Polynomials Test

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Polynomials Test - 5

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Weekly Quiz Competition

Question 1
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 2
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 3
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 4
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.

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Polynomials Test - 5

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Polynomials Test - 5

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Question 1

125x3 + 27y3 – 225x2y + 135xy2

125x3 – 27y3 + 225x2y + 135xy2

125x3 – 27y3 + 225x2y – 135xy2

125x3 – 27y3 – 225x2y + 135y2x

Solution

Question 2

– 7

7

– 5

– 2

Solution

Question 3

x2 – x3 + 2x + 1

x3 – x2 + 2x + 1

– x3 + x2 + 2x – 1

– x3 + x2 + 2x + 1

Solution

Question 4

1

2

3

0

Solution

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125x³ + 27y³ – 225x²y + 135xy²

125x³ – 27y³ + 225x²y + 135xy²

125x³ – 27y³ + 225x²y – 135xy²

125x³ – 27y³ – 225x²y + 135y²x

x² – x³ + 2x + 1

x³ – x² + 2x + 1

– x³ + x² + 2x – 1

– x³ + x² + 2x + 1