Remainder Theorem

Remainder Theorem

10th - 11th Grade

10 Qs

quiz-placeholder

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Remainder Theorem

Remainder Theorem

Assessment

Quiz

Mathematics

10th - 11th Grade

Practice Problem

Easy

CCSS
HSA.APR.B.2

Standards-aligned

Created by

judy sotelo

Used 8+ times

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10 questions

Show all answers

1.

DRAW QUESTION

3 mins • 5 pts

Find the remainder when

f(x)  =  x3 + 3x2 + 3x + 1

is divided by (x + 1).

Media Image

Answer explanation

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.

Tags

CCSS.HSA.APR.B.2

2.

DRAW QUESTION

3 mins • 5 pts

Find the remainder when

f(x)  =  x3 - 3x + 1

is divided by (2 - 3x).

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Answer explanation

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 - 3(2/3) + 1

f(2/3)  =  8/27 - 2 + 1

f(2/3)  =  8/27 - 1

f(2/3)  =  8/27 - 27/27

f(2/3)  =  (8 - 27)/27

f(2/3)  =  -19/27

So, the remainder is -19/27.

Tags

CCSS.HSA.APR.B.2

3.

DRAW QUESTION

3 mins • 5 pts

For what value of k is the polynomial

2x4 + 3x3 + 2kx2 + 3x + 6

is divisible by (x + 2).

Media Image

Answer explanation

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.

Tags

CCSS.HSA.APR.B.2

4.

DRAW QUESTION

3 mins • 5 pts

Show that (x + 2) is a factor of 

x3 - 4x2 - 2x + 20

Media Image

Answer explanation

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. 

Tags

CCSS.HSA.APR.B.2

5.

DRAW QUESTION

3 mins • 5 pts

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

Media Image

Answer explanation

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. 

Tags

CCSS.HSA.APR.B.2

6.

DRAW QUESTION

3 mins • 5 pts

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

2x3 - 6x2 + mx + 4

Media Image

Answer explanation

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. 

Tags

CCSS.HSA.APR.B.2

7.

DRAW QUESTION

3 mins • 5 pts

Media Image

Answer explanation

f(3)=3(3)^3−4(3)^2+(3)−2

= 3(27) - 4(9) + 3 - 2

= 81-36 + 1

= 46

Tags

CCSS.HSA.APR.B.2

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