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

10th Grade > Mathematics

POLYNOMIALS MCQs

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Total Questions : 79 | Page 5 of 8 pages

Question 41. Factor Theorem is equivalent to Remainder Theorem when remainder is

Discuss Question

Answer: Option C. -> 0
:
C
If f(x) is a polynomial and is divided by (x-a), then if f(a) = k, this k will be the remainder, as stated by the Remainder Theorem. If the value of k is 0, then (x-a) is the factor of f(x), as indicated by the Factor Theorem. Hence, factor theorem is equivalent to remainder theorem when remainder is 0.

Question 42.

If a polynomial (x^{2} + a x - c) cuts x-axis at 2 and -4, the value of a + c is

___

Discuss Question

:
Suppose

f (x)

is thepolynomial.
Now since it cuts x-axis at 2 and -4, the roots of

f (x)

are 2 and -4.
Sum of roots =

- a = 2 - 4 = - 2

\Rightarrow a = 2

Productof roots =

- c = 2 \times - 4 = - 8

\Rightarrow c = 8

Hence,

(a + c) = 10

Question 43. If f(x) = (x - a)(x - b) , then if a and b are the zeros of the polynomial f(x), f(a) = f(b).

True
False
0,0
-5, -5

Discuss Question

Answer: Option A. -> True
:
A
Since a and b are the zeroes of the polynomial f(x), both f(a) = 0 and f(b) = 0, i.e. a and b will be the roots of the equation f(x) = 0.
Thus,f(a) = f(b) = 0. Hence, the given statement is true.

Question 44. If

(x^{2} - y^{2}) = 18

and

(x - y) = 3

, find the value of

16 x^{2} y^{2}

Discuss Question

Answer: Option D. -> 729
:
D

(x^{2} - y^{2}) = (x - y) (x + y) = 18

(x - y) = 3 - - - - (i)

(x + y) = \frac{18}{3} = 6 - - - - (i i)

Adding equation (i) and (ii)we get

2 x = 9

x = \frac{9}{2}

Substituting the value of x in equation (ii) we get

y = \frac{3}{2}

16 x^{2} y^{2} = 16 \times (\frac{9}{2})^{2} \times (\frac{3}{2})^{2} = 729

Question 45. If

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

what is the value of x^{3} + y^{3} + z^{3}

3xyz
−3xyz
2xyz
−2xyz

Discuss Question

Answer: Option A. -> 3xyz
:
A
We know that

x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) - - - (1)

According toquestion,

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

∴

RHS will be zero in equation (1)
So,

x^{3} + y^{3} + z^{3} = 3 x y z

Question 46. The value of

(2 x - 3 y)^{3} + (x + 3 y)^{3} + 3 (2 x - 3 y)^{2} (x + 3 y) + 3 (2 x - 3 y) (x + 3 y)^{2} =

27x3
29x3+(x+3y)3
(2x−3y)3+(x+3y)3
0

Discuss Question

Answer: Option A. -> 27x3
:
A
Substitute

2 x - 3 y = a

and

(x + 3 y) = b

(2 x - 3 y)^{3} + (x + 3 y)^{3} + 3 (2 x - 3 y)^{2} (x + 3 y) + 3 (2 x - 3 y) (x + 3 y)^{2}

= a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2}

= (a + b)^{3}

= (2 x - 3 y + x + 3 y)^{3}

= (3 x)^{3}

= 27 x^{3}

Question 47. In

a^{2} x^{2} + a x

, if 'a' is constant and 'x' is a variable, this expression would become a polynomial in ___ variable(s). (Write in words)

Discuss Question

:
In this expression,

a^{2} x^{2}

and

a x

are the two terms. Since it is given that 'a'is a constant and 'x' is a variable, we can see that there is only one variable 'x' in the polynomial thus making it a polynomial in one variable.

Question 48.

p (x) = x^{3} + 8 x^{2} - 7 x + 12

and

g (x) = x - 1

. If

p (x)

is divided by

g (x)

, it gives

q (x)

and

r (x)

as quotient and remainder respectively. If

a

is the degree of

q (x)

and

b

is the degree of

r (x)

(a - b) = ?

Discuss Question

Answer: Option B. -> 2
:
B
Remainder

= r (x) = p (1) = 1^{3} + 8 \times 1^{2} - 7 \times 1 + 12 = 14

The degree of a polynomial is the highest degreeof its terms when the polynomial is expressed in its canonical form consisting of a linear combination of monomials.
So, Degree of

r (x) = b = 0

q (x) = \frac{p (x)}{g (x)}

Now ,

p (x) = x^{3} + 8 x^{2} - 7 x + 12

g (x) = (x - 1)

Performing long division

x^{2} + 9 x + 2 x - 1) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ x^{3} + 8 x^{2} - 7 x + 12 x^{3} - x^{2} (-) (+) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ 9 x^{2} - 7 x 9 x^{2} - 9 x (-) (+) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ + 2 x + 12 2 x - 2 (-) (+) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ 14

∴ q (x) = \frac{p (x)}{g (x)} = x^{2} + 9 x + 2

So, degree of

q (x) = a = 2

Hence,

a - b = 2 - 0 = 2

Question 49. If

(x - 1)

is a factor of

a x^{3} - b x^{2} + c x

, what is the value of

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

2abc
-2abc
3abc
-3abc

Discuss Question

Answer: Option D. -> -3abc
:
D

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

If (x - 1) is a factor then x = 1 is the zero of polynomial
p(1) = 0

p (1) = a - b + c = 0

It can be written as

a + (- b) + c = 0

We know that if

a + b + c = 0, t h e n a^{3} + b^{3} + c^{3} = 3 a b c

So,

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

Question 50.

$Find the zeros of the quadratic polynomial x^{2} - 8 x + 12.$

$2 and 6$
$8 and 4$
$10 and 2$
$4 and 5$

Discuss Question

Answer: Option A. ->

2 and 6

:
A

To find the zeroes, equate the polynomial to zero and then factorise.
$∴ x^{2} - 8 x + 12 = 0$
$\Rightarrow x^{2} - 6 x - 2 x + 12 = 0$
$\Rightarrow x (x - 6) - 2 (x - 6) = 0$
$\Rightarrow (x - 6) (x - 2) = 0$
$\Rightarrow x - 6 = 0 or x - 2 = 0 \Rightarrow x = 6; 2$
$2 and 6 are the zeros of the quadratic polynomial.$