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Polynomials Test - 14

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Polynomials Test - 14

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Question 1
1 / -0

Which of the following expressions represents the quotient when x³ + 1 is divided by x + 1?

A
x²+ x + 1

B
x²+ x – 1

C
x²– x + 1

D
x²– x – 1

Solution

x³ + 1 divided by x + 1:

Hence, quotient = x² – x + 1.
Question 2
1 / -0

Which of following expressions is the expanded form of (5x – 3y)³?

A
125x³ + 27y³ – 225x²y + 135xy²

B
125x³ – 27y³ + 225x²y + 135xy²

C
125x³ – 27y³ + 225x²y – 135xy²

D
125x³ – 27y³ – 225x²y + 135y²x

Solution

(5x – 3y)³ = (5x)³ – 3(5x)² 3y + 3(5x) (3y)² – (3y)³
= 125x³ – 225x²y + 135xy² – 27y³
Hence, 125x³ – 27y³ – 225 x²y + 135xy²is the answer.
Question 3
1 / -0

On dividing x⁴ + 2x² – 3x + 7 by a polynomial g(x), we get quotient and remainder as x² + 5 and –3x + 22, respectively. The polynomial g(x) is given as _______.

A
x – 3

B
x² – 3

C
x² + 3

D
x + 3

Solution

p(x) = q(x) g(x) + r(x)
x⁴ + 2x² – 3x + 7 = (x² + 5) g(x) + (–3x + 22)
(x² + 5) g(x) = x⁴ + 2x² – 3x + 3x + 7 – 22
(x² + 5) g(x) = x⁴ + 2x² – 15
g(x) =

g(x) = x² – 3
Question 4
1 / -0

Find the remainder when x² + 6x + 5 is divided by x + 1.

A
1

B
5

C
6

D
0

Solution

Divide x² + 6x + 5 by x + 1

Hence, R = 0
Question 5
1 / -0

One of the zeroes of the polynomial x² + 2x – 35 is 5. Which of the following options is the other zero?

A
– 7

B
7

C
– 5

D
– 2

Solution

On factorising x² + 2x – 35, we get
x² + 7x – 5x – 35
= x (x + 7) – 5 (x + 7)
= (x – 5) (x + 7)
x = 5, x = – 7
Hence, – 7 is the other zero.
Question 6
1 / -0

Which of the following expressions is the standard form of the polynomial x² – x³ + 1 + 2x?

A
x² – x³ + 2x + 1

B
x³ – x² + 2x + 1

C
– x³ + x² + 2x – 1

D
– x³ + x² + 2x + 1

Solution

The standard form of the polynomial x² – x³ + 1 + 2x is –x³ + x² + 2x + 1.
Question 7
1 / -0

Factorise: x² – 2x + 5

A
(x + )(x + )

B
(x – 2)(x + 3)

C
(x – )(x – )

D
None of these

Solution

On factorising x² – 2x + 5, we get
x² – x – x + 5
= x(x – ) – (x – )
= (x – )(x – )
Question 8
1 / -0

Find the remainder when x² + 4x + 7 is divided by x + 4.

A
0

B
4

C
7

D
1

Solution

On dividing x² + 4x + 7 by x + 4, we get

Hence, remainder is 7
Question 9
1 / -0

Find the remainder when x³ + 2x² + x + 1 is divided by x + 1.

A
1

B
2

C
3

D
0

Solution

On dividing x³ + 2x² + x + 1 by x + 1, we get
Remainder = 1
This can be calculated by putting x = -1 in the expression x³ + 2x² + x + 1; or we have -1 + 2 - 1 + 1 = 1 as the final remainder.
Question 10
1 / -0

Which of the following expressions represents the remainder when x² + 3x + 2 is divided by x + 2?

A
x + 1

B
x

C
0

D
x + 2

Solution

When x² + 3x + 2 is divided by x + 2, we get

Hence, the remainder = 0
Question 11
1 / -0

Which of the following expressions is the standard form of the polynomial x³ – x²?

A
x³ – x² + 0x

B
x³ + 0x – x²

C
x³ – x² + 0x + 0

D
x²– x³

Solution

Standard form of x³ – x²
= x³ – x² + 0x + 0
Question 12
1 / -0

One of the zeroes of the polynomial x² – 5x + 4 is 1. Which of the following options is the other zero?

A
3

B
- 2

C
4

D
- 4

Solution

The polynomial x² – 5x + 4 can be factorised as
x – 4x – x + 4
= (x – 4) (x – 1)
x = 4 or x = 1
Hence, 4 is the other zero of the polynomial x² – 5x + 4.
Question 13
1 / -0

Which of the following expressions represents the quotient when 3x² + 5x + 8 is divided by x + 4?

A
3x + 7

B
3x – 7

C
7x + 3

D
3x + 4

Solution

When 3x² + 5x + 8 is divided by x + 4, we get

Hence, the quotient is 3x – 7.
Question 14
1 / -0

Which of following expressions is the expanded form of (2u – v – w)²?

A
4u² + v² + w² – 4uw – 4uv + 2vw

B
4u² + v² + w² + 4uw + 4uv – 2vw

C
4u² + v² + w² + 4uw – 4uv + 2vw

D
4u² + v² + w² + 4uw – 4uv – 2vw

Solution

The expanded form of (2u – v – w)²
= 4u² + v² + w² – 4uv – 4uw + 2vw
Question 15
1 / -0

Which of the following expressions represents the remainder when 4x² + 6x + 2 is divided by 2x?

A
1

B
0

C
2

D
- 1

Solution

When 4x² + 6x + 2 is divided by 2x, we get

Hence, the remainder is 2
Question 16
1 / -0

Which of the following expressions is the standard form of the polynomial x + x²?

A
x⁴ + 0x³ + 0x² + 0 = 0

B
x² + x = 0

C
x⁴ + 0³ + 0x

D
x⁴ + 0x³ + 0x² + 0x + 0

Solution

The standard form of x + x²is :
x² + x = 0.
Question 17
1 / -0

Which of the following expressions results from the factorisation of x³ + 3x² + 3x + 2?

A
(x – 2)(x² – x + 1)

B
(x + 2)(x² + x + 1)

C
(x + 2)(x² – x + 1)

D
(x + 2)(x² + x – 1)

Solution

p(x) = x³ + 3x² + 3x + 2
Let x = –2
p (–2) = –8 + 12 – 6 + 2
= 0
Hence, x + 2 is a factor.
Now, dividing x³ + 3x² + 3x + 2 by x + 2, we get (x² + x + 1) as the remaining factor.
Factors = (x + 2)(x² + x + 1)
Question 18
1 / -0

Which of the following expressions represents the quotient when x³ + x² + 2x – 3 is divided by x – 1?

A
x²– 2x + 4

B
x²– 2x – 4

C
x²+ 2x – 4

D
x²+ 2x + 4

Solution

Dividing x³ + x² + 2x – 3 by x – 1, we get

Hence, Q = x² + 2x + 4
Question 19
1 / -0

Find the remainder when x³ + 2x² + x + 3 is divided by x + 2.

A
– 1

B
1

C
3

D
– 3

Solution

Remainder = 1
Question 20
1 / -0

Which of the following expressions represents the remainder when x³ + 1 is divided by x + 1?

A
0

B
1

C
– 1

D
x² – x + 1

Solution

When p(x) = x³ + 1 is divided by x + 1, then
remainder = p(–1)
Hence, p(–1) = (–1)³ + (1)
= – 1 + 1
= 0
Question 21
1 / -0

Which of the following expressions represents the remainder when x³ + 2x - 3 is divided by x + 1?

A
0

B
- 6

C
6

D
x - 27

Solution

When p(x) = x³ + 2x - 3 is divided by x + 1,
the remainder = p(-1)
So, p(-1) = (-1)³ + 2(-1) - 3
= - 1 - 2 - 3 = - 6
Question 22
1 / -0

On dividing x⁴ + x³ + 2x² + 3 by a polynomial g(x), we get the quotient x² + x + 7 and the remainder is 5x + 38. The polynomial g(x) is ______.

A
x² + 5

B
x² + 7

C
x² – 5

D
x² – 7

Solution

nbsp; x⁴ + x³ + 2x² + 3 = (x² + x + 7) g(x) + 5x + 38Hence, g(x) =
=

g(x) = x² – 5
Question 23
1 / -0

Which of the following expressions is the standard form of the polynomial (x + 2) (x – 1)?

A
x² + x – 2

B
– 2 + x + x²

C
x² – 2

D
– 2 + x²

Solution

The polynomial (x + 2) (x – 1)
= x² + 2x – x – 2
Standard form = x² + x – 2
Question 24
1 / -0

Which of the following expressions is the expanded form of (3x + 4y + 5z)²?

A
9x² + 16y² + 25z² + 24xy + 30xz + 40yz

B
9x² + 16y² + 25z² - 24xy + 30xz + 40yz

C
9x² + 16y² + 25z² + 40xy + 30xz + 24yz

D
9x² + 16y² + 25z² + 24xy + 40xxz + 30yz

Solution

Expanded form of (3x + 4y + 5z)²
= 9x² + 16y² + 25z² + 24xy + 40yz + 30xz
= 9x² + 16y² + 25z² + 24xy + 30xz + 40yz
Question 25
1 / -0

One of the zeroes of the polynomial 2x² + x – 1 is – 1. Which of the following options is the other zero?

A

B

C
3

D
– 2

Solution

The polynomial 2x² + x – 1 can be factorised as
2x² + x – 1 = 2x² + 2x – x – 1
= 2x (x + 1) – 1 (x + 1)
= (2x – 1) (x + 1)
x = and x = – 1
Therefore, x = is the other zero of the polynomial.
Question 26
1 / -0

Which of the following expressions is the standard form of the polynomial x⁴?

A
x⁴ + 0x³ + 0x² + 0

B
x⁴ + 0x² + 0

C
x⁴ + 0x³ + 0x

D
x⁴ + 0x³ + 0x² + 0x + 0

Solution

The standard form of x⁴
= x⁴ + 0x³ + 0x² + 0x + 0
Question 27
1 / -0

Find the remainder when 2x⁴ – 3x³ – 3x² + 6x – 2 is divided by 2x² – 3x + 1.

A
0

B
2

C
12

D
6

Solution

On dividing 2x⁴ – 3x³ – 3x² + 6x – 2 by 2x² – 3x + 1, we get

Hence, the remainder is 0
Question 28
1 / -0

Which of the following expressions is the factorisation of y³ – 23y² + 142y – 120?

A
(y – 1) (y + 2) (y – 10)

B
(y + 1) (y + 10) (y – 12)

C
(y + 1) (y – 10) (y – 12)

D
(y – 1) (y – 10) (y – 12)

Solution

Factorise y³ – 23y² + 142y – 120.
If y = 1, then
y³ – 23y² + 142y – 120 = (1)³ – 23(1)² + 142(1) – 120
= 1 – 23 + 142 – 120
= 0
Hence, (y – 1) is a factor.
On dividing y³ – 23y² + 142y – 120 by y – 1, we get

Now, y² – 22y + 120
= y² – 12y – 10y + 120
= y(y – 12) – 10(y – 12)
= (y – 10) (y – 12)
Therefore, (y – 1), (y – 10) and (y – 12) are the factors of y³ – 23y² + 142y – 120.
Question 29
1 / -0

Which of the following expressions represents the remainder when x⁴ + x³ + 2x² + 3x is divided by x² + x?

A
0

B
x

C
x²

D
x² + x

Solution

On dividing x⁴ + x³ + 2x² + 3x by x² + x, we get

Remainder = x
Question 30
1 / -0

Which of the following expressions is the expanded form of _?

A

B

C

D

Solution

=

=
Question 31
1 / -0

One of the zeroes of the polynomial x² + 2x – 3 is – 3. Which of the following options is the other zero?

A
– 1

B
3

C
2

D
1

Solution

The polynomial x² + 2x – 3 = x² + 3x – x – 3
= x(x + 3) (– 1) (x + 3)
= (x – 1) (x + 3)
x = 1 or x = – 3
Hence, 1 is the other zero of x² + 2x – 3.
Question 32
1 / -0

Which of the following expressions represents the quotient when x² + x + 1 is divided by x – 1?

A
3

B
x – 1

C
x + 2

D
x + 1

Solution

When x² + x + 1 is divided by x – 1, we get

Hence, the quotient is x + 2
Question 33
1 / -0

Factorise: x² – (y + z) x + yz

A
(x – y) (x + z)

B
(x + y) (z + x)

C
(x – y) (x – z)

D
(x – z) (z – y)

Solution

x² – (y + z) x + zy
= x² – xy – zx + zy
= x(x – y) – z (x – y)
= (x – y) (x – z)
Question 34
1 / -0

Which of the following expressions represents the quotient when x⁴ + x³+ 2x + 1 is divided by x² + 1?

A
x²+ x + 1

B
x²+ x – 1

C
x²– x + 1

D
x²– x – 1

Solution

When x⁴ + x³ + 2x + 1 is divided by x² + 1, we get

Hence, the quotient is x² + x – 1
Question 35
1 / -0

Find the quotient when x³ – 3x + 1 is divided by x + 2.

A
x³ + 2x – 7

B
x² + 2x + 7

C
x² + 2x – 7

D
x² – 2x + 1

Solution

x³ – 3x + 1 = (x + 2) g(x) – 1
g(x) =

Q = x² – 2x + 1
Question 36
1 / -0

Find the remainder when x⁵ + x³ + 3x² + 1 is divided by x³ + 1.

A
x²+ 1

B
2x²+ 1

C
2x²

D
x²+ x

Solution

On dividing x⁵ + x³ + 3x² + 1 by x³ + 1, we get

Hence, the remainder is 2x²
Question 37
1 / -0

One of the zeroes of the polynomial x² – 10x + 21 is 3. Which of the following options is the other zero?

A
- 3

B
21

C
7

D
- 10

Solution

Factorising x² – 10x + 21, we get
x² – 7x – 3x + 21
= x(x – 7) – 3(x – 7)
= (x – 7) (x – 3)
x = 7 or x = 3
Hence, the other zero is 7
Question 38
1 / -0

Find the quotient when x² + 2x – 3 is divided by x – 1.

A
x – 3

B
x + 3

C
3 – x

D
x + 1

Solution

Dividing x² + 2x – 3 by x – 1, we get

Hence, the quotient is x + 3
Question 39
1 / -0

Which of the following expressions is the expanded form of (2x – 1)²?

A
4x² – 2x + 1

B
4x² + 4x + 1

C
4x² – 4x + 1

D
4x² – 2x – 1

Solution

– 2 . 2x . 1 + (1)²
= 4x² – 4x + 1
Question 40
1 / -0

Factorise x² - 5x + 4.

A
(x - 4)(x + 1)

B
(x - 4)(x - 1)

C
(x + 4)(x + 1)

D
(x + 4)(x - 1)

Solution

Factors of x² - 5x + 4
= x² - 4x - x + 4
= x(x - 4) - 1(x - 4)
= (x - 4)(x - 1)

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