Polynomials Test

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Polynomials Test - 5

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Weekly Quiz Competition

Question 1
1 / -0

If quotient = 3x² – 2x + 1, remainder = 2x – 5 and divisor = x + 1, what is the dividend?

A
3x³ – x² + x – 3

B
3x³ – x² – x – 3

C
3x³ + x² – x – 3

D
3x³ + x² + x – 4

Solution

Dividend = Divisor Quotient + Remainder
= (x + 1) (3x²– 2x + 1) + (2x - 5)
= 3x³ + x²+ x – 4
Question 2
1 / -0

If x - = , then x² + =

A
2

B
4

C
6

D
9

Solution

If = , then find the value of x² + .
x - =
Squaring both sides, we get:

x² + - 2 x = 7
x² + - 2 = 7
x² + = 7 + 2
x² + = 9
Question 3
1 / -0

Factorise a^2x – b^2x.

A
(a^x + b^x)(a^x – b^x)

B
(a^x – b^x)²

C
(a^x + b^x)(a² – b²)

D
(a^x – b^x)(a² + b²)

Solution

By using the identity x² - y² = (x + y)(x - y),
We get
(a^x)² - (b^x)²= (a^x + b^x)(a^x – b^x)
Question 4
1 / -0

If x = , y = and z = , then the value of is

A
abc

B
a²b²c²

C
1

D
- 1

Solution

Applying compedendo and dividendo process, we get

Similarly, = , =

= 1
Question 5
1 / -0

If x² + = 62, then x + equals to which of the following values?

A

B
8

C

D
None of these

Solution

Given: x² + = 62
Adding 2 on both sides, we get
x² + + 2 = 62 + 2
= 64
Taking square roots on both sides, we get
x + = 8
Question 6
1 / -0

One factor of x² + ax – 6 = 0 is (x –1) and x² – 9x + b = 0. Find the value among the given options which is equal to a + b.

A
13

B
15

C
11

D
10

Solution

x² + ax – 6 = 0
1² + a(1) – 6 = 0
a = 5

x² – 9x + b = 0
1² – 9(1) + b = 0
b = 8
a + b = 5 + 8 = 13
Question 7
1 / -0

The product of x²y³and is equal to the quotient obtained when x² is divided by one of the following options. Choose the correct answer.

A
0

B
1

C
x

D

Solution

x²y³x = x³x2 = x³
Question 8
1 / -0

If 3x - 7y = 10 and xy = - 2, what is the value of 9x² + 49y²?

A
16

B
142

C
104

D
- 104

Solution

3x – 7y = 10…..(1) , xy = – 2 …. (2)
Squaring both sides of (1), we get:
= (10)²
9x² + 49y² – 2 3x 7y = 100
9x² + 49y² – 42xy = 100 ( xy = – 2)
9x² + 49y² – 42 (– 2) = 100
9x² + 49y² + 84 = 100
9x² + 49y² = 100 – 84
9x² + 49y² = 16
Question 9
1 / -0

If p = (2 - a), then a³ + 6 ap + p³ - 8 is

A
3

B
2

C
1

D
0

Solution

a + P + (- 2) = 0
a³ + p³ + (- 2)³ = 3 (a) (p) (-2)
a³ + p³ - 8 + 6ap
Question 10
1 / -0

If a + b+ c = 3x, find the value of (x - a)³ + (x - b)³ + (x - c)³ - 3 (x - a) (x - b) (x - c)

A
3

B
2

C
1

D
0

Solution

a + b+ c - 3x = 0 or
3x - a- b - c = 0
(x - a) + (x - b) + (x - c) = 0
(x - a)³ + (x - b)³ + (x - c)³
- 3 (x - a) (x - b) (x - c) = 0
Question 11
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 12
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 13
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 14
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 15
1 / -0

_If_{= 0, then}

A
P³ + Q³ + R³ = 0

B
P + Q + R = 27PQR

C
(P + Q + R)³ = 27PQR

D
P³ + Q³ - R³ = 27PQR

Solution

If a + b + c = 0, then a³+ b³+ c³ = 3abc

So, putting a, b and c as P^1/3, Q^1/3 and R^1/3, respectively we get
P + Q + R = 3P^1/3Q^1/3R^1/3
Taking cube on both sides, we get
(P + Q + R)³ = 27PQR
Question 16
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 17
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)
Question 18
1 / -0

Evaluate

A
0.52

B
1

C
0.01

D
0.1

Solution

=
= 0.76 + 0.24 = 1
Question 19
1 / -0

If a + b + c = 9 and ab + bc + ca = 26, then the value of a³ + b³ + c³ - 3abc is

A
27

B
29

C
495

D
729

Solution

a + b + c = 9 â¦. (1), ab + bc + ca = 26 â¦.(2)
(a³ + b³ + c³- 3abc) = (a + b + c)
= (9)
= 9 (81 - 78)
a³ + b³ + c³ - 3abc = 9 3 = 27
Question 20
1 / -0

If the polynomials 2x³ + ax² + 3x – 5 and x³ + x² – 4x + a leave the same remainder when divided by x - 2, then the value of a is

A

B

C

D

Solution

Using remainder theorem,
2(2)³ + a(2)² + 3(2) – 5 = (2)³ + (2)² – 4(2) + a
3a = –13 a = -