Polynomials Test

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Polynomials Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

The zero of the polynomial $$2x+1$$ is:

A
$$x=-\dfrac{1}{2}$$

B
$$x=-1$$

C
$$x=-2$$

D
None of the above

Solution

Let $$p(x)=2x+1$$

Consider $$p(x)=0$$

$$\Rightarrow 2x+1=0$$

$$\Rightarrow x=\dfrac{-1}{2}$$

Therefore, $$x=\dfrac{-1}{2}$$ is the zero of the polynomial $$2x+1=0$$
Question 2
1 / -0

If $$p(x) = 5x - 4$$, then $$p(2) = $$

A
$$1$$

B
$$6$$

C
$$0$$

D
$$3$$

Solution

The polynomial is $$p(x)=5x-4$$ we substitute $$x=2$$ in the polynomial:
$$p(2)=(5\times 2)-4=10-4=6$$
Hence, $$p(2)=6$$.
Question 3
1 / -0

A zero of a polynomial:

A
needs to be $$0$$.

B
can be any number including $$0$$.

C
needs to be an integer.

D
None of the above

Solution

Consider the polynomial, $$p(x) = ax^2 + bx +c$$.
To find the zero of a polynomial, we write $$p(x) = 0$$.
Hence, a real number $$c$$ is said to be a zero of the polynomial $$p(x)$$, if $$p(c)=0$$.

E.g. : Suppose, $$p(x)=x^2$$.
When $$x=0$$, we get $$p(x)=(0)^2=0$$.

Clearly, a zero of the polynomial can be any number including $$0$$.
Therefore, option $$B$$ is correct.
Question 4
1 / -0

The zero of the polynomial $$3x+5$$ is :

A
$$\dfrac{-5}{3}$$

B
$$\dfrac{5}{3}$$

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

D
$$\dfrac{-1}{2}$$

Solution

We find the zeros of the given polynomial using:
$$3x+5=0$$
$$3x=-5$$
$$x=\dfrac{-5}{3}$$.
Question 5
1 / -0

$$\left( 3-\sqrt { 7 } \right) \left( 3+\sqrt { 7 } \right) =$$?

A
$$4$$

B
$$2$$

C
$$6$$

D
$$8$$

Solution

Given, $$({3}-{\sqrt { 7 } })$$$$({3}+{\sqrt { 7 } })$$.

We know, $$(x-y)(x+y)$$ $$={x}^{2}-{y}^{2}$$.

Thus,
$$({3}-{\sqrt { 7 } })$$$$({3}+{\sqrt { 7 } })$$
$$={3}^{2}-{(\sqrt { 7 } )}^{2}$$
$$=9-7=2$$.

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

The value of $$(a+b)^2-2(a-b)^2+(a-b)(a+b)$$ is:

A
$$4ab-b^2$$

B
$$2ab-b^2$$

C
$$3ab-b^2$$

D
$$6ab-2b^2$$

Solution

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

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

Then,
$$(a+b)^2-2(a-b)^2+(a-b)(a+b)$$
$$=(a^2+b^2+2ab)-2(a^2+b^2-2ab)+(a^2-b^2)$$
$$=a^2+b^2+2ab-2a^2-2b^2+ 4ab+a^2-b^2$$
$$=2a^2+b^2+6ab-2a^2-3b^2$$
$$=6ab-2b^2$$.

Therefore, option $$D$$ is correct.
Question 7
1 / -0

If $$2$$ is a root of $$kx^4-11x^3+kx^2+13x+2$$, what is the value of $$k$$?

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

$$f(2) = k\times (2)^4-11\times (2)^3+k\times (2)^2+13\times 2+2$$
$$f(2)=0$$
$$16k-88+4k+26+2=0$$
$$ 20k -88+28=0$$
$$20k-60=0$$
$$20k =60$$
$$k=3$$
Question 8
1 / -0

If $$p(x) = 5x^2 - 4x + 2$$, then $$p(3) =$$

A
$$35$$

B
$$30$$

C
$$45$$

D
$$33$$

Solution

The given polynomial is $$p(x)=5x^2-4x+2$$ we substitute $$x=3$$ in the polynomial:
$$p(3)=5(3)^2-4(3)+2=5\times 9-12+2=45-12+2=47-12=35$$
Hence, $$p(3)=35$$.
Question 9
1 / -0

The zero of a polynomial $$P(x)$$ is:

A
A number $$c$$ such that $$P(c) = 0$$

B
A number $$c$$ such that $$P(-c) = 1$$

C
A number $$c$$ such that $$P(c)+P(-c) = 0$$

D
none of the above

Solution

Consider the polynomial, $$p(x) = ax^2 + bx +c$$.
To find the zero of a polynomial, we write $$p(x) = 0$$.
Hence, a real number $$c$$ is said to be a zero of the polynomial $$p(x)$$, if $$p(c)=0$$.
Therefore, option $$A$$ is correct.
Question 10
1 / -0

$$p(a) = 3a^2 + 4a - 4$$ is a polynomial in $$a$$ of degree

A
$$4$$

B
$$3$$

C
$$2$$

D
$$1$$

Solution

The polynomial $$ax^2+bx+c$$ is in standard form.On comparing $$3a^2+4a-4$$ with the standard polynomial we see, the power of the first term is $$2$$. Since the polynomial has the largest exponent, that is $$2$$ which is the degree of the polynomial.

Hence, the degree of the polynomial is $$2$$.

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Polynomials Test - 35

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Polynomials Test - 35

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Question 1

$$x=-\dfrac{1}{2}$$

$$x=-1$$

$$x=-2$$

None of the above

Solution

Question 2

$$1$$

$$6$$

$$0$$

$$3$$

Solution

Question 3

needs to be $$0$$.

can be any number including $$0$$.

needs to be an integer.

None of the above

Solution

Question 4

$$\dfrac{-5}{3}$$

$$\dfrac{5}{3}$$

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

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

Solution

Question 5

$$4$$

$$2$$

$$6$$

$$8$$

Solution

Question 6

$$4ab-b^2$$

$$2ab-b^2$$

$$3ab-b^2$$

$$6ab-2b^2$$

Solution

Question 7

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 8

$$35$$

$$30$$

$$45$$

$$33$$

Solution

Question 9

A number $$c$$ such that $$P(c) = 0$$

A number $$c$$ such that $$P(-c) = 1$$

A number $$c$$ such that $$P(c)+P(-c) = 0$$

none of the above

Solution

Question 10

$$4$$

$$3$$

$$2$$

$$1$$

Solution

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