Polynomials Test

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Polynomials Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

Solve for $$x$$: $$(502)^{2}$$

A
$$2,52,004$$

B
$$2,62,104$$

C
$$2,22,004$$

D
$$2,52,864$$

Solution

$$(502)^2$$
$$=(500+2)^2$$
using, $$(a+b)^2=a^2+2ab+b^2$$
$$=(500)^2+2(500)(2)+(2)^2$$
$$=250000+2000+4$$
$$=252004$$
Question 2
1 / -0

Use identities to evaluate: $$(101)^{2}$$

A
$$11,601$$

B
$$12,761$$

C
$$10,111$$

D
$$10,201$$

Solution

$$(101)^2$$
$$=(100+1)^2$$
using, $$(a+b)^2=a^2+2ab+b^2$$
$$=(100)^2+2(100)(1)+1^2$$
$$=10000+200+1$$
$$=10201$$
Question 3
1 / -0

The degree of $$5x^3+4x^2+7x$$ is

A
$$1$$

B
$$2$$

C
$$4$$

D
$$3$$

Solution

$$5x^3+4x^2+7x$$
The degree is the value of the greatest exponent of any expression (except the constant ) in the polynomial.
Here, term with greatest exponent $$= 5x^3$$
exponent of this term $$=3$$
Thus, degree $$=3$$
Question 4
1 / -0

Simplify $$a (a^2 + a + 1) + 5$$ and find its value for a = -1

A
0

B
4

C
2

D
6

Solution

$$a (a^2 + a + 1) + 5$$
$$=a^3+a^2+a+5$$
At $$a=-1$$
$$=(-1^3)+(-1^2)+(-1)+5$$
$$=-1+1-1+5$$
$$=4$$
Question 5
1 / -0

Substituting $$x= -3$$ in
$$x^2+5x +4$$ gives

A
$$-2$$

B
$$2$$

C
$$15$$

D
$$-15$$

Solution

For $$x=-3$$
$$x^2+5x+4$$
$$=(-3)^2+5(-3)+4$$
$$=9-15+4$$
$$=-2 \neq R.H.S$$
The correct statement is $$x^2+5x+4=-2$$
Question 6
1 / -0

The degree of $$5t-\sqrt 7$$ is

A
$$5$$

B
$$1$$

C
$$7$$

D
$$0$$

Solution

Clearly, the given the given expression $$5t-\sqrt7$$ is a linear polynomial, means the highest power of $$t$$ is 1.
$$\therefore$$ degree of the given polynomial is 1.
Question 7
1 / -0

Substituting $$x= -3$$ in
$$x^2-5x + 4$$

A
$$-2$$

B
$$28$$

C
$$2$$

D
$$-1$$

Solution

For $$x=-3$$
$$x^2-5x+4$$
$$=(-3)^2-5(-3)+4$$
$$=9+15+4$$
$$=28 \neq R.H.S$$
The correct statement is $$x^2+5x+4=28$$
Question 8
1 / -0

Simplify $$a (a^2 + a + 1) + 5$$ and find its value for a = 0

A
5

B
4

C
3

D
2

Solution

$$a (a^2 + a + 1) + 5$$
$$=a^3+a^2+a+5$$
At $$a=0$$
$$=0^3+0^2+0+5$$
$$=5$$
Question 9
1 / -0

Simplify $$a (a^2 + a + 1) + 5$$ and find its value for a = 1

A
6

B
5

C
8

D
7

Solution

$$a (a^2 + a + 1) + 5$$
$$=a^3+a^2+a+5$$
At $$a=1$$
$$=1^3+1^2+1+5$$
$$=1+1+1+5$$
$$=8$$
Question 10
1 / -0

Find the zero of the polynomial $$p(x)=x+5$$.

A
$$-5$$

B
$$5$$

C
$$0$$

D
$$3$$

Solution

Here, $$p(x)=x+5$$.
In order to find zero of $$p(x)$$, we equate $$p(x)$$ to $$0$$.
$$p(x)=0$$
$$x+5=0$$
$$\Rightarrow x=-5$$.
$$\therefore -5$$ is the zero of $$p(x)$$.

Therefore, option $$A$$ is correct.

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Polynomials Test - 20

Result

Polynomials Test - 20

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SHARING IS CARING

Question 1

$$2,52,004$$

$$2,62,104$$

$$2,22,004$$

$$2,52,864$$

Solution

Question 2

$$11,601$$

$$12,761$$

$$10,111$$

$$10,201$$

Solution

Question 3

$$1$$

$$2$$

$$4$$

$$3$$

Solution

Question 4

0

4

2

6

Solution

Question 5

$$-2$$

$$2$$

$$15$$

$$-15$$

Solution

Question 6

$$5$$

$$1$$

$$7$$

$$0$$

Solution

Question 7

$$-2$$

$$28$$

$$2$$

$$-1$$

Solution

Question 8

5

4

3

2

Solution

Question 9

6

5

8

7

Solution

Question 10

$$-5$$

$$5$$

$$0$$

$$3$$

Solution

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