Polynomials(10th Grade > Mathematics ) Questions and answers for exam Preparation

10th Grade > Mathematics

POLYNOMIALS MCQs

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Total Questions : 79 | Page 1 of 8 pages

Question 1. When we divide

3 t^{4} + 5 t^{3} - 7 t^{2} + 2 t + 2

t^{2} + 3 t + 1

, we get

3 t^{2} - 4 t + 2

as quotient.

True
False
6x2+3x+9
2x2+3x+1

Discuss Question

Answer: Option A. -> True
:
A
Lets divide

3 t^{4} + 5 t^{3} - 7 t^{2} + 2 t + 2 by t^{2} + 3 t + 1

3 t^{2} - 4 t + 2 t^{2} + 3 t + 1) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ 3 t^{4} + 5 t^{3} - 7 t^{2} + 2 t + 2_{-} 3 t^{4} \pm 9 t^{3} \pm 3 t^{2} - --------------- - - 4 t^{3} - 10 t^{2} + 2 t \mp 4 t^{3} \mp 12 t^{2} \mp 4 t - ----------------- - 2 t^{2} + 6 t + 2_{-} 2 t^{2} \pm 6 t \pm 2 - ------------ - 00

Thus, the given statement is true.

Question 2. Which of the following is a factor of

(x + y)^{3} - (x^{3} + y^{3})

x2−2xy+y2
xy2
3xy
x2+xy+y2

Discuss Question

Answer: Option C. -> 3xy
:
C

Let P (x) = (x + y) (x + y)^{2} - (x + y) (x^{2} - x y + y^{2})

= (x + y) (x^{2} + 2 x y + y^{2}) - (x + y) (x^{2} - x y + y^{2})

= (x + y) (x^{2} + 2 x y + y^{2} - x^{2} + x y - y^{2})

[Taking

(x + y)

as common]

= (x + y) (3 x y)

Hence, from the given options3xy is a factor ofthe given polynomial.

Question 3. The zeroes of

m n (x^{2} + 1) = (m^{2} + n^{2}) x

are:

mn and nm
−mn and nm
−mn and −nm
mn and −nm

Discuss Question

Answer: Option A. -> mn and nm
:
A
Given polynomial,

m n (x^{2} + 1) = (m^{2} + n^{2}) x

Let's factorise it.

m n x^{2} - (m^{2} + n^{2}) x + m n = 0

\Rightarrow m n x^{2} - m^{2} x - n^{2} x + m n = 0

\Rightarrow m x (n x - m) - n (n x - m) = 0

\Rightarrow (m x - n) (n x - m) = 0

\Rightarrow m x - n = 0 o r n x - m = 0

\Rightarrow x = \frac{m}{n} o r \frac{n}{m}

So, zeroes of the given polynomial are

\frac{m}{n}

and

\frac{n}{m}

Question 4. If

α, β

and

γ

are the zeroes of the polynomial

f (x) = a x^{3} + b x^{2} + c x + d

, then

\frac{1}{α} + \frac{1}{β} + \frac{1}{γ}

is _____.

cd
ad
−cd
−ba

Discuss Question

Answer: Option C. -> −cd
:
C
Given

α, β

and

γ

are the zeroes of the polynomial

f (x) = a x^{3} + b x^{2} + c x + d

\frac{1}{α} + \frac{1}{β} + \frac{1}{γ}

= \frac{β γ + α γ + α β}{(α β γ)}

= \frac{(\frac{c}{a})}{(- \frac{d}{a})}

= - \frac{c}{d}

Question 5. If

a + b + c = 8

and

a b + b c + c a = 20

, find the value of

a^{3} + b^{3} + c^{3} - 3 a b c

Discuss Question

Answer: Option B. -> 32
:
B
Since

{(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a)

a + b + c = 8 and a b + b c + c a = 20

{(8)}^{2} = a^{2} + b^{2} + c^{2} + 2 \times 20

64 = a^{2} + b^{2} + c^{2} + 40

a^{2} + b^{2} + c^{2} = 24

We now use the following identity:

a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - (a b + b c + c a))

a^{3} + b^{3} + c^{3} - 3 a b c = 8 \times (24 - 20) = 4 \times 8 = 32

Thus,

a^{3} + b^{3} + c^{3} - 3 a b c = 32

Question 6. If

{(x}^{4} + x^{2} y + y^{2})

is one of the factors of an expression which is the difference of two cubes, then the other factor is

x2−y
x2−y2
x−y2
x−y

Discuss Question

Answer: Option A. -> x2−y
:
A
Let two numbers be a and b.
The difference of their cubes is

a^{3} - b^{3}

and hence this is the expression we need.
This expression can be factorised as

a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})

.
Now, consider the given factor,

x^{4} + x^{2} y + y^{2} = {(x^{2})}^{2} + x^{2} y + {(y)}^{2}

This is in the form of

a^{2} + a b + b^{2}

where

a = x^{2}, b = y

, which is similar to the factor given.
Therefore, the other factor is of the form

a - b

.
Hence, other factoris

x^{2} - y .

Question 7. Find the roots of the equation: x-5=0

5
-5
0
Not defined.

Discuss Question

Answer: Option A. -> 5
:
A

x - 5 = 0

signifies that

(x - 5)

is a polynomial whose value is zero for a certain value of

x

and this

x

will be the root of the given equation

∴ at x=5 the value of expression x-5 is0.

Hence, the root of the given equation is 5.

Question 8. If

a x^{n}

is a zero degree polynomial in

x

, what can be said about

a

a is negative integer always
a is 1 always
a is 0 always
a is non zero

Discuss Question

Answer: Option D. -> a is non zero
:
D
A zero degree polynomial is essentially a constant.i.e.,any non-zero real number.
So, if

a x^{n}

is a zero degree polynomial, then

a x^{n}

a x^{0}

= a
So, n has to be zero, and

a

should not be zero.

Question 9. Factorize :

x^{2} - 7 x + 12

(x-3)(x-4)
(x+3)(x-4)
(x-3)(x+4)
(x+3)(x+4)

Discuss Question

Answer: Option A. -> (x-3)(x-4)
:
A
We have to break up

7 x

in thegiven expression into two parts in such a way that their sum is

7 x

and the product of their coefficients is

12

x^{2} - 7 x + 12

= x^{2} - 4 x - 3 x + 12

= x (x - 4) - 3 (x - 4)

= (x - 3) (x - 4)

Thus

x^{2} - 7 x + 12

can be factorised as

(x - 3) (x - 4)

Question 10.

If x + \frac{1}{x} = 2, find the value of x^{64} + \frac{1}{x^{64}}

Discuss Question

Answer: Option D. -> 2
:
D

x + \frac{1}{x} = 2

Squaring both sides, weget

x^{2} + \frac{1}{x^{2}} + 2. x . \frac{1}{x} = 4

So,

x^{2} + \frac{1}{x^{2}} = 4 - 2 = 2

Again Squaring both sides, weget

x^{4} + \frac{1}{x^{4}} + 2. x^{2} . \frac{1}{x^{2}} = 4

\Rightarrow x^{4} + \frac{1}{x^{4}} = 4 - 2 = 2

Similarly,

x^{8} + \frac{1}{x^{8}} = x^{32} + \frac{1}{x^{32}} = x^{64} + \frac{1}{x^{64}} = 2

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