Polynomials Test

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Polynomials Test - 40

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Weekly Quiz Competition

Question 1
1 / -0

The value of $$(7.2)^2$$ is (use an identity to expand):

A
$$49.9$$

B
$$14.4$$

C
$$51.84$$

D
$$49.04$$

Solution

We will use, $$(a+b)^2$$= $$a^2+2ab+b^2$$ to find the required result.

Substitute $$7$$ for $$a$$ and $$0.2$$ for $$B$$ in the above identity,
$$(7+0.2)^2=(7)^2+2(7)(0.2)+(0.2)^2$$
$$=49+2.8+0.04$$
$$=51.84$$

Hence, option $$C$$ is correct.
Question 2
1 / -0

The expansion of $$(2x - 3y)^2$$ is:

A
$$2x^2+3y^2+6xy$$

B
$$4x^2+9y^2-12xy$$

C
$$2x^2+3y^2-6xy$$

D
$$4x^2+9y^2+12xy$$

Solution

Given, $$(2x-3y)^2$$.

We know, $$(a-b)^2$$ $$ =$$ $$a^2-2ab+b^2$$.

Then,
$$(2x-3y)^2$$ $$ =$$ $$(2x)^2-2 (2x)(3y)+(3y)^2$$
$$=$$ $$4x^2-12xy+9y^2$$.

Therefore, option $$B$$ is correct.
Question 3
1 / -0

$$(a + b)^{2} - (a - b)^{2}=$$ _____.

A
$$4ab$$

B
$$2ab$$

C
$$a^{2} + 2ab + b^{2}$$

D
$$2(a^{2} + b^{2})$$

Solution

Given, $$(a+b)^{2}-(a-b)^{2}$$.

We know, $$(a+b)^{2}$$ $$=\left ( a^{2}+2ab+b^{2} \right )$$
and $$(a-b)^{2}$$ $$=(a^{2}-2ab+b^{2})$$.

Then,
$$(a+b)^{2}-(a-b)^{2}$$
$$=\left ( a^{2}+2ab+b^{2} \right )-(a^{2}-2ab+b^{2})$$
$$=a^{2}+2ab+b^{2}-a^{2}+2ab-b^{2}\\=4ab$$.

Therefore, the required answer is $$4ab$$.

Hence, option $$A$$ is correct.
Question 4
1 / -0

Find the value of $$p(x) = 4x^{2} - 3x + 7$$ at $$x = 1$$.

A
$$6$$

B
$$7$$

C
$$8$$

D
$$9$$

Solution

Given polynomial is $$p(x)=4x^{2}-3x+7$$
We need to find value of polynomial at $$x=1$$.
Then put the value of $$x=1$$ in the given polynomial, we get
$$p(x)=4(1)^{2}-3(1)+7$$
$$\Rightarrow p(x)= 4\times 1-3\times 1+7$$
$$\Rightarrow p(x)= 4-3+7$$
$$\Rightarrow p(x)= 11-3=8$$
So, value of polynomial is $$8$$.
Question 5
1 / -0

The number of real zeroes of polynomial $$P(x) = x^3 - x$$ is ..............

A
2

B
1

C
0

D
3

Solution

Given $$P(x)=x^3-x$$
To find the zeros of $$P(x)$$, let $$P(x)=0$$
$$\implies x^3-x=0$$
$$\implies x(x^2-1)=0$$
$$\implies x(x-1)(x+1)=0$$
$$\implies x=0$$ and $$x-1=0$$ and $$x+1=0$$
$$\implies x=0,1,-1$$ are the zeros of $$P(x)$$
Thus, the number of zeros of $$P(x)$$ is $$3$$.
Hence, the answer is $$3$$.
Question 6
1 / -0

Find the degree of the given algebraic expression:
$$ 3x-15$$

A
$$1$$

B
$$3$$

C
$$2$$

D
$$-15$$

Solution

The given algebraic expression $$3x-15$$ has two terms. The first one is $$3x$$ and the second is $$-15$$.

The exponent of the first term is $$1$$ because $$3x = 3x^1$$

The exponent of the second term is $$0$$ because $$15 = 15x^0$$

Since the highest exponent is $$1$$, therefore, the degree of $$3x-15$$ is $$1$$.

Hence, the degree of the algebraic expression $$3x-15$$ is $$1$$.
Question 7
1 / -0

The zero of the polynomial $$2x-5$$ is:

A
$$\dfrac{5}{2}$$

B
$$-\dfrac{5}{2}$$

C
$$\dfrac{2}{5}$$

D
$$-\dfrac{2}{5}$$

Solution

Given $$p(x)=2x-5$$
To find the zero, we put, $$p(x)=0$$
$$\Rightarrow 2x-5 =0$$
$$\Rightarrow x=\dfrac52$$
Hence, $$x=\dfrac52$$ is the zero of the polynomial $$p(x)$$
Hence, option $$A$$ is correct.
Question 8
1 / -0

Find the degree of the given algebraic expression $$ax^2+bx+c$$.

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

We know that the degree is the term with the greatest exponent.

Since the highest exponent is $$2$$, therefore, the degree of $$ax^2+bx+c$$ is $$2$$.

Hence, the degree of the algebraic expression $$ax^2+bx+c$$ is $$2$$.
Question 9
1 / -0

Find the product of $$\left(\dfrac{1}{2}x^2-\dfrac{1}{3}y^2\right)$$ and $$\left(\dfrac{1}{2}x^2+\dfrac{1}{3}y^2\right)$$.

A
$$\dfrac{1}{9}y^4-\dfrac{1}{4}x^4$$

B
$$\dfrac{1}{4}x^4-\dfrac{1}{9}y^4$$

C
$$4x^2-9y^4$$

D
$$\dfrac{1}{4}x^4+\dfrac{1}{9}y^4$$

Solution

Given, product of $$\Bigl(\dfrac{1}{2}x^2-\dfrac{1}{3}y^2\Bigr)$$ and $$ \Bigl(\dfrac{1}{2}x^2+\dfrac{1}{3}y^2\Bigr)$$
i.e. $$\Bigl(\dfrac{1}{2}x^2-\dfrac{1}{3}y^2\Bigr)\times \Bigl(\dfrac{1}{2}x^2+\dfrac{1}{3}y^2\Bigr)$$.

We know, the identity $$(a+b)(a-b)=a^2-b^2$$.

Then,
$$\Bigl(\dfrac{1}{2}x^2-\dfrac{1}{3}y^2\Bigr)\times \Bigl(\dfrac{1}{2}x^2+\dfrac{1}{3}y^2\Bigr)$$
$$=\Bigl(\dfrac{1}{2}x^2\Bigr)^2-\Bigr(\dfrac{1}{3}y^2\Bigl)^2$$
$$=\dfrac{1}{4}x^4-\dfrac{1}{9}y^4$$.

Therefore, option $$B$$ is correct.
Question 10
1 / -0

What is the degree of the given monomial $$7y$$?

A
$$0$$

B
$$1$$

C
$$2$$

D
$$7$$

Solution

We know that the degree is the term with the greatest exponent and,
To find the degree of a monomial with more than one variable for the same term, just add the exponents for each variable to get the degree.

Here, the given monomial $$7y$$ has one variable $$y$$ where the power of $$y$$ is $$1$$. Therefore degree of the monomial $$7y$$ is:

$$1$$ (The exponent of the variable)

Hence, the degree of the monomial $$7y$$ is $$1$$.

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Polynomials Test - 40

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Polynomials Test - 40

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Question 1

$$49.9$$

$$14.4$$

$$51.84$$

$$49.04$$

Solution

Question 2

$$2x^2+3y^2+6xy$$

$$4x^2+9y^2-12xy$$

$$2x^2+3y^2-6xy$$

$$4x^2+9y^2+12xy$$

Solution

Question 3

$$4ab$$

$$2ab$$

$$a^{2} + 2ab + b^{2}$$

$$2(a^{2} + b^{2})$$

Solution

Question 4

$$6$$

$$7$$

$$8$$

$$9$$

Solution

Question 5

2

1

0

3

Solution

Question 6

$$1$$

$$3$$

$$2$$

$$-15$$

Solution

Question 7

$$\dfrac{5}{2}$$

$$-\dfrac{5}{2}$$

$$\dfrac{2}{5}$$

$$-\dfrac{2}{5}$$

Solution

Question 8

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 9

$$\dfrac{1}{9}y^4-\dfrac{1}{4}x^4$$

$$\dfrac{1}{4}x^4-\dfrac{1}{9}y^4$$

$$4x^2-9y^4$$

$$\dfrac{1}{4}x^4+\dfrac{1}{9}y^4$$

Solution

Question 10

$$0$$

$$1$$

$$2$$

$$7$$

Solution

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