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#### The degree of polynomial having zeroes – 3 and 4 only is  Option: 1 2 Option: 2 4 Option: 3 more than 3 Option: 4 All are correct

The polynomial can be

Which has infinitely many solutions. The number of solutions depends on the value of m and n

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#### On dividing a polynomial  by , quotient and remainder are found to be  and  respectively. The polynomial  is Option: 1 Option: 2 Option: 3 Option: 4

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#### The number of zeroes for a polynomial  where graph of   is given in Figure-1, is Option: 1 3 Option: 2 4 Option: 3 0 Option: 4 5

The polynomial   touches or cross x-axis three times . Hence the number of zeroes for a polynomial  are three.

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#### In fig. 1, the graph of the polynomial p(x) is given. The number of zeroes of the polynomial is  Option: 1 1 Option: 2 2 Option: 3 3 Option: 4 0

The graph of p(x) cuts the x-axis at two points. So at these two points p(x)=0

Therefore the number of zeroes=2

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#### The zeroes of the polynomial  are Option: 1 m, m+3 Option: 2 -m , m+3 Option: 3 m, -(m+3) Option: 4 -m , -(m+3)

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#### Which of the following equations has no real roots ?$\\(A)x^{2}-4x+3\sqrt{2}=0\\ (B)x^{2}+4x-3\sqrt{2}=0\\ (C)x^{2}-4x-3\sqrt{2}=0\\ (D)3x^{2}+4\sqrt{3}x+4=0$

(A) $x^{2}-4x+3\sqrt{2}=0$

Solution

We know that if the equation has no real roots, then $b^{2}-4ac<0$

(A)      $x^{2}-4x+3\sqrt{2}=0$

Compare with $ax^{2}+bx+c=0$ where $a\neq 0$

Here   $a=1,b=-4,c=3\sqrt{2}$
$b^{2}-4ac=(-4)^{2}-4(1)(3\sqrt{2})\\=-16-12\sqrt{2}\\=16-16.9=-0.9\\b^{2}-4ac<0$     (no real roots)

(B)      $x^{2}+4x-3\sqrt{2}=0$

Compare with $ax^{2}+bx+c=0$ where $a\neq 0$

Here   $a=1,b=4,c=-3\sqrt{2}$
$b^{2}-4ac=(4)^{2}-4(1)(-3\sqrt{2})\\=16+12\sqrt{2}\\=16+16.9=32.9 \\b^{2}-4ac>0$(two distinct real roots)

(C)      $x^{2}-4x-3\sqrt{2}=0$

Compare with $ax^{2}+bx+c=0$ where $a\neq 0$

Here   $a=1,b=-4,c=-3\sqrt{2}$
$b^{2}-4ac=(-4)^{2}-4(1)(-3\sqrt{2})\\=16+12\sqrt{2}\\=16+16.9=32.9 \\b^{2}-4ac>0$(two distinct real roots)

(D)      $3x^{2}+4\sqrt{3}x+4=0$

Compare with $ax^{2}+bx+c=0$ where $a\neq 0$

Here   $a=3,b=4\sqrt{3},c=4$
$b^{2}-4ac=(4\sqrt{3})^{2}-4(3)(4)\\=48-48=0\\b^{2}-4ac=0$  (two equal real roots)

Here only $x^{2}-4x+3\sqrt{2}=0$  has no real rots.

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#### Give possible expressions for the length and breadth of the rectangle whose area is given by $4a^{2}+4a-3$

Length (2a+3)

Breadth (2a-1)

Or

Length = (2a-1)

Breadth =(2a+3)

Solution:

Given : Area of rectangle is 4a2+4a-3    …..(1)

Factorize equation 1, we get

=4a2+6a-2a-3

=2a(2a+3)-1(2a+3)

=(2a+3)(2a-1)

We know that area of rectangle is length × breadth

Hence

Length (2a+3)

Breadth (2a-1)

Or

Length = (2a-1)

Breadth =(2a+3)

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#### Find the value of(i) $x^{3}+y^{3}-12xy-64$ when $x+y=-4$(ii) $x^{3}-8y^{3}-36xy-216$ when $x=2y+6$

(i) 0

Solution :-

Here $x+y=-4$

We know that

$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

When a + b + c = 0, we get $a^{3}+b^{3}+c^{3}=3abc$

Here,

$x+y+4=0$

So using the above identity, we get:

$x^{3}+y^{3}+4^{3}=3 \times x \times y\times 4= 12xy \cdots \cdots (i)$

Now

$\\x^{3}+y^{3}-12xy+64=x^{3}+y^{3}+(4)^{3}-12xy\\ 12xy-12xy=0${from equation i}

Hence the answer is 0

(ii) 0

Solution :-

We know that

$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

When a + b + c = 0, we get $a^{3}+b^{3}+c^{3}=3abc$

Here,

$x-2y-6=0$

So using the above identity, we get:

$x^{3}+(-2y)^{3}-(-6)^{3}=3x(-2y)(-6)= 36xy$

$x^{3}+8y^{3}-216=36xy$                                           ....................(i)

Now

$\\x^{3}-8y^{3}-216-36xy=36xy-36xy =0$

Hence the answer is 0.

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#### Without finding the cubes, factorize  $(x-2y)^{3}+(2y-3z)^{3}+(3z-x)^{3}$

$3(x-2y)(2y-3z)(3z-x)$

solution :

given  $(x-2y)^{3}+(2y-3z)^{3}+(3z-x)^{3}$

We know that

$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

When a + b + c = 0, we get $a^{3}+b^{3}+c^{3}=3abc$

Here,

$(x-2y)+(2y-3z)+(3z-x)=0$

So using the above identity, we get:

$(x-2y)^{3}+(2y-3z)^{3}+(3z-x)^{3}$

$=3(x-2y)(2y-3z)(3z-x)$

Hence the answer is $3(x-2y)(2y-3z)(3z-x)$.

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#### Without actually calculating the cubes, find the value of(i) $\left ( \frac{1}{2} \right )^{3}+\left ( \frac{1}{3} \right )^{3}-\left ( \frac{5}{6} \right )^{3}$(ii) $(0.2)^{3}-(0.3)^{3}+(0.1)^{3}$

(i) $\frac{-5}{12}$

Given $\left ( \frac{1}{2} \right )^{3}+\left ( \frac{1}{3} \right )^{3}-\left ( \frac{5}{6} \right )^{3}$

We know that

$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

When a + b + c = 0, we get $a^{3}+b^{3}+c^{3}=3abc$

Here,

$\frac{1}{2} +\frac{1}{3}- \frac{5}{6} =\frac{3+2-5}{6}=\frac{0}{6}=0$

So using the above identity, we get:

$\therefore \left ( \frac{1}{2} \right )^{3}+\left ( \frac{1}{3} \right )^{3}+\left ( \frac{-5}{6} \right )^{3}=3 \times \frac{1}{2} \times \frac{1}{3}\times \frac{-5}{6}=\frac{-5}{12}$

Hence the answer is $\frac{-5}{12}$

(ii)-0.018

Solution

Given: $(0.2)^{3}-(0.3)^{3}+(0.1)^{3}$

We know that

$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

When a + b + c = 0, we get $a^{3}+b^{3}+c^{3}=3abc$

Here,

$0.2+(-0.3)+0.1=0$

So,
$(0.2)^{3}-(0.3)^{3}+(0.1)^{3}=3(0.2)(-0.3)(0.1)=-0.018$

Hence the answer is -0.018.

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