is the polynomial

2x4 + 3x3 + 2kx2 + 3x + 6

is divisible by (x + 2).

Problem 4 :

Using Factor Theorem, show that (x + 2) is a factor of

x3 - 4x2 - 2x + 20

Problem 5 :

Is (3x - 2) a factor of 3x3 + x2 - 20x + 12 ?

Problem 6 :

Find the value of m, if (x - 2) is a factor of the polynomial

2x3 - 6x2 + mx + 4

## Remainder Theorem and Factor Theorem Worksheet - Solutions

Problem 1 :

Using Remainder Theorem, find the remainder when

f(x)  =  x3 + 3x2 + 3x + 1

is divided by (x + 1).

Solution :

Here, the divisor is (x + 1).

Equate the divisor to zero.

x + 1  =  0

Solve for x.

x  =  -1

To find the remainder, substitute -1 for x into the function f(x).

f(-1)  =  (-1)3 + 3(-1)2 + 3(-1) + 1

f(-1)  =  -1 + 3(1) - 3 + 1

f(-1)  =  -1 + 3 - 3 + 1

f(-1)  =  0

So, the remainder is 0.

Problem 2 :

Using Remainder Theorem, find the remainder when

f(x)  =  x3 - 3x + 1

is divided by (2 - 3x).

Solution :

Here, the divisor is (2 - 3x).

Equate the divisor to zero.

2 - 3x  =  0

Solve for x.

-3x  =  -2

x  =  2/3

To find the remainder, substitute 2/3 for x into the function f(x).

f(2/3)  =  (2/3)3 - 2/3 + 1

f(2/3)  =  8/27 - 18/27 + 27/27

f(2/3)  =  (8 - 18 + 27) / 27

f(2/3)  =  17 / 27

So, the remainder is 17/27.

Problem 3 :

For what value of k is the polynomial

2x4 + 3x3 + 2kx2 + 3x + 6

is divisible by (x + 2).

Solution :

Let

f(x)  =  2x4 + 3x3 + 2kx2 + 3x + 6

Here, the divisor is (x + 2).

Equate the divisor to zero.

x + 2  =  0

Solve for x.

x  =  -2

To find the remainder, substitute -2 for x into the function f(x).

f(-2)  =  2(-2)4 + 3(-2)3 + 2k(-2)2 + 3(-2) + 6

f(-2)  =  2(16) + 3(-8) + 2k(4) - 6 + 6

f(-2)  =  32 - 24 + 8k - 6 + 6

f(-2)  =  8 + 8k

So, the remainder is (8 + 8k).

If f(x) is exactly divisible by (x + 2), then the remainder must be zero.

Then,

8 + 8k  =  0

Solve for k.

8k  =  -8

k  =  -1

Therefore, f(x) is exactly divisible by (x+2) when k  =  –1.

Problem 4 :

Using Factor Theorem, show that (x + 2) is a factor of

x3 - 4x2 - 2x + 20

Solution :

Let

f(x)  =  x3 - 4x2 - 2x + 20

Equate the factor (x + 2) to zero.

x + 2  =  0

Solve for x.

x  =  -2

By Factor Theorem,

(x + 2) is factor of f(x), if f(-2)  =  0

Then,

f(-2)  =  (-2)3 - 4(-2)2 - 2(-2) + 20

f(-2)  =  -8 - 4(4) + 4 + 20

f(-2)  =  -8 - 16 + 4 + 20

f(-2)  =  0

Therefore, (x + 2) is a factor of x3 - 4x2 - 2x + 20.

Problem 5 :

Is (3x - 2) a factor of 3x3 + x2 - 20x + 12 ?

Solution :

Let

f(x)  =  3x3 + x2 - 20x + 12

Equate the factor (3x + 2) to zero.

3x - 2  =  0

Solve for x.

3x  =  2

x  =  2/3

By Factor Theorem,

(3x - 2) is factor of f(x), if f(2/3)  =  0

Then,

f(2/3)  =  3(2/3)3 + (2/3)2 - 20(2/3) + 12

f(2/3)  =  3(8/27) + 4/9 - 40/3 + 12

f(2/3)  =  8/9 + 4/9 - 40/3 + 12

f(2/3)  =  8/9 + 4/9 - 120/9 + 108/9

f(2/3)  =  (8 + 4 - 120 + 108) / 9

f(2/3)  =  (120 - 120) / 9

f(2/3)  =  0

Therefore, (3x - 2) is a factor of 3x3 + x2 - 20x + 12

Problem 6 :

Find the value of m, if (x - 2) is a factor of the polynomial

2x3 - 6x2 + mx + 4

Solution :

Let

f(x)  =  2x3 - 6x2 + mx + 4

Equate the factor (x - 2) to zero.

x - 2  =  0

Solve for x.

x  =  2

By Factor Theorem,

(x - 2) is factor of f(x), if f(2)  =  0

Then,

f(2)  =  0

2(2)3 - 6(2)2 + m(2) + 4  =  0

f(2)  =  2(8) - 6(4) + 2m + 4  =  0

f(2)  =  16 - 24 + 2m + 4  =  0

f(2)  =  2m - 4  =  0

2m  =  4

m  =  2

Therefore (x - 2) is a factor of f(x), when m  =  2.

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