Polynomials Test

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Polynomials Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

A real number '$$\alpha$$' is said to be a ................. of the polynomial $$p(x)$$, if $$p(\alpha) = 0$$.

A
linear

B
zero

C
constant

D
cubic

Solution

If we replace the value of $$x$$ by a real number, say $$\alpha$$ in the polynomial $$p(x)$$, such that $$p(x)=0$$, i.e. $$p(\alpha)=0$$, then we say that, the real number '$$\alpha$$' is a zero of the polynomial $$p(x)$$.

E.g.: Suppose $$p(x)=x-1$$.
Here, the real number $$1$$ is a zero of the given polynomial as $$p(1)=1-1=0$$, i.e. $$p(1)=0$$.

Hence, we can say that, a real number $$\alpha$$ is said to be a zero of the polynomial $$p(x)$$, if $$p(\alpha)=0$$.

That is, option $$B$$ is correct.
Question 2
1 / -0

Using binomial identities, expand the following expression $$(t+3)^2$$

A
$$t^2-6t-9$$

B
$$t^2+6t-9$$

C
$$t^2+6t+9$$

D
$$t^2-6t+9$$

Solution

We need to expand $$(t+3)^2$$
We know that binomial identities for $$(a+b)^2=a^2+2ab+b^2$$
So, $$a = t, b = 3$$
$$=$$ $$t^2+2\times 3t+3^2$$
$$=$$ $$t^2+6t+9$$
Question 3
1 / -0

Find the value of $$(2x+5)(2x-5)$$.

A
$$4x-25$$

B
$$4x^2-25$$

C
$$2x^2-25$$

D
$$4x^2-5$$

Solution

We know that binomial identities for $$(x+y)(x-y)=x^2-y^2$$
$$\therefore (2x+5)(2x-5) = (2x)^2-5^2$$
$$=$$ $$4x^2-25$$
Question 4
1 / -0

Simplify the following expression: $$(s-r)^2-(s+r)^2$$

A
$$2sr$$

B
$$2s^2+2r^2$$

C
$$-4sr$$

D
$$2s^2-2r^2$$

Solution

We know that binomial identities for $$(x+y)^2=x^2+2xy+y^2$$
and $$(x-y)^2=x^2-2xy+y^2$$
So, $$(s-r)^2-(s+r)^2$$ $$=$$ $$s^2-2sr+r^2-s^2-2sr-r^2$$
$$=$$ $$-4sr$$
Question 5
1 / -0

Reduce the following expression using binomial identities: $$(3x+5)(3x+10)$$.

A
$$9x^2+45x+50$$

B
$$9x^2-45x+50$$

C
$$9x^2+45x-50$$

D
$$9x^2+45-50$$

Solution

We know that binomial identities for $$(x+y)(x+z)=x^2+(y+z)x+yz$$
So, $$x = 3x, y = 5, z = 10$$
$$=$$ $$(3x)^2+(5+10)3x+5\times 10$$
$$=$$ $$9x^2+45x+50$$
Question 6
1 / -0

What is the product of $$(x-2x)^2$$?

A
$$2x^2$$

B
$$x-2x^2$$

C
$$x^2$$

D
$$8x^2$$

Solution

$$(x-2x)^2 = (x-2x)(x-2x)$$
$$=$$ $$x.x-x.2x-2x.x+2x.2x$$
$$=$$ $$x^2-2x^2-2x^2+4x^2$$
$$=x^2$$
Question 7
1 / -0

If $$f(x) = x^{2}-1$$, and $$f(2a) = 35$$, then calculate the value of $$a$$.

A
$$-6$$

B
$$-3$$

C
$$1$$

D
$$2$$

E
$$6$$

Solution

Given, $$f\left( x \right) ={ x }^{ 2 }-1$$ and $$f\left( 2a \right) =35$$
Putting $$2a$$ in place of $$x$$
$$\Rightarrow f\left( 2a \right) ={ \left( 2a \right) }^{ 2 }-1=35$$
$$\Rightarrow { \left( 2a \right) }^{ 2 }-1=35$$
$$\Rightarrow 4{ a }^{ 2 }=36$$
$$\Rightarrow { a }^{ 2 }=9$$
$$\Rightarrow a=\sqrt { 9 } =\pm 3$$
Value of $$a$$ is $$+3$$ or $$-3$$
So, correct option is B.
Question 8
1 / -0

$$x^{3}(x^{2} - 5) = -4x$$
If $$x > 0$$, what is one possible solution to the equation above?

A
$$1$$ or $$2$$

B
only $$1$$

C
only $$2$$

D
none of these

Solution

$$x^3(x^2-5)=-4x$$
$$\Rightarrow x^5-5x^3+4x=0$$
$$\Rightarrow x(x^4-5x^2+4)=0$$
$$\Rightarrow x=0$$ or $$x^4-5x^2+4=0$$
But given $$x>0$$
$$\therefore x^4-5x^2+4=0 $$
$$ x^4-4x^2-x^2+4=0 $$
$$\Rightarrow (x^2-4)(x^2-1)=0$$
$$x^2-4=0$$ or $$x^2-1=0$$
$$x^2-4=0 \implies x = \pm 2$$
$$x^2-1=0 \implies x = \pm 1$$
Since it is given $$x > 0$$
$$x=1$$ or $$x=2$$
Question 9
1 / -0

If $$x\Box y=(x+y)^2-(x-y)^2$$. Then $$\sqrt 4\Box \sqrt 5=$$.

A
$$0$$

B
$$8\sqrt5$$

C
$$10\sqrt5$$

D
$$15$$

E
$$20$$

Solution
Question 10
1 / -0

If $$(3+\sqrt{2})$$ is a root of $${x}^{3}-11{x}^{2}+37x-35=0$$, find the rational root.

A
$$-3$$

B
$$5$$

C
$$2$$

D
None of these

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Polynomials Test - 39

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Polynomials Test - 39

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Question 1

linear

zero

constant

cubic

Solution

Question 2

$$t^2-6t-9$$

$$t^2+6t-9$$

$$t^2+6t+9$$

$$t^2-6t+9$$

Solution

Question 3

$$4x-25$$

$$4x^2-25$$

$$2x^2-25$$

$$4x^2-5$$

Solution

Question 4

$$2sr$$

$$2s^2+2r^2$$

$$-4sr$$

$$2s^2-2r^2$$

Solution

Question 5

$$9x^2+45x+50$$

$$9x^2-45x+50$$

$$9x^2+45x-50$$

$$9x^2+45-50$$

Solution

Question 6

$$2x^2$$

$$x-2x^2$$

$$x^2$$

$$8x^2$$

Solution

Question 7

$$-6$$

$$-3$$

$$1$$

$$2$$

$$6$$

Solution

Question 8

$$1$$ or $$2$$

only $$1$$

only $$2$$

none of these

Solution

Question 9

$$0$$

$$8\sqrt5$$

$$10\sqrt5$$

$$15$$

$$20$$

Solution

Question 10

$$-3$$

$$5$$

$$2$$

None of these

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