Polynomials Test

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Polynomials Test - 6

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Weekly Quiz Competition

Question 1
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 2
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 3
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 4
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 5
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 6
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

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Polynomials Test - 6

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Polynomials Test - 6

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Question 1

x3 – x2 + 0x

x3 + 0x – x2

x3 – x2 + 0x + 0

x2 – x3

Solution

Question 2

3

- 2

4

- 4

Solution

Question 3

4u2 + v2 + w2 – 4uw – 4uv + 2vw

4u2 + v2 + w2 + 4uw + 4uv – 2vw

4u2 + v2 + w2 + 4uw – 4uv + 2vw

4u2 + v2 + w2 + 4uw – 4uv – 2vw

Solution

Question 4

x4 + 0x3 + 0x2 + 0 = 0

x2 + x = 0

x4 + 03 + 0x

x4 + 0x3 + 0x2 + 0x + 0

Solution

Question 5

(x – 2)(x2 – x + 1)

(x + 2)(x2 + x + 1)

(x + 2)(x2 – x + 1)

(x + 2)(x2 + x – 1)

Solution

Question 6

0

1

– 1

x2 – x + 1

Solution

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x³ – x² + 0x

x³ + 0x – x²

x³ – x² + 0x + 0

x²– x³

4u² + v² + w² – 4uw – 4uv + 2vw

4u² + v² + w² + 4uw + 4uv – 2vw

4u² + v² + w² + 4uw – 4uv + 2vw

4u² + v² + w² + 4uw – 4uv – 2vw

x⁴ + 0x³ + 0x² + 0 = 0

x² + x = 0

x⁴ + 0³ + 0x

x⁴ + 0x³ + 0x² + 0x + 0

(x – 2)(x² – x + 1)

(x + 2)(x² + x + 1)

(x + 2)(x² – x + 1)

(x + 2)(x² + x – 1)

x² – x + 1