Polynomials Test

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Polynomials Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

Find the value of the polynomial $$(3x^{ 3 })\times (4x^{ 2 })+7{ x }^{ 5}$$ when $$x=3$$.

A
$$4617$$

B
$$4613$$

C
$$4611$$

D
$$4619$$

Solution

$$3{ x }^{ 3 }(4{ x }^{ 2 })+7{ x }^{ 5 }=12{ x }^{ 5 }+7{ x }^{ 5 }=19{ x }^{ 5 }$$.

To find the value of the expression for $$x = 3$$, we have to substitute $$x = 3$$.
So, $$19(3)^{ 5 }=19(243)=4617$$
Question 2
1 / -0

If $$p(x)=x^{2}-4x+3,$$. Evaluate $$p(2)-p(-1)+p\left ( \dfrac{1}{2} \right )$$.

A
$$\dfrac{-5}{4}$$

B
$$\dfrac{ 5}{4}$$

C
$$\dfrac{-31}{4}$$

D
$$\dfrac{ 31}{4}$$

Solution

After putting $$ x=2$$ in equation, we get $$-1$$.
After putting $$ x=-1$$, we get $$8$$
After putting $$x=\dfrac12$$, we get $$\dfrac54$$
So, $$p(2)-p(-1)+p\left(\dfrac12\right) = -1-8+\dfrac{5}{4}$$
$$ = -9+\dfrac{5}{4} = \dfrac{-36+5}{4} = \dfrac{-31}{4}$$
Question 3
1 / -0

If $$p(y) = (y + 2) (y - 2)$$. Find the value of $$p(1)$$.

A
$$2$$

B
$$0$$

C
$$-3$$

D
$$3$$

Solution

$$p(y) = (y+2)(y-2)$$

Putting the value of
$$y=1$$
$$p(y) = (1+2)(1-2) = 3\times-1 = -3$$

$$\textbf{Alternate:}$$
$$p(y) = (y+2)(y-2) = y^2 - 4$$ [using $$(a-b)(a+b) = a^2-b^2$$)
$$p(1) = 1^2-4 = -3$$
Question 4
1 / -0

Zero of the zero polynomial is:

A
$$0$$

B
$$1$$

C
Any real number

D
Not defined

Solution

Zeroes of a zero polynomial can be any real number.
That is, since a zero polynomial is defined as $$p(x)=0$$ for all $$x$$ in real numbers,
we can say that, its zero can be any real number.

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

The value of the polynomial $$5x^3- 4x^2 + 3$$ when $$x = -1$$ is

A
$$-6$$

B
$$6$$

C
$$2$$

D
$$-2$$

Solution

$$5x^{ 3 }-4x^{ 2 }+3$$
If $$x= -1$$, then replace $$x$$ with $$-1$$, we get
$$5\times(-1)^{ 3 }-4(-1)^{ 2 }+3$$
$$=-5-4+3=-6$$
Question 6
1 / -0

If $$p(x) = x + 3$$, then $$p(3) + p(-3)$$, is equal to

A
$$3$$

B
$$2$$

C
$$0$$

D
$$6$$

Solution

$$p(x) = x + 3$$

For $$x = 3$$, we get
$$p(3) = 3 + 3 = 6$$

For $$x = -3$$
$$p(-3) = -3 + 3 = 0$$

So,
$$p(3) + p(-3) = 6 + 0 = 6$$
Question 7
1 / -0

Find $$p(0)$$ for the following polynomial :
$$p(x) =10x -4x^2-3$$

A
$$3$$

B
$$-3$$

C
$$1$$

D
$$0$$

Solution

$$p(x)=10x-4x^{ 2 }-3$$
$$\Rightarrow p(0)=10(0)-4(0)^{ 2 }-3$$
$$\Rightarrow p(0)=0-0-3$$
$$\Rightarrow p(0)=-3$$
Question 8
1 / -0

The zero polynomial has :

A
One zero

B
two zeroes

C
infinite zeroes

D
No zero

Solution

Zeroes of the polynomial mean to find those values of $$x$$ for which the value of the polynomial becomes zero.
i.e. $$p(x)=0$$
Now, we have a zero polynomial $$p(x)=0$$.
This is the zero polynomial and will be true for all values of x belonging to complex numbers.
Hence, infinite no. of solutions.
Question 9
1 / -0

If $$f(x) =2x^{2} - x + 1$$ and $$g(x) = x^{3} - 3x + 1$$, then the value of $$f(1) + g(-1)$$ is

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

$$ f(x)= { 2x }^{ 2 }-x+1\\ \therefore f(1)=2\times { 1 }^{ 2 }-1+1\\ =2\\ g(x)={ x }^{ 3 }-3x+1\\ \therefore g(-1)=(-1)^{ 3 }-3\times (-1)+1\\ =-1+3+1\\ =3\\ \therefore f(1)+g(-1)=2+3\\ =5$$
Question 10
1 / -0

Simplify: $$(l + m)^2- 4lm$$

A
$$(l -m)^2$$

B
$$4l^2m^2$$

C
$$(l+2m)^2$$

D
$$(l-2m)^2$$

Solution

$$(l+m)^2-4lm$$
$$=l^2+2lm+m^2-4lm$$
$$=l^2-2lm+m^2$$
$$=(l-m)^2$$

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Re-Attempt Weekly Quiz Competition

Polynomials Test - 27

Result

Polynomials Test - 27

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SHARING IS CARING

Question 1

$$4617$$

$$4613$$

$$4611$$

$$4619$$

Solution

Question 2

$$\dfrac{-5}{4}$$

$$\dfrac{ 5}{4}$$

$$\dfrac{-31}{4}$$

$$\dfrac{ 31}{4}$$

Solution

Question 3

$$2$$

$$0$$

$$-3$$

$$3$$

Solution

Question 4

$$0$$

$$1$$

Any real number

Not defined

Solution

Question 5

$$-6$$

$$6$$

$$2$$

$$-2$$

Solution

Question 6

$$3$$

$$2$$

$$0$$

$$6$$

Solution

Question 7

$$3$$

$$-3$$

$$1$$

$$0$$

Solution

Question 8

One zero

two zeroes

infinite zeroes

No zero

Solution

Question 9

$$2$$

$$3$$

$$4$$

$$5$$

Solution

Question 10

$$(l -m)^2$$

$$4l^2m^2$$

$$(l+2m)^2$$

$$(l-2m)^2$$

Solution

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