Polynomials Test

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Polynomials Test - 42

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Question 1
1 / -0

The zeroes of the polynomial $$P(x)=x(x-2)(x+3)$$ is

A
$$0$$

B
$$0,2,3$$

C
$$0,2,-3$$

D
$$None\ of\ these$$

Solution

$$P\left(x\right)=x\left(x-2\right) \left(x+3\right)$$
$$\therefore \ $$To find zeroes,
$$P\left(x\right)=0$$
$$\therefore \left(x\right) \left(x-2\right) \left(x+3\right)=0$$
$$\therefore x=0$$ OR $$x-2=0$$ OR $$x+3=0$$
$$\therefore x=0$$ OR $$x=2$$ OR $$x=-3$$
$$\therefore$$ value of $$x$$ is $$ 0, 2, -3$$
Question 2
1 / -0

Degree of the polynomial $$13 + 11x + 12x^3 + 3x^2$$ is

A
$$2$$

B
$$1$$

C
$$3$$

D
$$4$$

Solution

The degree of an individual term of a polynomial is the exponent of its variable
$$13+11x+12{x}^{3}+3{x}^{2}=12{x}^{3}+3{x}^{2}+11x+13$$
The highest exponent of $$x$$ is $$3$$
$$\therefore\,$$ Degree$$ =3$$
Question 3
1 / -0

The roots of $$x^{3}-4x^{2}-8x+8$$ are :

A
$$1,2\pm \sqrt{3}$$

B
$$-2,3\pm \sqrt{5}$$

C
$$-2,2\pm \sqrt{3}$$

D
$$1,3\pm \sqrt{5}$$

Solution

\begin{array}{l} { x^{ 3 } }-4{ x^{ 2 } }-8x+8 \\ { { suppose } }, \\ y={ x^{ 3 } }-4{ x^{ 2 } }-8x+8 \\ and\, \, \, replace\, y\, with\, \, 0. \\ { x^{ 3 } }-4{ x^{ 2 } }-8x+8=0 \\ Now,\, factorizing\, the\, equ: \\ (x+2)\, ({ x^{ 2 } }-6x+4)=0 \\ \Rightarrow x+2=0 \\ \therefore \, \, \, x=-2 \\ and,\, \\ \, \, \, { x^{ 2 } }-6x+4=0\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ { u\sin g\, quadratic{ { } }formula=-\frac { { b\pm \sqrt { { b^{ 2 } }-4(ac) } } }{ { 2a } } } \right. \\ \, \, \, \, substitute\, the\, values\& \, find\, the\, value\, of\, x: \\ \, \, \, \, \, \, \, a=1,\, b=-6,\, c=4 \\ \, \Rightarrow x=\, \, \frac { { -6\pm \sqrt { { { (-6) }^{ 2 } }-4(1)\, (4) } } }{ { 2\, .1 } } \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\frac { { 6\pm \sqrt { 36-16 } } }{ { 2\, } } =\frac { { 6\pm \sqrt { 20 } } }{ 2 } =\frac { { 6\pm 2\sqrt { 5 } } }{ { 2\, } } =3\pm \sqrt { 5 } \\ \, \, \, \, \, \, \, \, \, \, \, \, \therefore \, \, \, \, x=3+\sqrt { 5 } \, ,3-\sqrt { 5 } \\ so,that\, we\, can\, say\, the\, value\, of\, \, x=-2,\, 3+\sqrt { 5 } \, ,3-\sqrt { 5 } \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, or\, \, x=-2,\, 3\pm \sqrt { 5 } \, \\ And\, the\, \, correct\, option\, is\, B. \end{array}
Question 4
1 / -0

Factorise : $${ (ax+by) }^{ 2 }+{ (2bx-2ay) }^{ 2 }-6abxy$$

A
$${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+2{ b }^{ 2 }{ x }^{ 2 }+2{ a }^{ 2 }{ y }^{ 2 }+6abxy$$

B
$${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+4{ b }^{ 2 }{ x }^{ 2 }+4{ a }^{ 2 }{ y }^{ 2 }-12abxy$$

C
$${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+{ b }^{ 2 }{ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }+4abxy$$

D
$${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+4{ b }^{ 2 }{ x }^{ 2 }+4{ a }^{ 2 }{ y }^{ 2 }+6abxy$$

Solution

$$\left( ax+by\right)^2+\left( 2bx-2ay\right)^2-babxy$$
Here we will use the following identity,
$$\Rightarrow \left( m+n\right)^2=m^2+n^2+2mn$$
and $$\left(m-n\right)^2=m^2+n^2-2mn$$

Using this we have :
$$= \left(ax\right)^2+\left(by\right)^2+2axby+4\left(bx\right)^2+4\left(ay\right)^2-8axby-6abxy$$
$$=a^2x^2+b^2y^2+4b^2x^2+4a^2y^2-14axby+2axby$$
$$=a^2x^2+b^2y^2+4b^2x^2+4a^2y^2-12axby$$
Hence, the answer is $$a^2x^2+b^2y^2+4b^2x^2+4a^2y^2-12axby.$$
Question 5
1 / -0

A polynomial of the degree two has _________________.

A
At least two zeros

B
At least one zero

C
At most one zeros

D
Exactly two zeros

Solution
Question 6
1 / -0

If one zero of a quadratic polynomial $$x^2+3x+k$$ is $$2$$. Find the value of $$k$$.

A
$$2$$

B
$$-10$$

C
$$-5$$

D
$$9$$

Solution

$$2$$ is a zero of $$x^2+3x+k$$.

Thus, $$(2)^2+3(2)+k=0$$

$$k+10=0$$

$$k=-10$$
Question 7
1 / -0

If $$xy=20$$ and $$(x+y)^{2}=70$$, then $$x^{2}+y^{2}$$ is equal to

A
$$25$$

B
$$35$$

C
$$30$$

D
$$50$$

Solution

$$(x+y)^2=x^2+y^2+2xy$$

It is given that $$(x+y)^2=70$$ and $$xy=20$$

$$\Rightarrow 70=x^2+y^2+(2\times 20)$$

$$x^2+y^2=70-40$$

$$=30$$
Question 8
1 / -0

The zeroes of the polynomial $${ p }({ x })={ x }({ x }-1)({ x }-2)$$ are

A
$$0$$

B
$$0 , - 1 , - 2$$

C
$$0,1 , - 2$$

D
$$0,1,2$$

Solution

$$\begin{array}{l} We\, have \\ p\left( x \right) =x\left( { x-1 } \right) \left( { x-2 } \right) \\ for\, the\, \, zero\, polynomial \\ p\left( x \right) =x \\ \therefore x\left( { x-1 } \right) \left( { x-2 } \right) =0 \\ Here\, we\, get \\ x=0\, \, \, or\, \, \left( { x-1 } \right) =0\, \, and\, \left( { x-2 } \right) =0 \\ \therefore x=0\, \, ,\, \, \, x=1\, \, and\, x=2 \\ So,\, \left( { 0,1,2 } \right) \, are\, the\, zero\, polynomila\, of\, p\left( x \right) \, \\ Hence,\, option\, D\, is\, the\, required\, asnwer. \end{array}$$
Question 9
1 / -0

The zero of the polynomial $$P\left( x \right) =\surd 5x-5$$ is ......

A
$$\surd 5$$

B
$$-\surd 5$$

C
$$-5$$

D
$$4$$

Solution

Given, $$p(x)=\sqrt{5}x-5$$.
Here, we find zero of a polynomial when $$p(x)=0$$.
Then,
$$\sqrt{5}x-5=0$$
$$\implies$$  $$\sqrt{5}x=5$$
$$\implies$$  $$x=\dfrac{5}{\sqrt{5}}$$
$$\implies$$  $$x=\dfrac{5}{\sqrt{5}}\times\dfrac{\sqrt{5}}{\sqrt{5}}$$
$$\implies$$  $$x=\dfrac{5\sqrt{5}}{5}=\sqrt{5}$$.

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

The degree of the polynomial $$\frac { 4 }{ 5 } { x }^{ 2 }-\frac { 7 }{ 5 } x+\frac { 2 }{ 3 } { x }^{ 3 }+6$$ is :

A
2

B
1

C
3

D
0

Solution

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Polynomials Test - 42

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Polynomials Test - 42

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SHARING IS CARING

Question 1

$$0$$

$$0,2,3$$

$$0,2,-3$$

$$None\ of\ these$$

Solution

Question 2

$$2$$

$$1$$

$$3$$

$$4$$

Solution

Question 3

$$1,2\pm \sqrt{3}$$

$$-2,3\pm \sqrt{5}$$

$$-2,2\pm \sqrt{3}$$

$$1,3\pm \sqrt{5}$$

Solution

Question 4

$${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+2{ b }^{ 2 }{ x }^{ 2 }+2{ a }^{ 2 }{ y }^{ 2 }+6abxy$$

$${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+4{ b }^{ 2 }{ x }^{ 2 }+4{ a }^{ 2 }{ y }^{ 2 }-12abxy$$

$${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+{ b }^{ 2 }{ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }+4abxy$$

$${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+4{ b }^{ 2 }{ x }^{ 2 }+4{ a }^{ 2 }{ y }^{ 2 }+6abxy$$

Solution

Question 5

At least two zeros

At least one zero

At most one zeros

Exactly two zeros

Solution

Question 6

$$2$$

$$-10$$

$$-5$$

$$9$$

Solution

Question 7

$$25$$

$$35$$

$$30$$

$$50$$

Solution

Question 8

$$0$$

$$0 , - 1 , - 2$$

$$0,1 , - 2$$

$$0,1,2$$

Solution

Question 9

$$\surd 5$$

$$-\surd 5$$

$$-5$$

$$4$$

Solution

Question 10

2

1

3

0

Solution

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