Polynomials Test

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Polynomials Test - 2

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Weekly Quiz Competition

Question 1
1 / -0

If a + b + c = 10 and a² + b² + c² = 40, then ab + bc + ca =

A
136

B
64

C
30

D
68

Solution

a + b + c = 10 …..(1)
a² + b² + c² = 40 ….(2)
Squaring both sides of (1), we get
(a + b + c)² = (10)²
a² + b² + c² + 2ab + 2bc + 2ca = 100
40 + 2(ab + bc + ca) = 100 (because a² + b² + c² = 40)
2(ab + bc + ca) = 60
ab + bc + ca = 30
Question 2
1 / -0

Choose the real factors of x² + 9 from the following:

A
(x² + 3) (x² - 3)

B
(x + 3) (x- 3)

C
does not exist

D
none of these

Solution

Real factors do not exist for x² + 4.
Question 3
1 / -0

For x² + 2x + 5 to be factor of x⁴ + px² + q, find the respective values of p and q out of the following options.

A
- 2, 5

B
5, 25

C
10, 20

D
6, 25

Solution

Let the other factor be x2 + ax + b.
we have
(x² + 2x + 5) (x² + ax + b)
= x⁴ + px² + q
Or, x4 + (a + 2)x3 + (2a + b + 5)x2 + (5a + 2b)x + 5b = x⁴ + px² + q

On comparing the coefficients, we get
2a + b + 5 = p ---- (1)
5b = q ----- (2)

2 + a = 0 ⇒ a = - 2

5a + 2b = 0 ⇒ b = 5

∴ p = 2a + b + 5 = 2 (-2) + 5 + 5 = 6

q = 5b = 5 (5) = 25
Question 4
1 / -0

Factorise: px² + (4p² - 3q) x - 12 pq

A
(x - 4p) (px - 3q)

B
(x + 4p) (px - 3q)

C
(x + 4p) (px + 3q)

D
(x - 4p) (px + 3q)

Solution

px² + 4p²x - 3qx - 12 pq

px(x + 4p) - 3q(x + 4p)

(x + 4p) (px -3q)
Question 5
1 / -0

Which of the following expressions is equivalent to (3x – 5y)³ + (2y – 5x)³ + (2x + 3y)³?

A
(3x – 5y) (5x – 2y) + (2x + 3y)

B
3 (3x – 5y) (5x – 2y) (2x + 3y)

C
3 (3x – 5y) (2y – 5x) (2x + 3y)

D
(8x – 3y)³ + (7x – y)³

Solution

(3x - 5y)³ + (2y - 5x)³ + (2x + 3y)³
If a + b + c = 0, then a3 + b³ + c³ = 3abc

a = 3x - 5y, b = 2y - 5x, c = 2x + 3y
a + b + c = (3x - 5y) + (2y - 5x) + (2x + 3y)

= 3x - 5x + 2x - 5y + 2y + 3y
= 5x - 5x - 5y + 5y = 0

Then, a³ + b³ + c³ = 3abc
(3x - 5y)³ + (2y - 5x)³ + (2x + 3y)³ = 3(3x - 5y) (2y - 5x) (2x + 3y)
Question 6
1 / -0

Which of the following is/are factor(s) of p³ (q - r)³ + q³ (r - p)³ + r³ ( p - q)³?

A
p - q

B
q - r

C
r - p

D
All of the above

Solution

[a (b - c)³] + [b (c - a)]³ + [c (a - b)]³
= 3a (b - c) b(c - a) c(a - b)
= 3abc(a - b) (b - c) (c - a)

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Re-Attempt Weekly Quiz Competition

Polynomials Test - 2

Result

Polynomials Test - 2

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SHARING IS CARING

Question 1

136

64

30

68

Solution

Question 2

(x2 + 3) (x2 - 3)

(x + 3) (x - 3)

does not exist

none of these

Solution

Question 3

- 2, 5

5, 25

10, 20

6, 25

Solution

Question 4

(x - 4p) (px - 3q)

(x + 4p) (px - 3q)

(x + 4p) (px + 3q)

(x - 4p) (px + 3q)

Solution

Question 5

(3x – 5y) (5x – 2y) + (2x + 3y)

3 (3x – 5y) (5x – 2y) (2x + 3y)

3 (3x – 5y) (2y – 5x) (2x + 3y)

(8x – 3y)3 + (7x – y)3

Solution

Question 6

p - q

q - r

r - p

All of the above

Solution

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(x² + 3) (x² - 3)

(x + 3) (x- 3)

(8x – 3y)³ + (7x – y)³