Polynomials Test

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Polynomials Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

Find roots of equation $$\displaystyle { a }^{ 4 }{ x }^{ 2 }-{ b }^{ 4 }=0$$

A
$$\displaystyle { { b }^{ 2 } }/{ { a }^{ 2 } },-{ { b }^{ 2 } }/{ { a }^{ 2 } }$$

B
$$\displaystyle { { b }^{ 2 } }/{ { a }^{ 2 } },{ { b }^{ 2 } }/{ { a }^{ 2 } }$$

C
$$\displaystyle { { a }^{ 2 } }/{ { { b }^{ 2 } } },\frac { { a }^{ 2 } }{ { { b }^{ 2 } } } $$

D
$$\displaystyle { { a }^{ 2 } }/{ { { b }^{ 2 } } },-\frac { { a }^{ 2 } }{ { { b }^{ 2 } } } $$

Solution

$$\displaystyle { a }^{ 4 }{ x }^{ 2 }-{ b }^{ 4 }=0$$
Since,$$(a^2-b^2)=(a-b)(a+b)$$
$$\displaystyle \left( { a }^{ 2 }x+{ b }^{ 2 } \right) \left( { a }^{ 2 }x-{ b }^{ 2 } \right) =0,{ a }^{ 2 }x={ b }^{ 2 }and \ ,{ a }^{ 2 }x=-{ b }^{ 2 }$$
$$\displaystyle x={ { b }^{ 2 } }/{ { a }^{ 2 } },x=-{ { b }^{ 2 } }/{ { a }^{ 2 } }$$
Question 2
1 / -0

$$p(x) = \dfrac{37}{15}$$ is a _______ polynomial.

A
linear

B
quadratic

C
cubic

D
none of these

Solution

The general form of a constant polynomial is $$p(x)=c$$ with constant $$c$$.

Since $$p(x)=\frac { 37 }{ 15 }$$ is a polynomial with constant term $$\frac { 37 }{ 15 }$$ and there is no variable in it.

Hence, $$p(x)=\frac { 37 }{ 15 }$$ is a constant polynomial.
Question 3
1 / -0

Is $$\displaystyle x=-10$$ is a solution of $$\displaystyle { x }^{ 2 }+5x-9=0$$?

A
Yes

B
No

C
None

D
Can't say

Solution

Put $$\displaystyle x=-10$$ in equation $$\displaystyle { x }^{ 2 }+5x-9=0$$.
$$\displaystyle { \left( -10 \right) }^{ 2 }+5\left( -10 \right) -9=100-50-9$$
$$\displaystyle =\quad 100-59=41\neq 0$$.
$$\displaystyle \therefore $$ The equation does not have $$x=-10$$, solution.
Therefore, option $$B$$ is correct.
Question 4
1 / -0

Is $$\displaystyle x=\dfrac { 5 }{ 2 } $$ a solution of $$\displaystyle \left( 6x+16 \right) \left( 4x+10 \right) =0$$?

A
Yes

B
No

C
Can't say

D
None

Solution

Put $$\displaystyle x=\dfrac { 5 }{ 2 } $$ in equation $$\displaystyle \left( 6x+16 \right) \left( 4x+10 \right) =0$$
$$\displaystyle \left( 6\times \frac { 5 }{ 2 } +16 \right) \left( 4\times \frac { 5 }{ 2 } +10 \right) =\left( \frac { 30 }{ 2 } +16 \right) \left( \frac { 20 }{ 2 } +10 \right) $$
$$\displaystyle =\left( 15+16 \right) \left( 10+10 \right) $$
$$\displaystyle =\left( 31 \right) \left( 20 \right) \neq 0$$
$$\displaystyle \therefore \quad x=\frac { 5 }{ 2 } $$ is not a solution of equation $$\displaystyle \left( 6x+16 \right) \left( 4x+10 \right)$$

Therefore, option $$B$$ is correct.
Question 5
1 / -0

Is $$\displaystyle x=-4$$ is a solution of equation $$\displaystyle { x }^{ 2 }-x-20=0$$?

A
Yes

B
No

C
Can't say

D
None

Solution

Put $$x=-4$$ in equation $$\displaystyle { x }^{ 2 }-x-20=0$$
$$\displaystyle { \left( -4 \right) }^{ 2 }-\left( -4 \right) -20=16+4-20=0$$
$$\displaystyle \therefore \quad x=-4$$ is a solution of equation $$\displaystyle { x }^{ 2 }-x-20=0$$.

Therefore, option $$A$$ is correct.
Question 6
1 / -0

Which of the following is INCORRECT?

A
$$p(x) = 5x + 5, $$ degree $$= 1$$

B
$$p(x) = 4x^4 + 4, $$ degree $$= 4$$

C
$$p(x) = x^8 , $$ degree $$= 8$$

D
$$p(x) = 9, $$ degree $$= 9$$

Solution

A polynomial is a function of the form $$f(x) = a_{n}x^{n} + a_{n−1}x^{n−1} + ... + a_{2}x^{2} + a_1x + a_0 $$
where $$a_{n}, a_{n−1} , ... a_{2} ,a_1 , a_0 $$ . are contants
and $$n$$ is a natural number.
The degree of a polynomial is the highest power of $$x$$ in its expression.
In option A, $$p(x)=5x+5$$
Highest power of $$x$$ is $$1$$
So, degree of $$p(x)$$ is $$1$$.

In option B, $$p(x)=4x^{4}+4$$
Highest power of $$x$$ is $$4$$
So, degree of $$p(x)$$ is $$4$$.

In option C, $$p(x)=x^{8}$$
Highest power of $$x$$ is $$8$$
So, degree of $$p(x)$$ is $$8$$.

Looking at Option $$D$$
$$p(x) = 9=9.x^{0}$$
Highest power of $$x$$ in $$p(x) $$ is $$=0$$
So, degree of the polynomial $$p(x)$$ is $$0$$.
Thus, it is incorrect.
Hence, the answer is option D.
Question 7
1 / -0

Is $$x=9$$ a solution of $$\displaystyle { x }^{ 2 }-63-2x=0$$?

A
Yes

B
No

C
Can't say

D
None

Solution

Put $$x=9$$ in equation $$\displaystyle { x }^{ 2 }-63-2x=0$$.
$$\displaystyle= { \left( 9 \right) }^{ 2 }-63-2\left( 9 \right) =81-63-18$$
$$\displaystyle = 81-81$$
$$\displaystyle =0$$.
$$\displaystyle \therefore x=9$$ is solution of equation $$\displaystyle { x }^{ 2 }-63-2x=0$$.
Therefore, option $$A$$ is correct.
Question 8
1 / -0

Is $$x=-1$$ a solution of $$\displaystyle { x }^{ 2 }+x+1=0$$?

A
Yes

B
No

C
Can't say

D
None

Solution

Put $$x=-1$$ in equation $$\displaystyle { x }^{ 2 }+x+1=0$$.
$$\displaystyle f(-1)= { \left( -1 \right) }^{ 2 }+\left( -1 \right) +1$$
$$\displaystyle =1-1+1\ne0$$.
$$\displaystyle \therefore \quad x=-1$$ is not a solution of $$\displaystyle { x }^{ 2 }+x+1=0$$.
Therefore, option $$B$$ is correct.
Question 9
1 / -0

$$\displaystyle x=\frac { 4 }{ 3 } $$ is a root of equation $$\displaystyle { x }^{ 2 }-\frac { 16 }{ 9 } =0$$

A
Yes

B
No

C
Can't possible

D
None

Solution

$$\displaystyle { x }^{ 2 }-\dfrac { 16 }{ 9 } =0,\\\left( x+\dfrac { 4 }{ 3 } \right) \left( x-\dfrac { 4 }{ 3 } \right) =0$$
Since,$$\displaystyle \left[ \left( a+b \right) \left( a-b \right) ={ a }^{ 2 }-{ b }^{ 2 } \right] $$
$$\displaystyle x=\dfrac { -4 }{ 3 }\ and\ \dfrac { 4 }{ 3 } $$ are the roots of $$\displaystyle { x }^{ 2 }-\dfrac { 16 }{ 9 } =0$$
Question 10
1 / -0

$$p(x) = 25 $$ is a _______ polynomial.

A
linear

B
quadratic

C
constant

D
cubic

Solution

The general form of a constant polynomial is $$p(x)=c$$ with constant $$c$$.

Since $$p(x)=25$$ is a polynomial with constant term $$25$$ and there is no variable in it.

Hence, $$p(x)=25$$ is a constant polynomial.

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Polynomials Test - 23

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Polynomials Test - 23

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Question 1

$$\displaystyle { { b }^{ 2 } }/{ { a }^{ 2 } },-{ { b }^{ 2 } }/{ { a }^{ 2 } }$$

$$\displaystyle { { b }^{ 2 } }/{ { a }^{ 2 } },{ { b }^{ 2 } }/{ { a }^{ 2 } }$$

$$\displaystyle { { a }^{ 2 } }/{ { { b }^{ 2 } } },\frac { { a }^{ 2 } }{ { { b }^{ 2 } } } $$

$$\displaystyle { { a }^{ 2 } }/{ { { b }^{ 2 } } },-\frac { { a }^{ 2 } }{ { { b }^{ 2 } } } $$

Solution

Question 2

linear

quadratic

cubic

none of these

Solution

Question 3

Yes

No

None

Can't say

Solution

Question 4

Yes

No

Can't say

None

Solution

Question 5

Yes

No

Can't say

None

Solution

Question 6

$$p(x) = 5x + 5, $$ degree $$= 1$$

$$p(x) = 4x^4 + 4, $$ degree $$= 4$$

$$p(x) = x^8 , $$ degree $$= 8$$

$$p(x) = 9, $$ degree $$= 9$$

Solution

Question 7

Yes

No

Can't say

None

Solution

Question 8

Yes

No

Can't say

None

Solution

Question 9

Yes

No

Can't possible

None

Solution

Question 10

linear

quadratic

constant

cubic

Solution

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