Remainder Theorem
Quiz
•
Mathematics
•
10th - 11th Grade
•
Practice Problem
•
Easy
Standards-aligned
judy sotelo
Used 8+ times
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10 questions
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1.
DRAW QUESTION
3 mins • 5 pts
Find the remainder when
f(x) = x3 + 3x2 + 3x + 1
is divided by (x + 1).
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).
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).
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
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 ?
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
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
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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