Polynomials Test

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Polynomials Test - 21

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Question 1
1 / -0

Suppose a, b denote the distinct real roots of the quadratic polynomial $$x^2+20x-2020$$ and suppose c, d denote the distinct complex roots of the quadratic polynomial $$𝑥^2−20𝑥+2020$$. Then the value of $$ac(a-c)+ad(a-d)+bc(b-c)+bd(b-d)$$ is

A
0

B
8000

C
8080

D
16000

Solution

Given that,
$$a,\ b$$ are distinct real roots of the quadratic polynomial $$x^2+20x-2020$$.
$$c,\ d$$ are distinct complex roots of the quadratic polynomial $$x^2-20x+2020$$.

To find out,
The value of $$ac(a-c)+ad(a-d)+bc(b-c)+bd(b-d)$$

We know that, for a quadratic polynomial of the form $$ax^2+bx+c$$,
Sum of roots $$=\dfrac{-\text{ Coefficient of }x}{\text{Coefficient of }x^2}=\dfrac{-b}{a}$$,

And product of roots $$=\dfrac{\text{ Constant term}}{\text{Coefficient of }x^2}=\dfrac{c}{a}$$

Hence, for the first quadratic polynomial, $$x^2+20x-2020$$,
$$a+b=-20\quad \quad \ \ \ \ \ \ \ ....(1)$$
And $$ab=-2020\quad \quad ....(2)$$

And, for the first quadratic polynomial, $$x^2-20x+2020$$,
$$c+d=20\quad \quad \ \ \ \ \ \ \ \ ....(3)$$
And $$cd=2020\quad \quad \ ....(4)$$

Now, simplifying the expression $$ac(a-c)+ad(a-d)+bc(b-c)+bd(b-d)$$ by multiplying and combing terms that make the sum and product of the roots of the given polynomials.

$$ac(a-c)+ad(a-d)+bc(b-c)+bd(b-d)$$

$$\Rightarrow a^2c-ac^2+a^2d-ad^2+b^2c-bc^2+b^2d-bd^2$$

$$\Rightarrow a^2c+a^2d+b^2c+b^2d-ac^2-ad^2-bc^2-bd^2$$

$$\Rightarrow a^2(c+d)+b^2(c+d)-[a(c^2+d^2)+b(c^2+d^2)]$$

$$\Rightarrow (a^2+b^2)(c+d)-[(c^2+d^2)(a+b)]$$

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

Using the above relation, we get:

$$ \{(a+b)^2-2ab\}(c+d)-[\{(c+d)^2-2cd\}(a+b)]$$

Substituting the values of $$(a+b),\ ab,\ (c+d)$$ and $$CD$$ from $$(1),\ (2),\ (3)$$ and $$(4)$$, we get:
$$\{(-20)^2-2(-2020)\}(20)-[\{(20)^2-2(2020)\}(-20)]$$

$$\Rightarrow \{400+4040\}(20)-[\{400-4040\}(-20)]$$

$$\Rightarrow \{4440\}(20)-[\{-3640\}(-20)]$$

$$\Rightarrow 88800-72800$$

$$\Rightarrow 16000$$

Hence, $$ac(a-c)+ad(a-d)+bc(b-c)+bd(b-d)=16000$$.
Question 2
1 / -0

Degree of the polynomial $$4x^4 + 5x + 7$$, is

A
4

B
5

C
2

D
7

Solution

Given polynomial is $$4x^4 + 5x + 7$$
The highest power of $$x$$ having a non zero coefficient is 4.
Hence, the degree is 4.
Question 3
1 / -0

Which type of polynomial is $$2-x^2+x^3$$ ?

A
Quadratic Polynomial

B
Cubic Polynomial

C
Binomial

D
None of these

Solution

$$ 2 - { x }^{ 2 }+ { x }^{ 3 }$$ is a cubic polynomial as the highest power of $$x$$ is $$3$$.
Question 4
1 / -0

Which type of polynomial is $$3x^3$$ ?

A
Cubic Polynomial

B
Linear Polynomial

C
Square Polynomial

D
None of These

Solution

$$3{ x }^{ 3 }$$ is a cubic polynomial as the highest power of $$x$$ is $$3$$.
Question 5
1 / -0

The degree of the polynomial $$p(x) = x^{3}-9x+3x^{5}$$ is

A
$$3$$

B
$$0$$

C
$$5$$

D
$$2$$

Solution

Given polynomial is $$p(x) = x^{3}-9x+3x^{5}$$.

We know that, the degree of a polynomial is the highest power of any of its variables.

Here, in $$p(x) = x^{3}-9x+3x^{5}$$,the highest power of variable $$x$$ is $$ 5$$.

So, the degree is $$ 5$$.

Hence, the degree of $$p(x) = x^{3}-9x+3x^{5}$$ is $$5$$.
Question 6
1 / -0

The degree of the polynomials $$p(y) = y^{3}, q(y) = (1-y^{4})$$ are

A
$$7,0$$

B
$$0,4$$

C
$$3,4$$

D
$$4,1$$
Question 7
1 / -0

Which type of polynomial is $$4 5y^2$$ ?

A
Linear Polynomial

B
Bi-quadratic polynomial

C
Cubic Polynomial

D
Quadratic Polynomial

Solution

$$45{ y }^{ 2 }$$ is a quadratic polynomial, as the highest power of $$y$$ is $$2.$$
Question 8
1 / -0

The degree of the polynomial $$2x-1$$ is

A
$$0$$

B
$$\dfrac{1}{2}$$

C
$$-1$$

D
$$1$$

Solution

For a polynomial the degree is the value of highest power of the variable.
For $$2x-1$$
Highest power of variable $$x$$ is $$ 1$$

$$\therefore$$ Degree $$= 1$$
Question 9
1 / -0

Which type of polynomial is $$5t-\sqrt{7}$$ ?

A
Linear Polynomial

B
Quadratic polynomial

C
Cubic Polynomial

D
None of above

Solution

$$5t-\sqrt 7$$ is a polynomial in variable $$t$$ and the highest power of variable $$t$$ is $$1.$$

A polynomial of degree $$1$$ is called a linear polynomial.

Hence, $$5t-\sqrt 7$$ is a linear polynomial.
Question 10
1 / -0

Degree of the polynomial $$p(x) = -10$$ is:

A
$$-10$$

B
$$10$$

C
$$0$$

D
$$1$$

Solution

Degree of a polynomial is the value of highest power of any variable.
For $$p(x) = -10$$
There is no variable in the equation.
So, highest degree of variable is zero.
$$\therefore $$ Degree $$= 0$$

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Polynomials Test - 21

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Polynomials Test - 21

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Question 1

0

8000

8080

16000

Solution

Question 2

4

5

2

7

Solution

Question 3

Quadratic Polynomial

Cubic Polynomial

Binomial

None of these

Solution

Question 4

Cubic Polynomial

Linear Polynomial

Square Polynomial

None of These

Solution

Question 5

$$3$$

$$0$$

$$5$$

$$2$$

Solution

Question 6

$$7,0$$

$$0,4$$

$$3,4$$

$$4,1$$

Question 7

Linear Polynomial

Bi-quadratic polynomial

Cubic Polynomial

Quadratic Polynomial

Solution

Question 8

$$0$$

$$\dfrac{1}{2}$$

$$-1$$

$$1$$

Solution

Question 9

Linear Polynomial

Quadratic polynomial

Cubic Polynomial

None of above

Solution

Question 10

$$-10$$

$$10$$

$$0$$

$$1$$

Solution

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