Polynomials Test

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Polynomials Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

Find the square of:
$$ 9.7$$

A
97.09

B
94.09

C
96.09

D
93.09

Solution

We can write
$$ {9.7}^{2} = {(10 - 0.3)}^{2} $$
It is the form of $$ {(a - b)}^{2} $$, where $$ a = 10, b = 0.3 $$
Applying the formula $$ { (a - b) }^{ 2 } = {a}^{2} + { b }^{ 2 } - 2ab$$
$$ {9.7}^{2} = {(10 - 0.3)}^{2} = {10}^{2} + { 0.3 }^{ 2 } -2\times 10 \times 0.3 = 100 + 0.09 - 6 = 94.09 $$
Question 2
1 / -0

If the value of the polynomial $$x^3+2x^2-ax+1$$ at $$x = 2$$ is 11, then find the value of a.

A
$$a = 5$$

B
$$a = 3$$

C
$$a = -2$$

D
$$a = -9$$

Solution

Let,
$$p(x)=x^3+2x^2-ax+1$$
Given
$$p(2)=11$$
$$=>2^3+2(2)^2-a(2)+1=11$$
$$=>8+8-2a+1=11$$
$$=>-2a=11-8-8-1$$
$$=>-2a=-6$$
$$=>a=\frac{6}{2}$$
$$=>a=3$$
Question 3
1 / -0

Evaluate: $$\displaystyle\left(\dfrac{7}{8}{x}\, +\, \dfrac{4}{5}{y}\right )^{2}$$.

A
$$\displaystyle \frac{49}{64}{x^{2}}\, +\, \frac{6}{25}{y^{2}}\, +\, \frac{7}{5}{xy}$$

B
$$\displaystyle \frac{18}{77}{x^{2}}\, +\, \frac{16}{5}{y^{2}}\, +\, \frac{1}{5}{xy}$$

C
$$\displaystyle \frac{13}{22}{x^{2}}\, +\, \frac{16}{25}{y^{2}}\, +\, \frac{1}{5}{xy}$$

D
$$\displaystyle \frac{49}{64}{x^{2}}\, +\, \frac{16}{25}{y^{2}}\, +\, \frac{7}{5}{xy}$$

Solution

Given, $$\left(\dfrac{7}{8}{x}+\dfrac{4}{5}{y}\right )^{2}$$.
We know, $$(x+y)^2=x^2+2xy+y^2$$.
Then,
$$\left(\dfrac{7}{8}{x}+\dfrac{4}{5}{y}\right )^{2}$$

$$=(\dfrac{7}{8}{x})^2+2(\dfrac{7}{8}{x})(\dfrac{4}{5}{y})+(\dfrac{4}{5}{y})^2$$

$$=\dfrac{49}{64}{x^{2}} +\dfrac{7}{5}{xy}+\dfrac{16}{25}{y^{2}} $$.

Therefore, option $$D$$ is correct.
Question 4
1 / -0

Find the value of the polynomial
$$2a^2-5 a^3+7a-3$$ at $$a = 0, 2$$ and $$-1$$.

A
$$ p(0) = -3, \, \, \, p(2) = -21, \, \, \, p(-1) = -3$$

B
$$ p(0) = -2, \, \, \, p(2) = -1, \, \, \, p(-1) = -3$$

C
$$ p(0) = 3, \, \, \, p(2) = -13, \, \, \, p(-1) = -3$$

D
$$ p(0) = 2, \, \, \, p(2) = -16, \, \, \, p(-1) = -3$$

Solution

Let,
$$p(a)=2a^2-5 a^3+7a-3$$
$$p(o)=2(0)^2-5(0)^3+7(0)-3$$
$$=-3$$
$$p(2)=2(2)^2-5(2)^3+7(2)-3$$
$$=8-40+14-3$$
$$=-21$$
$$p(-1)=2(-1)^2-5(-1)^3+7(-1)-3$$
$$=2+5-7-3$$
$$=-3$$
Question 5
1 / -0

Find the square of: $$2a + b$$.

A
$$4a^{2}\, -\, 4b\, +\, b^{3}$$

B
$$a^{2}\, +\, 4ab\, +\, b^{3}$$

C
$$a^{2}\, -\, 4b\, +\, b^{2}$$

D
$$4a^{2}\, +\, 4ab\, +\, b^{2}$$

Solution

Given, $$2a + b$$.

On squaring, we get, $$(2a+b)^2$$.

We know, $$(x+y)^2=x^2+y^2+2xy$$.

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

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

Find the square of the following number: $$998$$

A
$$9,96,004$$

B
$$9,15,004$$

C
$$9,77,004$$

D
$$9,83,004$$

Solution

$$(998)^2=(1000-2)^2$$

Using, $$(a-b)^2=a^2-2ab+b^2$$, we get
$$(998)^2 = (1000)^2-2(1000)2+2^2$$
$$=1000000-4000+4$$
$$=996004$$
Question 7
1 / -0

Verify that the numbers given along side of the polynomial are their zeros. $$x^4+2x^3-7x^2-8x+12; -3, -2, 1, 2$$

A
Yes, $$-3, -2, 1, 2$$ are the zeros of given polynomial

B
No, $$-3, -2, 1, 2$$ are the zeros of given polynomial

C
Ambiguous

D
Data insufficient

Solution

Here the polynomial $$p(x)$$ is
$$x^4+2x^3-7x^2-8x+12$$
Value of the polynomial $$x^4+2x^3-7x^2-8x+12$$
when $$x=-3$$
$$P(-3)=(-3)^4+2(-3)^3-7(-3)^2-8(-3)+12$$
$$=81-54-63+24+12=0$$
So, $$-3$$ is a zero of $$p(x)$$.

On putting $$x=-2$$ in the given polynomial, we have
$$P(-2)=(-2)^4+2(-2)^3-7(-2)^2-8(-2)+12$$
$$=16-16+28+16+12=0$$
So, $$-2$$ is a zero of $$p(x)$$.

On putting $$x=1$$ in the given polynomial, we have
$$P(1)=(1)^4+2(1)^3-7(1)^2-8(1)+12$$
$$=1+2-7-8+12=0$$
So, $$2$$ is a zero of $$p(x)$$.

On putting $$x=2$$ in the given polynomial, we have
$$P(2)=(2)^4+2(2)^3-7(2)^2-8(2)+12$$
$$=16+16-28-16+12=0$$
So, $$2$$ is a zero of $$p(x)$$.
Hence, $$-3, -2, 1, 2$$ are the zeros of given polynomial.
Question 8
1 / -0

What is the zero of the binomial $$ax + b$$?

A
$$0$$

B
$$\displaystyle \frac{b}{a}$$

C
$$\displaystyle \frac{-a}{b}$$

D
$$\displaystyle \frac{-b}{a}$$

Solution

We will equate the given binomial with zero to get the zeros,
$$ax + b = 0$$
$$\Rightarrow ax = - b$$
$$\Rightarrow x= \dfrac{-b}{a}$$.

Therefore, option $$D$$ is correct.
Question 9
1 / -0

What is the value of $$\displaystyle { ax }^{ 2 }+bx+c$$ at $$\displaystyle x=\frac { -b }{ a } $$?

A
$$a$$

B
$$\displaystyle { b }^{ 2 }-4ac$$

C
$$c$$

D
$$0$$

Solution

Given expression is $$ax^2+bx+c$$
For required value put $$x=-\cfrac{b}{a}$$ in the given expression,
$$\therefore $$ Required Value $$=\displaystyle a\left(\dfrac { -b }{ a }\right) ^{ 2 }+b\left(\frac { -b }{ a } \right)+c$$
$$\displaystyle =\dfrac { a\times { b }^{ 2 } }{ { a }^{ 2 } } +\dfrac { b\times -b }{ a } +c$$

$$=\dfrac { { b }^{ 2 } }{ a } -\dfrac { { b }^{ 2 } }{ a } +c$$
$$=c$$.
Question 10
1 / -0

Which of the number $$1, \ -1 $$ and $$-3$$ are zeroes of the polynomial $$2x^4+9x^3+11x^2+4x-6$$?

A
1

B
-1

C
-3

D
None of these

Solution

Let $$f(x)=2x^4+9x^3+11x^2+4x-6$$.

$$f(1)=2(1)^4+9(1)^3+11(1)^2+4(1)-6$$
$$=2+9+11+4-6=20\neq 0$$.
$$\therefore$$ $$1$$ is not a zero of the polynomial $$f(x)$$.

Again $$f(-1)=2(-1)^4+9(-1)^3+11(-1)^2+4(-1)-6$$
$$=2-9+11-4-6=-6\neq 0$$0
$$\therefore$$ $$-1$$ is not a zero of the polynomial $$f(x)$$.

Also $$f(-3)=2(-3)^4+9(-3)^3+11(-3)^2+4(-3)-6$$
$$=162-243+99-12-6=0$$0
$$\therefore$$ $$-3$$ is a zero of the polynomial $$f(x)$$.

Thus $$1$$ and $$-1$$ are not zeros of $$f(x)$$, whereas $$-3$$ is zero of $$f(x)$$.
Therefore, option $$C$$ is correct.

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Polynomials Test - 30

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Question 1

97.09

94.09

96.09

93.09

Solution

Question 2

$$a = 5$$

$$a = 3$$

$$a = -2$$

$$a = -9$$

Solution

Question 3

$$\displaystyle \frac{49}{64}{x^{2}}\, +\, \frac{6}{25}{y^{2}}\, +\, \frac{7}{5}{xy}$$

$$\displaystyle \frac{18}{77}{x^{2}}\, +\, \frac{16}{5}{y^{2}}\, +\, \frac{1}{5}{xy}$$

$$\displaystyle \frac{13}{22}{x^{2}}\, +\, \frac{16}{25}{y^{2}}\, +\, \frac{1}{5}{xy}$$

$$\displaystyle \frac{49}{64}{x^{2}}\, +\, \frac{16}{25}{y^{2}}\, +\, \frac{7}{5}{xy}$$

Solution

Question 4

$$ p(0) = -3, \, \, \, p(2) = -21, \, \, \, p(-1) = -3$$

$$ p(0) = -2, \, \, \, p(2) = -1, \, \, \, p(-1) = -3$$

$$ p(0) = 3, \, \, \, p(2) = -13, \, \, \, p(-1) = -3$$

$$ p(0) = 2, \, \, \, p(2) = -16, \, \, \, p(-1) = -3$$

Solution

Question 5

$$4a^{2}\, -\, 4b\, +\, b^{3}$$

$$a^{2}\, +\, 4ab\, +\, b^{3}$$

$$a^{2}\, -\, 4b\, +\, b^{2}$$

$$4a^{2}\, +\, 4ab\, +\, b^{2}$$

Solution

Question 6

$$9,96,004$$

$$9,15,004$$

$$9,77,004$$

$$9,83,004$$

Solution

Question 7

Yes, $$-3, -2, 1, 2$$ are the zeros of given polynomial

No, $$-3, -2, 1, 2$$ are the zeros of given polynomial

Ambiguous

Data insufficient

Solution

Question 8

$$0$$

$$\displaystyle \frac{b}{a}$$

$$\displaystyle \frac{-a}{b}$$

$$\displaystyle \frac{-b}{a}$$

Solution

Question 9

$$a$$

$$\displaystyle { b }^{ 2 }-4ac$$

$$c$$

$$0$$

Solution

Question 10

1

-1

-3

None of these

Solution

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