Polynomials Test

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Polynomials Test - 6

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Weekly Quiz Competition

Question 1
1 / -0

The value of m for which (x + 3) is a factor of (x + 2)⁶ – (3x + m)⁴ is ____.

A
-8

B
10

C
12

D
16

Solution

Let f(x) = (x + 2)⁶ - (3x + m)⁴
Since (x + 3) is a factor of f(x), by factor theorem:
f(-3) = 0
(-3 + 2)⁶ - [3(-3) + m]⁴ = 0(-1)⁶ - (-9 + m)⁴= 0
1 - (-9 + m)⁴ = 0 ±1 = -9 + m
m = 10 or 8
Therefore, only option 2 is the correct answer.
Question 2
1 / -0

The remainder obtained when x⁶ - y⁶ is divided by x + y is _____.

A
0

B
x⁶ + y⁶

C
x⁶ - y⁶

D
1

Solution

Let f(x) = x⁶ - y⁶
g(x) = x + y
x⁶ - y⁶ = (x +y)(x - y)(x⁴+ y⁴ + x²y²)
It is clear that g(x) is a factor of f(x).
So, the remainder is 0.
Question 3
1 / -0

When p(x) = x ³- ax² + 3x – a is divided by (x – a), the remainder obtained is _____.

A
-3a

B
-2a

C
2a

D
3a

Solution

We have:
p(x) = x³ - ax² + 3x – a

By remainder theorem, we know that p(x) is divided by (x – a), and the remainder is p(a).
Now, p(a) = a³- aa²+ 3a – a
Or, p(a) = a³- a³+ 2a
Or, p(a) = 2a
Option 3 is correct.
Question 4
1 / -0

x¹⁸– y¹⁸can also be written as

A
[(x + y) (x² - xy + y²) (x⁶ + x³y³ + y⁶)] [(x + y) (x² – xy + y²) (x⁶ - x³y³+ y⁶)]

B
[(x – y) (x² + xy + y²) (x⁶ + x³y³ + y⁶)] [(x + y) (x² – xy + y²) (x⁶ - x³y³ + y⁶)]

C
[(x – y) (x² + xy - y²) (x⁶ + x³y³ + y⁶)] [(x + y) (x² – xy + y²) (x⁶ - x³y³ - y⁶)]

D
[(x – y) (x² + xy + y²) (x⁶ + x³y³ + y⁶)] [(x - y) (x² – xy + y²) (x⁶ + x³y³ + y⁶)]

Solution

x¹⁸ – y¹⁸
= (x⁹)² – (y⁹)²
= (x⁹ – y⁹) (x⁹ + y⁹)
= [(x³)³ – (y³)³] [(x³)³ + (y³)³]
= [(x³ – y³) (x⁶ + x³y³ + y⁶)] [(x³ + y³) (x⁶ - x³y³ + y⁶)]
= [(x – y) (x² + xy + y²) (x⁶ + x³y³ + y⁶)] [(x + y) (x² – xy + y²) (x⁶ - x³y³ + y⁶)]
Question 5
1 / -0

If m = , n = , o = , then the value of is ______.

A
-3

B
-2

C
-1

D
1

Solution

We have:

m = , n = , o =

Now, 1 + m = 1 +

On simplifying, we get:

1 + m =

Similarly, 1 + n =

1 + o =

And 1 – m = 1 -

On simplifying, we get

1 – m =

1 – n =

1 – o =

Putting the values, we get:

=

= -1
Question 6
1 / -0

If (x + 3) and (x + 4) are the factors of (2x³+ 5x²+ ax + b), then find the values of a and b.

A
-27, -102

B
-39, -108

C
-39, 108

D
-67, -111

Solution

Let f(x) = (2x³ + 5x²+ ax + b)
Here, (x + 3) and (x + 4) are the factors of f(x).
Therefore, by factor theorem:
f(-3) = 0 and f(-4) = 0
(2(-3)³ + 5(-3)²+ a(-3) + b) = 0
And (2(-4)³ + 5(-4)²+ a(-4) + b) = 0
(-54 + 45 - 3a + b) = 0 and (-128 + 80 – 4a + b) = 0
(-9 – 3a + b) = 0
b – 3a = 9 ... (i)
and
(-48 + b – 4a) = 0
b - 4a = 48 .... (ii)

Solving further:
b – 3a – b + 4a = 9 – 48
a = -39
Putting (a) in (i), we get:
b – 3(-39) = 9
b + 117 = 9 – 117 = -108
Question 7
1 / -0

x = 3 is a solution of x³+ 2x² – 9x – 18 = 0. What is/are the other solution(s)?

A
x = 2, 3

B
x = -2, -3

C
x = -2, 3

D
x = 2, -3

Solution

Let f(x) = x³ + 2x² – 9x – 18, g(x) = x – 3

By long division method, we get:

(x – 3) (x²+ 5x + 6) = 0
For other solutions, (x²+ 5x + 6) = 0
x²+ 2x + 3x + 6 = 0
Or, (x + 2) (x + 3) = 0
Or, x = -2 or -3
Question 8
1 / -0

If (x + m) is a common factor of f(x) = (x² - ax - b) and g(x) = (x²+ px - q), then m is equal to

A

B

C

D

Solution

(x + m) is a common factor of f(x) = (x² - ax - b) and g(x) = (x²+ px – q), so f(-m) = 0 and g(-m) = 0.
(m² + a(-m) - b) = 0 and (m² + p(-m) - q) = 0

Or, -am - b = -pm - q

Or, -am + pm = -q + b

m = =
Question 9
1 / -0

The product of polynomials [(x + y)² - 2xy] and (x⁴ + y⁴ – x²y²) is equal to __________.

A
x⁶ - y⁶

B
x⁶ + y⁶

C
x³ + y³

D
x³ - y³

Solution

[(x + y)² - 2xy] (x⁴ + y⁴ - x²y²)
Now, we know:
[(x + y)² - 2xy] = (x² + y²)
(x² + y²) (x⁴ + y⁴ – x²y²) = [x²(x⁴ + y⁴ – x²y²)] + [y²(x⁴ + y⁴ – x²y²)]
x⁶ + x²y⁴- x²y⁴+ x⁴y² + y⁶ + x²y⁴=x⁶+y⁶
Question 10
1 / -0

If (x - m)³ + (x - n)³ + (x - o)³ = 3(x - m) (x - n) (x - o), then what is the value of 3x - m - n - o?

A
-1

B
0

C
2

D
3

Solution

We have:
(x - m)³ + (x - n)³ + (x - o)³ = 3 (x - m) (x - n) (x - o)
Here, x - m = a
x - n = b
x - o = c
Using the identity a³+ b³+ c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ca):
If a + b + c = 0, then a³ + b³ + c³= 3abc. ,
(x - m) + (x - n) + (x – o) = 0
3x - m - n - o = 0
Question 11
1 / -0

If , then what is the value of R?

A
-2

B
-1

C
0

D
1

Solution

48 - 49 + a² - a² + 14a - 14a = 2R - R

R = -1
Question 12
1 / -0

When x³ - 3x² + gx - h is divided by x² - x - 12, the remainder obtained is 0. What is the value of ?

A
6

B
4

C
3

D
2

Solution

f(x) = x³ - 3x² + gx - h
x² - x - 12 = (x + 3)(x – 4)
Thus, f(-3) = 0 and f(4) = 0
f(-3) = (-3)³ - 3(-3)² + g(-3) - h = 0
-27 - 27 - 3g - h = 0
3g + h = -54 ... (i)
f(4) = (4)³ - 3(4)² + g(4) - h = 0
4g – h = -16 ... (ii)
Adding equations (i) and (ii), we get: g = -10
Thus, using g = -10 in equation (i), we get h = -24.
= -2 + 4 = 2
Question 13
1 / -0

What will be the remainder obtained when the expression x³ + 7x² + 14x + 14 is divided by x + 4?

A
-6

B
0

C
6

D
8

Solution

Using remainder theorem:
f(x) = x³ + 7x² + 14x + 14
When f(x) is divided by (x + 4), the remainder obtained will be f(-4).
f(-4) = (-4)³ + 7(-4)² + 14(-4) + 14
= -64 + 112 - 56 + 14
= 6
Question 14
1 / -0

Which of the following is true if x² - 25 is a factor of px⁴ + qx³ + rx² + sx + t?

A
25q + s = 0

B
25s + q = 0

C
625p + 25r + 6s = 0

D
625p + 25r - t = 0

Solution

f(x) = px⁴ + qx³ + rx² + sx + t

x² - 25 = (x – 5) (x + 5)

f(-5) = f(5) = 0

f(-5) = p(-5)⁴ + q(-5)³ + r(-5)² + s(-5) + t

= 625p - 125q + 25r - 5s + t (Equation I)

f(5) = p(5)⁴ + q(5)³ + r(5)² + s(5) + t

= 625p + 125q + 25r + 5s + t (Equation II)

Equating equations (I) and (II), we get:

625p - 125q + 25r - 5s + t = 625p + 125q + 25r + 5s + t

250q + 10s = 0

25q + s = 0
Question 15
1 / -0

If and a + b + c = 3, then find the value of a² + b² + c².

A
3

B
6

C
9

D
12

Solution

Upon simplifying, we get:
= 0
2ab + 2bc + 2ca = 0
According to an algebraic identity:
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
3² = a² + b² + c² + 0
a² + b² + c² = 9
Question 16
1 / -0

There are (3a + 4b) biscuits in a pack and (4b - 3a) packs in a box. Now, if there are 16b² + 9a² boxes in a store, then find the total number of biscuits in the store.

A
256b⁴ + 81a⁴

B
256b⁴ - 81a⁴

C
256a² + 81b²

D
256a² - 81b²

Solution

Total biscuits in a pack = (4b + 3a)
Total biscuits in a box = (4b - 3a)(4b + 3a) = 16b² - 9a²(Using identity: (x + y) (x - y) = x² - y²)
Total biscuits in the store = (16b² + 9a²) (16b² - 9a²) = 256b⁴ - 81a⁴(Using identity: (x + y) (x - y) = x² - y²)
Question 17
1 / -0

The area of a triangular garden is (3x² + 2.5x - 3) m². Which among the following can be a possible combination of height and base of the triangle?

A
(2x + 3), (3x - 2)

B
(2x – 3), (3x + 2)

C
(x – 3), (x + 2)

D
(2x – 1), (2x + 3)

Solution

Area = × Base × Height

Base × Height = 2 (Area)
= 2(3x² + 2.5x – 3)
= 6x² + 5x – 6

On factorising, we get:
6x² - 4x + 9x – 6
= 2x(3x - 2) + 3(3x - 2)
= (2x + 3) (3x - 2)

So, the possible values of height and base are (2x + 3) and (3x - 2).
Question 18
1 / -0

Devinder had x³ - 2x² - 12x + 16 candies. He distributed them equally among (x – 4) friends. If he gave each friend the maximum possible number of candies while dividing equally among them, then find the number of leftover candies.

A
4

B
0

C
28

D
54

Solution

f(x) = x³ - 2x² - 12x + 16
g(x) = x – 4
The remainder obtained will give us the number of candies left over.
Using remainder theorem:
Remainder = f(4)
f(4) = (4)³ - 2 (4)² - 12 (4) + 16
= 64 - 32 - 48 + 16
= 0
Question 19
1 / -0

Side of an equilateral triangle ism. Length and width of a rectangle are (x²+ 4) and (x²- 4). Find the sum of the areas of the equilateral triangle and the rectangle.

A
2x⁴ + 4x² - 12

B
2x⁴ + 4x² + 12

C
2x⁴ + 4x + 12

D
2x⁴ + 4x - 12

Solution

Side of equilateral triangle = m
Area =
Area of equilateral triangle =
Area of equilateral triangle = (x²+ 2)²= (x⁴ + 4x² + 4) m²

Area of rectangle = Length × Width
= (x²+ 4) (x²– 4)
= (x⁴ – 16) m²

Sum of areas = (x⁴ + 4x² + 4) + (x⁴ – 16) = 2x⁴ + 4x² - 12
Question 20
1 / -0

Kavya has (x³ + x² - 2ax - 2b) sweets. In order for no sweets to be left over, she can equally share them among either (x – 2) children or (x + 1) children. If x = 5, then find the number of sweets she has.

A
78

B
100

C
126

D
140

Solution

f(x) = x³ + x²- 2ax - 2b
In case no sweet is left over, the remainder is 0.
f(2) = 2³ + 2²- 4a - 2b = 0
2a + b = 6 ...(i)

Also,
f(-1) = (-1)³ + (-1)² + 2a - 2b = 0
a = b ...(ii)

Solving (i) and (ii), we get:
a = b = 2

If x = 5, then
f(5) = 5³ + 5²- 2(2)(5) - 2(2)
= 125 + 25 – 20 – 4 = 126

She has 126 sweets.
Question 21
1 / -0

Which of the following statements is INCORRECT?

A
p(x) is the remainder obtained when polynomial p(y) is divided by y – x.

B
y – b is a factor of polynomial p(y), if b is a zero of p(y).

C
Every real number is a zero of the polynomial.

D
A term with no variable in polynomial is called a constant.

Solution

1. p(x) is the remainder obtained when polynomial p(y) is divided by y – x.

Explanation: Based on remainder theorem, here p(y) is an unnamed polynomial and y - x is a linear factor.
y - x = 0
So, y = x
p(y) = p(x)
So, the remainder obtained is p(x).

2. y – b is a factor of polynomial p(y), if b is a zero of p(y).

Explanation: Using the factor theorem, p(y) is a polynomial of degree n > 1, b is any real number and p(b) = 0, then y - b is a factor of p(y).

3. Every real number is a zero of the polynomial.

Explanation: Every polynomial has distinct roots. Every real number cannot be a root of every polynomial.

4. A term with no letter in polynomial is called a constant.

Explanation: It is because it does not contain any modifiable variables in which values can be substituted to get a new answer.
Question 22
1 / -0

If (4x² + 6x + 1)² – (3x² + 2x + 3)² is divided by (x² + x + 1), then what will be the quotient and the remainder obtained, respectively?

A
(7x² + 29x -14) and (-15x + 6)

B
(7x² - 29x + 14) and (-15x + 6)

C
(7x² + 29x - 14) and (15x + 6)

D
(7x² + 29x + 14) and (15x - 6)

Solution

Expanding the expression, we get:
(4x² + 6x + 1)² – (3x² + 2x + 3)²
= (16x⁴ + 36x²+ 1 + 48x³ + 12x + 8x²) – (9x⁴ + 4x²+ 9 + 12x³+ 12x + 18x²)
= (7x⁴ + 36x³+ 22x²– 8)

Thus, the quotient is 7x² + 29x - 14 and the remainder is -15x + 6.
Question 23
1 / -0

Select the correct statement(s):

(A) If x =, then the value ofis.
(B) Every rational number terminates as a decimal expansion.
(C) There are five real numbers between -3 and 2.
(D) When x² - 3x + 5 is divided by x - 2, the remainder obtained is 0.

A
Only (A)

B
Only (A), (B) and (C)

C
Only (A) and (D)

D
None of them

Solution

(A) x =
Simplifying:

=
x =

Also,= =

Using x = :
==

Thus, statement A is false.

(B) Some rational numbers do not terminate as decimal expansions. For example:

Thus, statement B is false.

(C) There are infinite real numbers between two numbers. Thus, statement C is false.

(D) f(x) = x² - 3x + 5
f(2) = 4 – 6 + 5 = 3

Thus, the remainder is 3, not 0.

Therefore, none of the statements is correct.

Question 24

1 / -0

Match the following:

Column - I	Column - II
(p) If f(x) = x³+ 25x²+ 9x – 16, then f(3) =	(I) 263
(q) If f(x) = 2x³+ 3x²+ 19x – 25, then f(-1) =	(II) -43
(r) If x = is a root of f(x) = 10x³- 5x²+ ax - 12, then a =	(III)
(s) If x = 1 is a root of f(x) = x¹⁵²+ 2x⁸⁷– m, then m =	(IV) 3

(p) - (II), (q) - (I), (r) - (III), (s) - (IV)

(p) - (I), (q) - (II), (r) - (III), (s) - (IV)

(p) - (I), (q) - (III), (r) - (II), (s) - (IV)

(p) - (I), (q) - (II), (r) - (IV), (s) - (III)

Solution

(p) Given: f(x) = x³ + 25x² + 9x – 16
f(3) = (3)³ + 25 × (3)² + 9 × (3) – 16
= 27 + 225 + 27 – 16 = 263
f(3) = 263

(q) Given: f(x) = 2x³ + 3x² + 19x – 25
f(-1) = 2(-1)³ + 3(-1)² + 19(-1) – 25
= -2 + 3 - 19 - 25 = -43

(r) Given: x = is a root of f(x) = 10x³ - 5x² + ax - 12.
f(x) = 10x³ - 5x² + ax - 12

Upon simplifying, we get:
– + – 12 = 0

Or, a =
(s) Given: x = 1 is a root of f(x) = x¹⁵²+ 2x⁸⁷- m.
Or, f(1) = 1¹⁵²+ 2(1)⁸⁷- m
= 1 + 2 - m = 0
Or, m = 3

Question 25
1 / -0

Study the given equations:

Equation I:

Equation II:

A
Both equations I and II are true.

B
Equation I is true, while equation II is false.

C
Equation I is false, while equation II is true.

D
Both equations I and II are false.

Solution

Equation I:
LHS:

=

=

=

=

= (a + b) (a – b) = RHS

Thus, equation I is true.

Equation II:
LHS:

=

=

=

Thus, equation II is false.