Polynomials Test

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Polynomials Test - 7

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Weekly Quiz Competition

Question 1
1 / -0

What is the value of 'a' for which (z + 2) is a factor of the polynomial -2z⁴ - 7z³ - 3z² - az - 10?

A
4

B
-3

C
-1

D
1

Solution

Let p(z) = -2z⁴ - 7z³ - 3z² - az - 10
Since (z + 2) is a factor of the polynomial p(z);
p(-2) = 0
-2(-2)⁴ - 7(-2)³ - 3(-2)² - a(-2) - 10 = 0
-2(16) - 7(-8) - 3(4) - a(-2) - 10 = 0
-32 + 56 - 12 + 2a - 10 = 0
2a + 2 = 0
a = -1
Question 2
1 / -0

a⁸- b⁸ equals

A
(a + b)(a - b)(a² + b²)(a⁴+ b⁴)

B
(a - b)(a - b)(a² + b²)(a⁴+ b⁴)

C
(a + b)(a + b)(a² + b²)(a⁴+ b⁴)

D
(a + b)(a - b)(a² + b²)(a⁴- b⁴)

Solution

a⁸ - b⁸
= (a⁴)² - (b⁴)²
= (a⁴ - b⁴)(a⁴ + b⁴)
= [(a²)² - (b²)²)](a⁴+ b⁴)
= (a² - b²)(a² + b²)(a⁴+ b⁴)
= (a + b)(a - b)(a² + b²)(a⁴+ b⁴)
Question 3
1 / -0

What is the remainder when a³ - b³ is divided by (a - b)?

A
a² - b²

B
0

C
a + b

D
a - b

Solution

Let f(a) = a³ - b³
We know that when f(a) is divided by a - b, then by remainder theorem, the remainder will be f(b).
So, f(b) = b³ - b³ = 0
Question 4
1 / -0

If a = , b = and c = , then the value of is

A

B

C

D

Solution

Given: a = ,b = ,c =

Now, =
=
And = =

So, =
=
Question 5
1 / -0

What is the remainder when p(y) = y³ + ay² - 2y + a is divided by (y + a)?

A
-a

B
2a

C
a

D
3a

Solution

We have p(y) = y³ + ay² - 2y + a
From Remainder Theorem, we know that when p(y) is divided by (y + a), the remainder is p(-a).
Now, p(-a) = (-a)³ + a(-a)² - 2(-a) + a = -a³ + a³ + 2a + a = 3a
Question 6
1 / -0

If 3x³ - ax² - 2bx - 7 leaves a remainder of 5 when divided by (x - 2), then which of the following is the value of a + b?

A
3

B
4

C
-3

D
1

Solution

Let p(x) = 3x³ - ax² - 2bx - 7 and g(x) = x – 2
On division of p(x) by g(x), remainder = 5; this implies that p(x) - 5 is divisible by g(x).
p(2) - 5 = 0 (By Remainder Theorem)
3(2)³ - a(2)² - 2b(2) - 7 = 5
3 × 8 - 4a - 4b - 7 = 5
24 - 4(a + b) - 7 = 5
17 - 4(a + b) = 5
-4(a + b) = 5 - 17
a + b = 3
Question 7
1 / -0

What is the value of x + y if (a + 1) and (a + 2) are the factors of p(a) = xa³ + a² - 2a + y?

A
-1/7

B
10/7

C
0

D
-11/7

Solution

Let p(a) = xa³ + a² - 2a + y
Since (a + 1) and (a + 2) are the factors of p(a);
p(-1) = 0 and p(-2) = 0
Therefore, p(-1) = x(-1)³ + (-1)² - 2(-1) + y = 0
-x + 1 + 2 + y = 0
x - y = 3 ... (i)
Again, p(-2) =
-8x + 4 + 4 + y = 0
-8x + y = -8 ... (ii)
On solving equations (i) and (ii), we get x = 5/7 and y = -16/7.
Now, x + y = (5/7) + (-16/7) = -11/7
Question 8
1 / -0

Let x and y be real numbers such that (x² – y²)(x² – 2xy + y²) = 3 and x – y = 1. What is the value of xy?

A
2

B
1 +

C
1 –

D
1

Solution

(x² – y²)(x² – 2xy + y²) = 3
Or, (x – y)(x + y)(x² – 2xy + y²) = 3
Or, (x – y)(x + y)(x – y)² = 3
x + y = 3
We have, x + y = 3 and x – y = 1
Adding two, we get 2x = 4
i.e. x = 2
and y = 1
Value of xy = 2
Question 9
1 / -0

What will be value of 'l' if (x - l) is a common factor of f(x) = x² + ax + b and g(x) = x² + cx + d?

A

B
(c + d)

C

D
(a - c)

Solution

Since (x - l) is a common factor of f(x) = x² + ax + b and g(x) = x² + cx + d;
f(l) = 0 and g(l) = 0
l² + la + b = 0 and l² + lc + d = 0
l² = - (la + b) ... (i)
l² = -(lc + d) ... (ii)
From (i) and (ii),
la + b = lc + d
la - lc = d - b
l =
Question 10
1 / -0

What will be the value of (a² + 9b²) (a + 3b) (a - 3b)?

A
a⁴ + 9b⁴

B
a⁴ - 81b⁴

C
a³ - 27b³

D
a² - 9b²

Solution

(a² + 9b²)(a + 3b)(a - 3b)
= (a² + 9b²)(a² - 9b²)
= (a²)² - (9b²)²
= a⁴ - 81b⁴
Question 11
1 / -0

If 4t³ - at² + 7 and t² + at + 8 leave the same remainder when divided by (t - 1), then what is the value of a?

A
0

B
1

C
-3

D
2

Solution

Let f(t) = and g(t) = t² + at + 8
Since f(t) and g(t) leave the same remainder when divided by (t - 1);
f(1) = g(1)
4 - a + 7 = 1 + a + 8
11 - 9 = 2a
2 = 2a
a = 1
Question 12
1 / -0

What is the value of k if = (x - 1) + ?

A
-32

B
-22

C
-10

D
-25

Solution

= (x - 1) +

(x² - 6x - 27) = ((x - 5)(x - 1) + k)
(x² - 6x - 27) = x² - 6x + 5 + k
-27 - 5 = k
-32 = k
k = -32
Question 13
1 / -0

Find the remainder when the expression 2x³ + 3x² - 14x – 20 is divided by (x – 4).

A
150

B
76

C
200

D
100

Solution

Let p(x) = 2x³ + 3x² - 14x - 20
When the expression p(x) is divided by (x - 4), then by Remainder Theorem, the remainder will be p(4).
So, the remainder is:
p(4) = 2(4)³ + 3(4)² - 14(4) - 20
= 128 + 48 - 56 - 20
= 176 - 76
= 100
Question 14
1 / -0

Find the value of k if (x + 2) is a factor of .

A
-2

B
2

C
-3

D
3

Solution

We have

Also, we know that if (x + 2) is a factor of the given expression, then on putting the value of x = -2, the expression will reduce to 0.

On putting the values and solving for the value of k, we get

= 0

-8 + 12 - 6 + k = 0

-14 + 12 + k = 0

-2 + k = 0

k = 2
Question 15
1 / -0

The length, breadth and height of a cuboidal tank are (2p + q) cm, (2p – q) cm, and (4p²+ q²) cm, respectively. Find the volume of the tank.

A
(16p² – q²) cm³

B
(8p⁴– 4q²) cm³

C
(16p⁴– q⁴) cm³

D
(16q⁴– p⁴) cm³

Solution

Volume of the cuboid = Length × breadth × height
= (2p + q) × (2p - q) × (4p² + q²)
= (4p² – 2pq + 2pq – q²)(4p² + q²)
= (4p² - q²) × (4p² + q²)
= (16p⁴- q⁴) cm³
Question 16
1 / -0

The length and breadth of a rectangle are (4x + 3) m and (2x – 5) m, respectively. What will be the area of the rectangle?

A
(16x² – 28x + 30) m²

B
(8x² – 14x – 15) m²

C
(8x²+ 14x + 15) m²

D
(4x² + 20x + 15) m²

Solution

Area of the rectangle = Length × breadth
= (4x + 3) × (2x – 5)
= 8x² – 20x + 6x – 15
= (8x² – 14x – 15) m²
Question 17
1 / -0

What must be added to x⁴ + 2x³ – 2x² + x – 1, so that the result is exactly divisible by x² + 2x – 3?

A
– x + 2

B
x + 2

C
x – 2

D
– x – 2

Solution

Let f(x) = x⁴ + 2x³ - 2x² + x - 1 is divided by x²+ 2x - 3.
x⁴+ 2x³ - 2x² + x -1 = {(x² + 2x -3) (x²+ 1)} - x + 2
i.e. we get remainder a= -x + 2.
Hence, x - 2 must be added to x⁴ + 2x³ - 2x² + x - 1, so that the result is exactly divisible by x² + 2x - 3.
i.e x - 2 should be added.
Question 18
1 / -0

^{A farmer has to pack x³- 4x² + 2x + 5 apples in trays. Each tray can hold (x – 2) apples. How many apples would remain in the end after filling the trays?}

A
4 apples

B
5 apples

C
9 apples

D
1 apple

Solution

To find the number of apples that are left in the end, we have to divide the total number of apples with the number of apples in each tray.

So, in the end, there would be 1 apple left after filling (x²- 2x - 2) trays.
Question 19
1 / -0

Rahul has Rs. (x³ + 5x² + px + q). With this money he can buy exactly (x - 2) balls or (x + 2) gloves with no money left. What would be the value of 'q'?

A
-20

B
-6

C
10

D
12

Solution

Amount of money Rahul has = Rs. (x³ + 5x²+ px + q)

Now, he can buy exactly(x - 2) balls or (x + 2) gloves.

Therefore,(x - 2) and (x + 2) are the factors of (x³ + 5x² + px + q).

Therefore, (2)³ + 5(2)² + 2p + q = 0

8 + 20 + 2p + q = 0

2p + q = -28 --- (1)

And (-2)³ + 5(-2)² - 2p + q = 0

-8 + 20 - 2p + q = 0

-2p + q = -12 --- (2)

On adding (1) and (2), we get

2q = -40

q = -20
Question 20
1 / -0

If the polynomial x³ - px² - 4x + 12 leaves the same remainder on dividing by (x - 3) and (x - 2), find the value of 'p'.

A
1

B
2

C
3

D
4

Solution

From Remainder Theorem,
At x = 2, the remainder is 8 - 4p - 8 + 12 = 12 - 4p.
At x = 3, the remainder is 27 - 9p - 12 + 12 = 27 - 9p.
As the remainders are same;
12 - 4p = 27 - 9p
5p = 15
p = 3
Question 21
1 / -0

R₁ and R₂ are remainders when x³ + 2x² - 5ax - 7 and x³ + ax² - 12x + 6 are divided by (x + 1) and (x) - 1, respectively. If 2R₁ + R₂ = 6, find the value of a.

A
23/12

B
23/11

C
23/13

D
None of these

Solution

Since (x + 1) and (x - 1) are the factors of x³ + 2x² - 5ax - 7 and x³ + ax² - 12x + 6, we will put x = -1 and x =1 in the equations to get values the remainders R₁ and R₂ respectively.

(-1)³ + 2(-1)² - 5a(-1) - 7 = R₁R₁ = - 6 + 5a
(1)³ + a(1)² - 12(1) + 6 = R₂R₂ = -5 + a

2R₁ + R₂ = 6
2(5a - 6) + (a - 5) = 6
a = 23/11
Hence, option 2 is correct.
Question 22
1 / -0

PART (i) REJECTED

Which of the following statements is/are correct?

(i) If y = , then the value of y² - 4y + 1 is 0.

(ii) The degree of the zero polynomial is zero.

(iii) If m² + n² + 2m + 1 = 0, then the value of m³¹ + n³⁵ is 1.

A
Only statement (i) is correct.

B
Statements (i) and (ii) are correct.

C
Statements (ii) and (iii) are correct.

D
None of the statements is correct.

Solution

(i) Given: y =

On rationalising, we get y = 2 +
y² - 4y + 1
= (2 + )² - 4(2 + ) + 1
= 4 + 3 + 4 - 8 - 4 + 1
= 0

(ii) The degree of the zero polynomial is not defined.
So, this statement is false.

(iii) m² + 2m + 1 + n² = 0
(m + 1)² + n² = 0
m + 1 = 0, n = 0
m = -1, n = 0
m³¹ + n³⁵ = (-1)³¹ + (0)³⁵= -1

Question 23

1 / -0

PART (A) REJECTED

Match the following:

Column - I	Column - II
(A) If f(x) = , then the value of f(-3) is	(i) 0
(B) If h(x) = , then equals	(ii) -110
(C) If p(x) = x²⁰¹ + 5, then {p(8) + p(-8)} is equal to	(iii) 1
(D) is a polynomial of degree	(iv) -10

(A) = (i), (B) = (iii), (C) = (iv), (D) = (ii)

(A) = (ii), (B) = (iv), (C) = (i), (D) = (iii)

(A) = (ii), (B) = (iv), (C) = (iii), (D) = (i)

(A) = (ii), (B) = (i), (C) = (iii), (D) = (iv)

Solution

(A) f(x) =
On putting in the value of the function, we get

Taking 5 common from each,

Solving further,

= 5(-22)

= -110

(B) Given: h(x) =

Put x = 3/2 = 1.5 in the above expression.

h(3/2) =

=

= -13.5 + 10.5 - 7

= -10

(C)

p(8) = (8)²⁰¹ + 5

p(-8) = (-8)²⁰¹+ 5 = -8²⁰¹ + 5

p(8) + p(-8) = 10

So,

(D) is a polynomial of degree 0.

Question 24
1 / -0

If , find the value of ^.

A
527

B
227

C
427

D
327

Solution

Squaring both sides,

Squaring both sides again,
Question 25
1 / -0

Directions: Consider the two statements and choose the correct option accordingly.

Statement 1: If (x + 6) and (x – 4) are factors of the expression 2x³+ 2ax²+ 2bx – 24, then the value of 'b' is -23.
Statement 2: One of the factors of the expression (2x³– 134x + 252) is (2x - 14). On factorising the expression, the other two factors are (x – 2) and (x + 9).

A
Both the statements are false.

B
Both the statements are true.

C
Statement 1 is true and statement 2 is false.

D
Statement 1 is false and statement 2 is true.

Solution

Statement 1:
(x + 6) and (x – 4) are factors of the expression 2x³ + 2ax²+ 2bx – 24.
f(x) = 2x³ + 2ax²+ 2bx – 24
Put x = 4 in the equation.
f(4) = 2(4)³+ 2a(4)² + 2b(4) – 24 = 0
128 + 32a + 8b – 24 = 0
32a + 8b + 104 = 0
4a + b + 13 = 0 ... (I)

Now, put x = -6 in the given equation.
2(-6)³+ 2a(-6)² + 2b(-6) – 24 = 0
-432 + 72a - 12b - 24 = 0
-36 + 6a - b – 2 = 0
6a – b – 38 = 0 ... (II)

Adding equations (I) and (II),
4a + b +13 + 6a – b – 38 = 0
10a – 25 = 0
2a – 5 = 0, a =
Put the value of 'a' in equation (I).
4a + b + 13 = 0
4(2.5) + b +13 = 0
10 + b + 13 = 0
b = -23
So, statement 1 is true.

Statement 2:
2x³ – 134x + 252 ... (i)
As we know, (2x – 14) is a factor and we need to find two more factors for the expression, we will break 134x in such a way that it is easy to factorise.
134x = 98x + 36x
Also, adding and subtracting 14x²in the expression (i),
2x³ + 14x²–14x²– 98x - 36x + 252
= (2x³ –14x²) + (14x²– 98x) – (36x – 252)
= x²(2x – 14) + 7x(2x – 14) – 18(2x – 14)
= (2x – 14)(x²+ 7x – 18)
= (2x – 14)(x² + 9x – 2x – 18)
= (2x – 14)[x(x + 9) - 2(x + 9)]
= (2x – 14)(x – 2)(x + 9)

So, both the statements are true.