Maths M.C.Q

$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\text{x}+\frac{1}{\text{x}}=3$ (given)
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=(3)^2-2$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=7\ ...(1)$
Cubing both side of equation (1). we have
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^3=(7)^3$
$\Rightarrow(\text{x}^2)^3+\Big(\frac{1}{\text{x}^2}\Big)^3+3(\text{x}^2)\frac{1}{\text{x}^2}\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=7^3$
$\Rightarrow\text{x}^6+\frac{1}{\text{x}^6}+3(7)=7^3$
$\Rightarrow\text{x}^6+\frac{1}{\text{x}^6}=343-21$
$\Rightarrow\text{x}^6+\frac{1}{\text{x}^6}=322$
Hence, correct option is $(d).$

$x^2-9$
$x^2-3^2$
$=(x+3)(x-3)\left[\text { Using identity } a^2-b^2=(a+b)(a-b)\right]$
Then the zeroes are $x + 3 = 0$ and $x - 3 = 0$
$\Rightarrow x = -3$ and $x = 3$

 It is given $(x+2)$ and $(x-1)$ are the factors of the polynomial $f(x)=x^3+10 x^2+m x+n$
i.e., $f(-2)=0$ and $f(1)=0$
New,
$f(-2)=(-2)^3+10(-2)^2+\mathrm{m}(-2)+\mathrm{n}=0$
$-8+40-2 \mathrm{~m}+\mathrm{n}=0$
$\Rightarrow-2 \mathrm{~m}+\mathrm{n}=-32$
$\Rightarrow 2 \mathrm{~m}-\mathrm{n}=32 \ldots . \text { (i) }$
$\mathrm{f}(1)=(1)^3+10(1)^2+\mathrm{m}(1)+\mathrm{n}=0$
$1+10+\mathrm{m}+\mathrm{n}=0$
$\mathrm{~m}+\mathrm{n}=-11 \ldots \text { (ii) }$
Solving equation $(i)$ and $(ii)$ we get,
$\mathrm{m}=7 \text { and } \mathrm{n}=-18$

 $(x-2)^2-(x+2)^2$
$= (x - 2 + x + 2) (x - 2 - x - 2) [$Using identity $a^2-b^2 = (a + b) (a - b)]$
$= (2x) (-4)$
$= -8x$
Then the zero is,
$-8 = 0$
$⇒ x = 0$

 Let $p(x) = 2 x^2+k x$
Since, $(x + 1)$ is a factor of $p(x)$, then
$p(-1) = 0$
$2(-1)^2 + k(-1) = 0$
$⇒ 2 - k = 0$
$⇒ k = 2$

 $P(x) = 5x - 4x^2 + 3$
$⇒ p(-1) = 5(-1) - 4(-1)^2 + 3$
$= -5 - 4 + 3$
$= -6$

 The general form of a polynomial is $a_n x^n$, where n is a natural number.
For zero polynomial $a =0 $.
Since the largest value of $n$ for which an is non-zero is negative infinity $($all the integers are bigger than negative infinity$).$
Therefore, the degree of zero polynomials is not defined.

 Given: $f(x) =px^3 + x^2 - 2x + q$
If $x + 1$ is a factor of $ f(x).$
Then $f(-1) = 0$
$p(-1)^3 + (-1)^2 - 2(1) + q = 0$
$-p + 1 + 2 + q = 0$
$-p + q = -3$
$p - q = 3 ......(i)$
Also, if $x - 1$ is a factor of $f(x),$ then
$p(1)^3 + (1)^2 - 2(1) + q = 0$
$p + 1 - 2 + q = 0$
$p + q = 1 ......(ii)$
$2p = 4$
$p = 2$
subtracting eq.$(ii)$ from eq. $(i),$
we get,
$-2q = 2$
$q = -1$
$q = -1$
Therefore, $p = 2, q = -1$

 $2x^2 + 7x - 4$
$= 2x^2 + 8x - x - 4$
$= 2x(x + 4) - 1(x + 4)$
$= (2x - 1)(x + 4)$
$2x - 1 = 0$ and $x + 4 = 0$
$\text{x}=\frac{1}{2}$ and $x = -4$
Therefore, one zero of the given polynomial is $\frac{1}{2}$

 $4x^3$ because monomial means only one term in an expression.

$ p(x) = x(x - 2) (x + 3)$
$⇒ x = 0$ and $x - 2 = 0$ and $x + 3 = 0$
$⇒ x = 0, x = 2$ and $x = -3$
Therefore, the zeroes are $0, 2, -3$

 In a polynomial, the power of the variable of each term should be a whole number.
Here, the power of variable $x$ is $3,$ which is a whole number.
Therefore, it is a polynomial.

 The given expression to be factorized is $a^2 - 1 - 2x - x^2$
Take common -1 from the last three terms and then we have
$a^2 - 1 - 2x - x^2 = a^2 - (1 + 2x + x^3)$
$= a^2 - {(1)^2 + 2.1 × x + (x)^2}$
$= a^2 - (1 + x)^2$
$= (a)^2- (1 + x)^2$
$= {a + (1 + x)} {a - (1 + x)}$
$= (a + 1 + x) (a - 1 - x)$
$= (a + x + 1) (a - x - 1)$

 A polynomial of degree $1$ is called a linear polynomial.
Options $(a),$ and $(c)$ have degree $2,$
so ther are quadratic polynomials.
option $(d)$ has a negative power, so it is not a polynomial.
The degree of $x + 1$ is $1,$ so it is a linear polynomial.

 As $(x^2 - 1)$ Is a factor of polynomial
$f(x^2) = ax^4 + bx^3 + cx^2 + dx + e$
Therefore,
$f(x) = 0$
And,
$f(1) = 0$
$a(1)^2 + b(1)^3 + c(1)^2 + d(1) + e = 0$
$⇒ a + b + c + d + e = 0$
And,
$f(-1) = 0$
$a(-1)^4 + b(-1)^3 + c(-1)^2 + d(-1) + e = 0$
$a - b - c - d + e = 0$
Hence, $a + c + e = b + d$

 By we know that $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
If $a + b + c = 0,$ then
$a^3 + b^3 + c^3 = 3abc$
In given expression,
Let $a - b = A, b - c = B, c - a = C$
Now, $a - b + b - c + c - a = 0$
i.e. $A + B + C = 0$
$⇒ A^3 + B^3 + C^3 = 3ABC$
$⇒ (a - b)^3 + (b - c)^3 + (c - a)^3 = 3(a - b)(b - c)(c - a)$
Hence, correct option is $(b).$

If both $x - a$ and $\text{x}-\frac{1}{2}$ are the factors of $\text{px}^2 + 5\text{x} +\text{r},$ than $f(2) = 0$
$\Rightarrow\text{p}(2)^2+5(2)+\text{r}=0$
$\Rightarrow4\text{p}+10+\text{r} = 0$
$\Rightarrow4\text{p}+\text{r}=-10\ ...(\text{i})$
Also, $\text{f}\Big(\frac{1}{2}\Big)=0$
$\Rightarrow\text{p}\Big(\frac{1}{2}\Big)^2+5\Big(\frac{1}{2}\Big)+\text{r}=0$
$\Rightarrow\frac{\text{p}}{4}+\frac{5}{2}+\text{r}=0$
$\Rightarrow\text{p}+10+4\text{r}=0$
$\Rightarrow\text{p}+4\text{r}=-10\ ...{\text{ii}}$
From eq. $(i)$ and eq. $(ii),$ we get
$4\text{p}+\text{r}=\text{p}+4\text{r}$
$\Rightarrow3\text{p}=3\text{r}$
$\Rightarrow\text{p}=\text{r}$

$ (x - a)^3 + (x - b)^3 + (x - c)^3 - 3(x - a) (x - b) (x - c)$
$= [x - a + x - b + x - c]$
$[(x - a^2) + (x - b^2) + (x - c^2) - (x - a) (x - b) - (x - b) (x - c) - (x - c) (x - a)]$
$= [3x - ({a + b + c})]$
$[(x - a)^2 + (x - b)^2 + (x - c)^2 - (x - a) (x - b) - (x - b) (x - c) - (x - c) (x -a)]$
$= 3x - 3x$
$[(x - a)^2 + (x - b)^2 + (x - c)^2 - (x - a) (x - b) - (x - b) (x - c) - (x - c) (x - a)]$
$= 0$
$[0] [(x - a)^2 + (x - b)^2 + (x - c)^2 - (x - a) (x - b) - (x - b) (x - c) - (x - c) (x - a)]$

$\sqrt{2}$ is a constant polynomial. The only term here is $\sqrt{2}$ which can be written as $\sqrt{2}\text{x}^\circ.$ So, the exponent of $x$ is zero. Therefore, the degree of the polynomial is $0.$

$p(x) = 5x - 4x^2 + 3$
Putting $x = -1$ in $p(x),$ we get
$p(-1) = 5 × (-1)-4 × (-1)^2 + 3 = -5 - 4 + 3 = -6$

Let $p(x) = 2x^2 + 7x - 4$
$= 2x^2 + 8x - x- 4 [$by splitting middle term$]$
$= 2x(x + 4) -1(x + 4)$
$=(2x - 1)(x + 4)$
For zeroes of $p(x),$ put $p(x) = 0 ⇒ (2x - 1)(x + 4) = 0$
$⇒ 2x - 1 = 0$ and $x + 4 = 0$
$\Rightarrow\text{x}=\frac{1}{2}$ and $\text{x}=-4$
Hence, one of the zeroes of the polynomial $p(x)$ is $\frac{1}{2}.$

 Zero of the zero polynomial is any real number.
e.g., Let us consider zero polynomial be $0(x - k),$ where $k$ is a real number.
For determining the zero, put $x - k = 0 ⇒ x = k$ Hence, zero of the zero polynomial be any real number.

 $\frac{(2.3)^3-0.027}{(2.3)^2+0.69+0.09}$
$=\frac{(2.3)^3-(0.3)^3}{(2.3)^2+(0.3)^3+(2.3)(0.3)}$
$=\frac{(2.3 - 0.3)\{(2.3)^2+(0.3)^2+(2.3)(0.3)\}}{((2.3)^2+(0.3)^2+(2.3)(0.3))}$
$=2.3 - 0.3$
$=2$
Hence, correct option is $(a).$

 $\frac{(0.87)^3+(0.13)^3}{(0.87)^2-(0.87\times0.13)+(0.13)^2}$
$=\frac{(0.87+0.13)[(0.87)^2-(0.87\times0.13)+(0.13)^2]}{(0.87)^2-(0.87\times0.13)+(0.13)^2}$
$=0.87+0.13$
$=1$

 Let $p(x)$ be a polynomial. If $\text{p}(\alpha)=0,$ then we say that $\alpha$ is a zero of a polynomial.
$p(x) = x^2 - 3x$
Now, $p(x) = 0$
$⇒ x^2 - 3x$
$⇒ x(x - 3) = 0$
$⇒ x = 0$ or $(x - 3) = 0$
$⇒ x = 0$ or $x = 3$
$\therefore 0$ and $3$ are the zeroes of the polynomial $p(x).$

 $p(x) = x^3 - ax^2 + x$
$x - a = 0 ⇒ x = a$
By the remainder theorem, we know that when $p(x)$ is divided by $(x - a)$, the remainder is $p(a).$
Now, $p(a) = a^3 - ax^2 + a$
$= a^3 - a^3 + a$
$= a$

 The given polynomial is $p(x) = 2x^2 + 5x - 4$
Putting $\text{x}=\frac{1}{2}$ in $p(x),$ we get
$\text{p}\Big(\frac{1}{2}\Big)=2\times\Big(\frac{1}{2}\Big)^2+5\times\frac{1}{2}-3$
$=\frac{1}{2}+\frac{5}{2}-3=3-3=0$
Putting $x = -3$ in $p(x),$ we get
$\text{p}(-3)=2\times(-3)^2+5\times(-3)-3$
$= 18-15-3$
$= 0$
Therefore, $x = -3$ is a zero of the polynomial $p(x)$
Thus, $\frac{1}{2}$ and $-3$ are the zeros of the given polynomial $p(x).$

 $\sqrt{2}$ is a constant term. Therefore, the degree of $\sqrt{2}$ is $0.$

 Let
$a - b = A$
$b - c = B$
$c - a = C$
Now $(A + B + C)^3 = A^3 + B^3 + C^3 + 3(A + B)(B + C)(C + A)$
$⇒ A^3 + B^3 + C^3 = (A + B + C)^3- 3(A + B)(B + C)(C + A)$
Now putting values of $A, B$ and $C.$ we get
$(\text{a} - \text{b})^3 + (\text{b} - \text{c})^3 + (\text{c} - \text{a})^3\\=(\not\text{a}-\not\text{b}+\not\text{b}-\not\text{c}+\not\text{c}-\not\text{a})^3\\-3(\text{a}-\not\text{b}+\not\text{b}-\text{c})(\text{b}-\not\text{c}+\not\text{c}-\text{a})(\text{c}-\not\text{a}+\not\text{a}-\text{b})$
$⇒ (a - b)^3+ (b - c)^3 + (c - a)^3 = 0 - 3 (a - c)(b - a)(c - b)$
$⇒ (a - b)^3+ (b - c)^3 + (c - a)^3 = 3(a - b)(b - c)(c - a)$
Hence, correct option is $(c).$

$(102)^3 = (100 + 2)^3$
$= (100)^3 + (2)^3 + 3 × 100 × 2(100 + 2)$
$= 1000000 + 8 + 60000 + 1200$
$= 1061208$

$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
If $a + b + c = 0,$ then
$a^3 + b^3 + c^3 - 3abc = 0$
$⇒ a^3 + b^3 + c^3 = 3abc ...(1)$
Now, consider $\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}$
Multiplying dividing by $a. b.$ and $c$ in $\frac{\text{a}^2}{\text{bc}}.\frac{\text{b}^2}{\text{ca}} and \frac{\text{c}^2}{\text{ab}}$ respectively. we get
$\frac{\text{a}^3}{\text{abc}}+\frac{\text{b}^3}{\text{bca}}+\frac{\text{c}^3}{\text{cab}}$
$=\frac{\text{a}^3+\text{b}^3+\text{c}^3}{\text{abc}}$
$=\frac{3\text{abc}}{\text{abc}} ....[$From $(1)]$
$=3$
Hence, correct option is $(d).$

 $\text{p}(\text{x})=\text{ax}^2+2\text{x}+\text{b}$
$\Rightarrow\text{a}(-2)^2 + 2(-2) + b = 0$
$\Rightarrow4\text{a} - 4+\text{b}=0\ .... (\text{i})$
Also,
$\text{p}\Big(\frac{-1}{2}\Big)=0$
$\Rightarrow\text{a}\Big(\frac{-1}{2}\Big)^2+2\Big(\frac{-1}{2}\Big)+\text{b}=0$
$\Rightarrow\frac{\text{a}}{4}-1+\text{b}=0$
$\Rightarrow\text{a}-4+4\text{b}=0\ ...(\text{ii})$
Subtracting eq. $(ii)$ from eq. $(i),$ we get
$3\text{a}+0-3\text{b}=0$
$\Rightarrow3(\text{a}-\text{b})=0$
$\Rightarrow\text{a}-\text{b}=0$

 The given expression to be factorized is $x^3 - 1 + y^3 + 3xy$
This can be written in the form
$x^3 - 1 + y^3 + 3xy = (x)^2 + (-1)^3 + (y)^3 - 3(x) × (-1).(y)$
Recall the formula $a^3 + b^3 + c^3 - 3abc = (a + b + c) (a^2 + b^2 + c^2 - ab - bc - ca)$
Using the above formula, we have,
$x^3 - 1 + y^3 + 3xy$
$= (x + (-1) + y) {(x)^2 + (-1)^2 + (y)^2 - (x) × (-1) - (-1) × (y) - (y) × (x)}$
$= (x - 1 + y) (x^2 + 1 + y^2 + x + y - xy)$

 Given: $P(x) = (x - 1) (x + 1),$ then$p(2) + p(1) - p(0)$
$= (2 - 1) (2 + 1) + (1 - 1) (1 + 1) - (0 - 1) (0 + 1)$
$= 1 × 3 + 0 × 2 - (-1) × 1$
$= 3 + 0 + 1$
$= 4$

 $p(x) = x^4 + 2x^3 - 3x^2 + x - 1$
$x - 2 = 0 ⇒ x = 2$
By the remainder theorem, we know that when $p(x)$ is divided by
$(x - 2)$, the remainder is $p(2).$
Now, $p(2) = x^4 + 2x^3 - 3x^2 + x - 1$
$= (2)^4 + 2(2)^3 - 3(2)^2 + 2 - 1$
$= 16 +16 - 12 + 2 - 1$
$= 21$

 Now, $(x + y)^3 = x^3+ 3^3= 3.x.3(x + 3)$
$[$Using identity, $(a + b)^3 = a^3 + b^3 + 3ab(a + b)]$
$= x^3 + 27 + 9x(x + 3)$
$= x^3 + 27 + 9x^2 + 27x$
Hence, the coefficient of $x$ in $(x + 3)^3$ is $27.$

 If both $x - 2$ and $\text{x}-\frac{1}{2}$ are the factors of $f(x) = px^2 + 5x + r,$ then $f(2) = 0$
$⇒ p(2)^2 + 5(2) + r = 0$
$⇒ 4p + 10 + r = 0$
$⇒ 4p + r = -10 .......(i)$
Also, $\text{f}\Big(\frac{1}{2}\Big)=0$
$\Rightarrow\text{p}\Big(\frac{1}{2}\Big)^2+5\Big(\frac{1}{2}\Big)+\text{r}=0$
$\Rightarrow\text{p}+10+4\text{r}=0$
$\Rightarrow\text{p}+4\text{r}=-10\ .....(\text{ii})$
From eq. $ (i)$ and eq. $(ii),$ we get
$4\text{p}+\text{r}=\text{p}+4\text{r}$
$\Rightarrow3\text{p}=3\text{r}$
$\Rightarrow\text{p}=\text{r}$

 $\text{x}^2+\frac{1}{\text{x}^2}=102,$
$\Rightarrow(\text{x}^2)+\Big(\frac{1}{\text{x}^2}\Big)-2\times\text{x}\times\frac{1}{\text{x}}=102-2\times\text{x}\times\frac{1}{\text{x}}$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=100$
$\Rightarrow\text{x}-\frac{1}{\text{x}}=\sqrt{100}=10$

By remainder theorem, when $p(x) = x^3 - ax^2 + x$ is divided by $(x - a),$ then the remainder $= p(a)$
Putting $x = a $ in $p(x),$ we get
$p(a) = a^3 - a × a^2 + a = ^3 - a^3 + a = a$
$\therefore$ Remainder $= a$

$-8$
$f(x) = x^3 + 6x^2 + 4x + k$
$f(-2) = 0$
$\therefore (-2)^3 + 6(-2)^2 + 4(-2) + k = 0$
$\therefore -8 + 6(4) + (-8) + k = 0$
$24 - 16 + k = 0$
$k + 8 = 0$
$k = -8$

$x + 1$ is a factor of $p(x) = 2x^2 + kx$
Then, $p(-1) = 0$
$i.e. 2(-1)^2 + k(-1) = 0$
$2 - k = 0$
$k = 2$

$(249)^2 - (248)^2 = (249 + 248)(249 - 248) [(a)^2 - (b)^2 = (a + b)(a - b)]$
$= (497)(1) = 497$

Let: $\text{p(x)} = 3\text{x}^2 - 1$
To find the zeroes of $p(x),$
We have:
$\text{p(x)}=0\Rightarrow3\text{x}^2-1=0$
$\Rightarrow3\text{x}^2=1$
$\Rightarrow\text{x}^2=\frac{1}{3}$
$\Rightarrow\text{x}=\pm\frac{1}{\sqrt{3}}$
$\Rightarrow\text{x}=\frac{1}{\sqrt{3}}\text{ and }\text{x}=\frac{-1}{\sqrt{3}}$

Given: $p(x) = (x - 1) (x + 1),$ then
$p(2) + p(1) - p(0)$
$= (2 - 1) (2 + 1) + (1 - 1) (1 + 1) - (0 - 1) (0 + 1)$
$= 1 × 3 + 0 × 2 -(-1) × 1$
$= 3 + 0 + 1$
$= 4$

Let $f(x) = px^2 + 5x + r$
Now,
If $x - 2$ and $\text{x}-\frac{1}{2}$ are factors of $f(x).$
Then at $x = 2$ and $\text{x}-\frac{1}{2}, f(x) = 0.$
So, $f(2) = 0, \text{f}\Big(\frac{1}{2}\Big)=0$
$⇒ p(2)^2 + 5(2) + r = 0$
And, $\text{p}\Big(\frac{1}{2}\Big)^2+5\Big(\frac{1}{2}\Big)+\text{r}=0$
$⇒ 4p + r + 10 = 0 ...(1)$
$And 4r + p + 10 = 0 ...(2)$
Subtracting equation $(2)$ from $(1),$ we have
$3p - 3r = 0$
$⇒ p = r$

$ \frac{\text{a}}{\text{b}}+\frac{\text{b}}{\text{a}}=-1$
$\Rightarrow\frac{\text{a}^2+\text{b}^2}{\text{ab}}=-1$
$⇒ a^2 + b^2 = -ab$
$⇒ a^2 + b^2 + ab = 0$
Thus, we have:
$(a^3 - b^3) = (a - b)(a^2 + b^2 + ab)$
$= (a - b) × 0$
$= 0$

Given: $\text{x}^3-\Big(\frac{1}{\text{x}^3}\Big)=14$
Let $x = a$ and $\frac{1}{\text{x}}=\text{b}$
Say, $\text{x}-\frac{1}{\text{x}}=\text{A}$
Then, $\text{a}^3-\text{b}^3=14$
$\Rightarrow(\text{a}-\text{b})(\text{a}^2+\text{ab}+\text{b}^2)=14$
$\Rightarrow(\text{a}-\text{b})(\{(\text{a}-\text{b})^2+2\text{ab}\}+2\text{ab})=14$
$\Rightarrow(\text{a}-\text{b})\{(\text{a}-\text{b})^2+3\text{ab}\}=14$
$\Rightarrow(\text{a}-\text{b})\{(\text{a}-\text{b})^2+3\}=14$
$\Rightarrow\text{A}(\text{A}^2+3)=14$
$\Rightarrow\text{A}(\text{A}^2+3)=14$
$\Rightarrow\text{A}^3+3\text{A}-14=0$
$\Rightarrow\text{A}^3-2\text{A}^2+2\text{A}^2-4\text{A}+7\text{A}-14=0$
$\Rightarrow\text{A}^2(\text{A}-2)+2\text{Y}(\text{Y}-2)+7(\text{Y}-2)=0$
$\Rightarrow(\text{A}-2)(\text{A}^2+2\text{A}+7)=0$
$\Rightarrow\text{A}-2=0,$
$\Rightarrow\text{A}=2$
$\Rightarrow\text{x}-\frac{1}{\text{x}}=2$

$3$ terms of not more than the power of $3$

Now, $a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - ab - be - ca) + 3abc$
$[$Using identity, $a^3 + b^3 + c^3 – 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - be - ca)] = 0 + 3abc$
$\therefore a + b + c = 0,$ given$a^3 + b^3 + c^3 = 3abc$

On cubing we get.
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=\text{x}^3+\Big(\frac{1}{\text{x}^3}\Big)+3\times\text{x}\times\frac{1}{\text{x}}\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow27=\text{x}^3+\Big(\frac{1}{\text{x}^3}\Big)+3\times\text{3}$
$\Rightarrow\text{x}^3+\Big(\frac{1}{\text{x}^3}\Big)=27-9$
$\Rightarrow\text{x}^3+\Big(\frac{1}{\text{x}^3}\Big)=18$
Now, $\Big(\text{x}^3+\frac{1}{\text{x}^3}\Big)^2=\text{x}^6+\Big(\frac{1}{\text{x}^6}\Big)+2\times\text{x}^3\times\frac{1}{\text{x}^3}$
$\Rightarrow18^2=\text{x}^6+\Big(\frac{1}{\text{x}^6}\Big)+2$
$\text{x}^6+\Big(\frac{1}{\text{x}^6}\Big)=324-2=322$

$p(-1) = (-1)^3 + 3(-1)^2 + 3(-1) + 1$
$= -1 + 3 × 1 - 3 + 1$
$= -1 + 3 - 3 + 1$
$= 0$

Let: $p(x) = (x + 4)$
$\therefore p(-x) = (x + 4)$
$= -x + 4$
Thus, we have: $p(x) + p(-x) =\{(x + 4) + (-x + 4)\}$
$= 4 + 4$
$= 8$

$(x + 3)^3$
$= x^3 + (3)^3 + 3 × x × 3(x + 3)$
$= x^3 +27 + 9x^2 +27x$
$= x^3 + 9x^2 +27x + 27$
Therefore, the coefficient of $x,$ in the expansion of $(x + 3)^3$ is $27.$

$x - 3$ is a factor of $x^2 - ax - 15,$
then at $x = 3,$
$x^2 - ax - 15 = 0$
$i.e. (3)^2 - a(3) - 15 = 0$
$9 - 3a - 15 = 0$
$a = -2$

Given that,
$(3x - 1)^7 = a_7x^2 + a_5x^5 + \_\_\_ + a_1x + a_0$
Putting $x = 1$
We get$(3 × 1 - 1)^7 = a_6(1)^5 + a_5(1)^5 \_\_\_\_ + a_1(1) + a_0$
$⇒ a_6 + a_5 + \_\_\_\_ + a_1 + a_0 = 2^7 = 128$

If $p(x)$ is divided by $x + a,$ then the remainder is $p(-a).$
Here $p(x) = x^{51} + 51$ is divided by $x + 1,$ then
$x = -1$
Remainder $= p(-1) = (-1)^{51} + 51 = 50 = -1 + 51 = 50$

Finding a zero of $p(x)$ is the same as solving an equation $p(x) = 0.$
Now, $p(x) = 0 ⇒ 2x + 5 = 0,$
$2x = -5$
Which give us $\text{x}=-\frac{5}{2}.$
Therefore, $-\frac{5}{2}$ is the zero of the polynomial.

Volume of a cuboid of side $a, b$ and $c = abc$
Now, Volume $= 3x^2 - 27 ($given$)$
$abc = 3(x^2 - 9)$
$abc = 3(x - 3)(x + 3)$
So, possible dimensions are $3, x - 3$ and $x + 3$
Hence, correct option is $(b).$

$x^4 + 4$
$= x^4 + 4 + 4x^2 - 4x^2$
$= (x^4 + 4x^2 + 4) - 4x^2$
$= (x^2 + 2)^2 - (2x)^2$
$= (x^2 + 2 - 2x)(x^2 + 2 + 2x)$
$= (x^2 + 2x + 2)(x^2 - 2x + 2)$
Hence, correct option is $(a).$

We know, $(a - b)^2 = a^2 + b^2 + 2ab$
Here, $a = 75$ and $b = 25$
$⇒ (75 × 75) + (2 × 75 × 25) + (25 × 25)$
$= 75^2 + 2 × 75 × 25 + 25^2$
$= (75 + 25)^2$
$= (100)^2$
$= 10000$

 Assume $a = 0.013$ and $v = 0.007.$
Than the given expression can be rewritten as $\frac{\text{a}^3+\text{b}^3}{\text{a}^2-\text{ab}+\text{b}^2}$
Recall the formula for sum of two cubes
$a^3+b^3=(a+b)\left(a^2-a b+b^2\right)$
$$Using the above formula, the expression becomes $\frac{\text{(a}+\text{b)}(\text{a}^2-\text{ab}+\text{b}^2)}{(\text{a}^2-\text{ab}+\text{b}^2)}$
Note that both $a$ and $b$ are positive. So, neither $a^3 + b^3$ nor any factor of it can be zero.
Therefore we can cancel the term $(a^2 - ab + b^2)$ from both numerator and denominator. then the expression becomes
$\frac{(\text{a}+\text{b})(\text{a}^2-\text{ab}+\text{b}^2)}{\text{a}^2-\text{ab}+\text{b}^2}=\text{a}+\text{b}$
$=0.013+0.007$
$=0.02$

 A coefficient is a multiplicative factor in a term of a polynomial or any expression.
Therefore, the coefficient of $x^2$ in the polynomial $2 x^3+4 x^2+3 x+1$ is $4.$

 $x^2 + 5x + 4$ is a polynomial of deree $2.$
So, it is a quadratic polynomial.

 Given expression is $\left(x^2-1\right)\left(x^4+x^2+1\right)$
Let ${x}^2={A}$ and $1=\mathrm{B}$
Then, we have
$(A-B)\left(A^2+A B+B^2\right)$
$= A^3-B^3$
$= (X^2)^3-(1)^3$
$= X^6-1$
Hence, correct option is $(c).$

 Put $a = x + y$ and $b = x - y, $ then
$(x+y)^3-(x-y)^3=a^3-b^3$
$=(a-b)\left(a^2+b^2+a b\right)$
$=(x+y-x+y)\left[(x+y)^2+(x-y)^2+(x-y)(x+y)\right]$
$=2 y\left[2\left(x^2+y^2\right)+\left(x^2-y^2\right)\right]$
$=2 y\left[3 x^2+y^2\right]$

 Given: $x + y = 5 ⇒ x = 5 - y$
$x^3+y^3+15 x y-125$
$\text { Putting the value of } x \text {, we get }$
$(5-y)^3+y^3+15(5-y) y-125$
$=125-y^3-3 \times 5 \times y(5-y)+y^3+15(5-y) y-125$
$=125-y^3-75 y+15 y^2+y^3+75 y-15 y^2-125$
$=0$

 $x^2+4 y^2+4 y-4 x y-2 x-8$
$=x^2+(2 y)^2-2 \times x(2 y)+4 y-2 x-8$
$=(x-2 y)^2+4 y-2 x-8 \ldots(1)$
Now making eq$(1)$ a perfect square by adding $1$ and $-1$
$(x-2 y)^2+4 y-2 x-8=(x-2 y)^2+4 y-2 x-8+1-1$
$=(x-2 y)^2+(1)^2-2 \times(1) \times(x-2 y)-9$
$=(x-2 y-1)^2-(3)^2$
$=[(x-2 y-1)-3][x-2 y-1+3]$
$=(x-2 y-4)(x-2 y+2)$
Hence, correct option is $(a).$

 $(x-y)(x+y)=x^2-y^2\left[\text { by identity }(a+b)(a-b)=a^2-b^2\right]$
$\left(x^2-y^2\right)\left(x^2+y^2\right)=x^4-y^4$
$\left(x^4-y^4\right)\left(x^4+y^4\right)=x^8-y^8$
$\text { Now, }$
$(x-y)(x+y)\left(x^2+y^2\right)\left(x^4+y^4\right)$
$=\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)$
$=\left(x^4-y^4\right)\left(x^4+y^4\right)$
$=x^8-y^8$
Hence, correct option is $(b).$

$(x^2 - 4x - 21) = x^2 - 7x + 3x - 21$
$= x(x - 7) + 3(x - 7)$
$= (x - 7) (x + 3)$

 $(x+3)^3$
$=x^3+3^3+9 x(x+3)$
$=x^3+27+9 x^2+27 x$
So, the coefficient of $x$ in $(x+3)^3$ is $27.$

$a + b + c = 0 ⇒ a^3 + b^3 + c^3 = 3abc$
Thus, we have:
$\Big(\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}\Big)=\frac{\text{a}^3+\text{b}^3+\text{c}^3}{\text{abc}}$
$=\frac{3\text{abc}}{\text{abc}}$
$=3$

$ (102)^3 = (100 + 2)^3$
$= (100)^3 + (2)^3 + 3 × 100 × 2(100 + 2)$
$= 1000000 + 8 + 60000 + 1200$
$= 1061208$

 Put $x - 3 = 0,$ then $x = 3$
Therefore, value of $x^2 - ax - 15$ at $x = 3$ is zero
$⇒ 3^2 - 3a - 15 = 0$
$⇒ -6 - 3a = 0$
$⇒ a = -2$

 $\sqrt{3}$ is a constant term, so it is a polynomial of degree $0.$

$ (249)^2 - (248)^2$
$= (249 + 248) (249 - 248) [$Using identity $a^2 - b^2 = (a + b) (a - b)]$
$= 497 × 1$
$= 497$

$ p(x) = 2x^3 - kx^2 + 3x + 10$
$x + 2 = 0 ⇒ x = -2$
By the factor theorem, we know that when $p(x)$ is divided by $(x + 2),$ the remainder is $p(-2).$
Now, $p(-2) = 2(-2)^3 + k(-2)^2+ 3(-2) + 10$
$⇒ 0 = -16 - 4k - 6 + 10$
$⇒ 0 = -12 - 4k$
$⇒ 4k = -12$
$⇒ k = -3$

 Let $p(x)=x^3-3 x^2 a+2 a^2 x+b$
$(\mathrm{x}-\mathrm{a})$ is a factor of $\mathrm{p}(\mathrm{x})$.
So,
$p(a)=0$
$a^3-3 a^2 a+2 a^2 a+b=0$
$a^3-3 a^3+2 a^3+b=0$
$3 a^3-3 a^3+b=0$
$b=0$

 $(3 x-5)^3$
$=(3 x)^3-(5)^3-3 \times 3 x \times 5(3 x-5)$
$=27 x^3-125-135 x^2+225 x$
$=27 x^3-135 x^2+225 x-125$

 Let $f(x) = x^3 + 10x^2 + mx + n$
Now, $x + 2 = 0 ⇒ x = -2$
and $x - 1 = 0 ⇒ x = 1$
By factor theorem,
$f(-2) = 0$
$⇒ (-2)^3 + 10(-2)^2 + m(-2) + n$
$⇒ -8 + 40 - 2m + n = 0$
$⇒ 2m - n = 32 ...(i)$
By factor theorem,
$f(1) = 0$
$⇒ (1)^3 + 10(1)^2 + m(1) + n = 0$
$⇒ m + n = -11 ...(ii)$
Adding $(i)$ and $(ii),$ we get
$3m = 21$
$⇒ m = 7$
Substituting in $(ii),$ we get
$n = -18$

 Let $p(x) = 4x^3 + 3x^2 - 4x + k$
Now,
if $(x - 1)$ is a factor of $p(x),$ then at $x = 1, p(x) = 0$
So, $p(1) = 0$
$⇒ 4(1)^3 + 3(1)^2 - 4(1) + k = 0$
$⇒ 4 + 3 - 4 + k = 0$
$⇒ k = -3$

$\text { Given: } x+y+z=9 \text { and } x y+y z+z x=23$
$x^3+y^3+z^3-3 x y z=(x+y+z)\left(x^2+y^2+z^2-x y-y z-z x\right)$
$=(x+y+y)\left[(x+y+z)^2-2 x y-2 y z-2 z x-x y-y z-z x\right]$
$=(x+y+z)\left[(x+y+z)^2-3 x y-3 y z-3 z x\right]$
$=(x+y+z)\left[(x+y+x)^2-3(x y+y z+x x)\right]$
$=(9)\left[(9)^2-3(23)\right]$
$=9 \times[(81-69)]$
$=9 \times 12$
$=108$

$ x^3 + 2x^2 - x - 2$
$= x^2 (x + 2) - 1(x + 2)$
$= (x^2 - 1) (x + 2)$
$= (x + 1) (x - 1) (x + 2)$

 Clearly, $\sqrt2\text{x}^2-\sqrt3\text{x}+6$ is a polynomial in one variable because it has only non-negative integral powers of $x.$

Given, $a+b+c=9$
Hence, $(a+b+c)^2=81$
So, $a^2+b^2+c^2+2 a b+2 b c+2 c a=81$
i.e. $a^2+b^2+c^2+2(a b+b c+c a)=81$
i.e. $a^2+b^2+c^2+2(23)=81$
i.e. $a^2+b^2+c^2=81-46=35$
Now, $a^3+b^3+c^3-3 a b c$
$=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
$=(a+b+c)\left[\left(a^2+b^2+c^2\right)-(a b+b c+c a)\right]$
$=(9)[35-23]$
$=9 \times 12$
$=108$
Hence, correct option is $(a).$

The given polynomial is $p(x) = x^2 + x - 6$
Putting $x = 2$ in $p(x),$ we get
$p(2) = 2^2 + 2 - 6 = 4 + 2 - 6 = 0$
Therefore, $x = 2$ is a zero of the polynomial $p(x).$
Putting $x = -3$ in $p(x),$ we get
$p(-3) = (-3)^2 - 3 - 6 = 9 - 9 = 0$
Therefore, $x = -3$ is a zero of the polynomial $p(x)$
Thus, $2$ and $-3$ are the zeros of the given polynomial $p(x).$

Let $p(x)$ be a polynomial. If $\text{p}(\alpha)=0,$ then we say that $\alpha$ is a zero of a polynomial.
A polynomial consisting of one term, namely zero only, is called a zero polynomoial.
So, the zero of a zero polynomial is not defined.

$(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3$
Here,$a^2-b^2+b^2-c^2+c^2-a^2=0$
Therefore,
$(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3=3(a^2-b^2)(b^2-c^2)(c^2-a^2)$
${[\text { Since } x^3+y^3+z^3=3 x y z \text { if } x+y+z=0]}$
$(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3=$
$3(a+b)(b+c)(c+a)(a-b)(b-c)(c-a)$

$x^2 + kx -3 = (x - 3) (x + 1),$
$⇒ x^2 + kx - 3 = x^2 + (-3 + 1) x + (-3) × 1$
$⇒ x^2 + kx - 3 = x^2 = 2x - 3$
On comparing the term, we get $= -2$

Let $p(x) = x^2 + 3ax - 2a$ be the given polynomial.
$x - 2$ is a factor of $p(x).$
Thus,
$p(2) = 0$
$(2)^2 + 3a × 2 - 2a = 0$
$4 + 4a = 0$
$a = -1$

 For making perfect square to $x^4+x^2+25$
We add $+10 x^2$ and $-10 x^2$ to it.
$=x^4+x^2+25$
$=x^4+x^2+25+10 x^2-10 x^2$
$=[x^4+10 x^2+25]-9 x^2$
$=(x^2+5)^2+(3 x)^2$
$=[(x^2+5)+3 x][(x^2+5)-3 x]$
$=(x^2+3 x+5)(x^2-3 x+5)$
Hence, correct option is $(a).$

 $\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)=4 ($given$)$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=(4)^2-2=16-2=14\ ...(1)$
Squaring equation $(1)$
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=(14)^2$
$\Rightarrow(\text{x}^2)^2+\Big(\frac{1}{\text{x}^2}\Big)^2+2(\text{x}^2)\frac{1}{\text{x}^2}=196$
$\Rightarrow\text{x}^4+\frac{1}{\text{x}^4}=196-2$
$\Rightarrow\text{x}^4+\frac{1}{\text{x}^4}=194$
Hence, correct option is $(b).$

 A polynomial containing two non-zero terms is called a binomial.
Example: $3x + 4, 5y + 1, x^2 + x$

$ 2x^2 + 7x - 4$
$= 2x^2 + 8x - x - 4$
$= 2x(x + 4) -1(x + 4)$
$= (2x - 1) (x + 4)$
$2x - 1 = 0$ and $x + 4 = 0$
$\text{x}=\frac{1}{2}$ and $\text{x}-4$
Therefore, one zero of the given polynomial is $\frac{1}{2}$

$ (6x^2) + 17x + 5) = 6x^2 + 15x + 2x + 5$
$= 3x(2x + 5) + 1(2x + 5)$
$= (2x + 5) (3x + 1)$

 $\text{x}^3-\Big(\frac{1}{\text{x}^3}\Big)=110$
$\text{x}^3+\Big(\frac{1}{\text{x}^3}\Big)+3\text{x}\times\frac{1}{\text{x}}\Big(\text{x}+\frac{1}{\text{x}}\Big)=110+\text{3}\text{x}\times\frac{1}{\text{x}}\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)63=110+3\Big(\text{3}+\frac{1}{\text{x}}\Big)$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^3-3\Big(\text{x}+\frac{1}{\text{x}}\Big)-110=0$
Let $\text{x}+\frac{1}{\text{x}}=\text{a}$
$\Rightarrow\text{a}^3-3\text{a}-110=0$
$\Rightarrow\text{a}^3-5\text{a}^2+5\text{a}^2-25\text{a}+22\text{a}-110=0$
$\Rightarrow\text{a}^2(\text{a}-5)+5\text{a}(\text{a}-5)+22(\text{a}-5)=0$
$\Rightarrow(\text{a}-5)(\text{a}^2+5\text{a}+22)=0$
$\Rightarrow\text{a}-5=0$ or $\text{a}^2+5\text{a}+22=0$ neglected
$\Rightarrow\text{a} = 5$
$\Rightarrow\text{x}+\frac{1}{\text{x}}=5$

The degree of any constant term $5 ($say$)$
We can write $5$ as $5 \times 1 = 5x^\circ [$Since $a^\circ = 1]$
Therefore, the degree of any constant term is $0$

$(x + y - z)^2 = (x)^2 + (y)^2 + (-z)^2 + 2 × x × y + 2 × y × (-z) + 2 × (-z) ×x$
$= x^2 + y^2 + z^2 + 2xy - 2yz - 2zx$

$(x+y)^3-\left(x^3+y^3\right)$
$=x^3+y^3+3 x y(x+y)-\left(x^3+y^3\right)$
$=3 x y(x+y)$
Thus, the factors of $(x+y)^3-\left(x^3+y^3\right)$ are $3 x y$ and $(x+y)$

$x^2+k x+6=(x+2)(x+3)$
$\Rightarrow x^2+k x+6=x^2+(2+3) x+2 \times 3$
$\Rightarrow x^2+k x+6=x^2+5 x+6$
On comparing the terms,
We get $k=5$

We know that,
$(x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 z x$
$2 a^2+b^2+4+4 a b+8 a+4 b+4$
$=4 a^2+b^2+4+4 a b+8 a+4 b$
$=\left(2 a^2\right)+b^2+2^2+2(2 a) b+2(2 a)(2)+2(2 b)$
$=(2 a+b+2)^2$

$ x^{31} - 31$
Using remainder theorem.
$= (-1)^{31}- 31$
$= -1 - 31$
$= -32$

 When the polynomial $p(x)$ is divided by $q(x)$ i. e. $(\text{x}\pm\alpha)$ then $\text{p}(\mp\alpha)$ is the remainder.
If $\text{x}\pm\alpha$ is the factor of polynomial, then remainder is $'0'.$
So,
If $x^{51} + 51$ is divided $x + 1$.
Remainder $= (-1)^{51} + 51 = -1 + 51 = 50.$

 Now, $4x^2 + 8x + 3 = 4x^2  + 6x + 2x + 3 [$by splitting middle term$]$
$= 2x(2x + 3) + 1(2x + 3)$
$= (2x + 3) (2x + 1)$

 $\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=\text{x}^3+\frac{1}{\text{x}^3}+3\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^3-3\Big(\text{x}+\frac{1}{\text{x}}\Big)=\text{x}^3+\frac{1}{\text{x}^3}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^3-3\Big(\text{x}+\frac{1}{\text{x}}\Big)=110$
Let $\text{x}+\frac{1}{\text{x}}=\text{t}$
$\Rightarrow\text{t}^3-3\text{t}-110=0$
t =5 is one of it's solution which is real, other two solutions are imaginary
$\Rightarrow\text{x}+\frac{1}{\text{x}}=5$
Hence, correct option is $(a).$

If $x+1$ is a factor of $x^n+1$
then, at $x=-1, x^n+1=0$
$(-1)^n+1=0$
$(-1)^n=-1$
$(-1)^{\mathrm{n}}$ will be equal to $-1$ if and only if $n$ is an odd integer.
If $n$ is even, then $(-1)^{\mathrm{n}}=1$
So, $n$ should be an odd integer.

$x^3-x^2 y-x y^2+y^3=x^3+y^3-x y(x+y)$
Now by identity $x^3+y^3=(x+y)\left(x^2+y^2-x y\right)$, we have
$x^3-x^2 y-x y^2+y^3=(x+y)\left(x^2+y^2-x y\right)-x y(x+y)$
$=(x+y)\left(x^2+y^2-x y-x y\right)$
$=(x+y)\left(x^2+y^2-2 x y\right)$
$=(x+y)(x-y)^2$
Hence, correct option is $(d).$

$(x+3)^3=x^3+(3)^3+3 \times x \times 3(x+3)=x^3+27+9 x^2+27 x=x^3+9 x^2+27 x+27$
Therefore, the coefficient of $x$. in the expansion of $(x+3)^3$ is $27.$

 $(x+3)^4$
$=(x+3)^2 \times(x+3)^2$
$=\left[x^2+6 x+9\right]\left[x^2+6 x+9\right]$
$=x^4+36 x^2+81+12 x^3+108 x+18 x^2$
$=x^4+12 x^3+54 x^2+108 x+81$
Therefore, the coefficient of $x^2$ is $54 .$

$\text{p}(\text{x})=\text{x}^2-2\sqrt2\text{x}+1$
$\text{p}(2\sqrt2)=(2\sqrt2)^2-2\sqrt2(2\sqrt2)+1$
$=8-8+1$
$=1$

Let $p(x) = 5x - 4x^2 + 3 ... (i)$
On putting $x= -1$ in eq. $(i),$ we get
$p(-1) = 5(-1) - 4(-1)^2 + 3 = -5 - 4 + 3 = -6$

$\text{a}^{\frac{1}{3}}+\text{b}^{\frac{1}{3}}+\text{c}^{\frac{1}{3}}=0$
$\Rightarrow\text{a}^{\frac{1}{3}}+\text{b}^{\frac{1}{3}}=-\text{c}^{\frac{1}{3}}$
$\Rightarrow\Big[(\text{a}^{\frac{1}{3}})(\text{b}^{\frac{1}{3}})\Big]^3=\Big(-\text{c}^{\frac{1}{3}}\Big)^3$
$\Rightarrow\text{a}+\text{b}+\Big[3\times\text{a}^{\frac{1}{3}}\times\text{b}^{\frac{1}{3}}\Big(\text{a}^{\frac{1}{3}}+\text{b}^{\frac{1}{3}}\Big)\Big]=-\text{c}$
$\Rightarrow\text{a}+\text{b}+3\times\text{a}^{\frac{1}{3}}\times\text{b}^{\frac{1}{3}}\Big(-\text{c}^{\frac{1}{3}}\Big)=-\text{c}$
$\Rightarrow\text{a}+\text{b}+\text{c}=3\times\text{a}^{\frac{1}{3}}\times\text{b}^{\frac{1}{3}}\times\text{c}^{\frac{1}{3}}$
$\Rightarrow(\text{a}+\text{b}+\text{c})^3=\Big(3\times\text{a}^{\frac{1}{3}}\times\text{b}^{\frac{1}{3}}\times\text{c}^{\frac{1}{3}}\Big)$
$\Rightarrow(\text{a}+\text{b}+\text{c})^3=27\text{abc}$

 $4 a^2+4 a-3$
To find the length and breadth, we will factorize the given polynomial.
$=4 a^2-6 a-2 a-3$
$=2 a(a+3)-1(2 a+3)$
$=(2 a+3)(2 a-1)$
Therefore, the Possible expressions for the length and breadth of the rectangle whose area is given by $4 a^2+4 a-3$ is ( $2 a+3$ ) and $(2 a-1)$.

 The zero of the polynomial $p(x)$ can be obtained by putting $p(x)$
$p(x) = 0$
$⇒ 2x + 5 = 0$
$⇒ 2x = -5$
$\Rightarrow\text{x}=\frac{-5}{2}$

 Now, $4 x^2+8 x+3=4 x^2+6 x+2 x+3 [$by splitting middle term$]$
$= 2x(2x + 3) + 1 (2x + 3)$
$= (2x + 3)(2x + 1)$

 $(a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a$
$\Rightarrow(9)^2=a^2+b^2+c^2+2(a b+b c+c a)$
$\Rightarrow(9)^2=a^2+b^2+c^2+2(23)$
$\Rightarrow a^2+b^2+c^2=81-46=35$
as we know that $a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
$\Rightarrow a^3+b^3+c^3-3 a b c=9 \times(35-23)$
$\Rightarrow a^3+b^3+c^3-3 a b c=108$

 A polynomial of degree $3$ is called a cubic polynomial.
Its general form is $ax^3 + bx^2 + cx + d.$

 The linear polynomial $(x - 1)$ is a factor of $xn + 1,$ only if,
$f(-1) = (-1)^n+ 1 = 0$ 
If n is odd integer, then $f(-1) = -1 + 1 = 0$

 $x^3+x^2+x+1=x^3(x+1)+1(x+3)=x^3+27~9 x^2+27 x=x^3+9 x^2+27 x+27$
therefore, the coefficient of $x$, in the expansion of $(x+3)^3$ is $27$ .

 $\Big(\text{x}-\frac{1}{\text{x}}\Big)^3=\text{x}^3-\frac{1}{\text{x}^3}-3\not\text{x}\frac{1}{\not\text{x}}\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\text{x}^3-\frac{1}{\text{x}^3}=\Big(\text{x}-\frac{1}{\text{x}}\Big)^3+3\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^3+3\Big(\text{x}-\frac{1}{\text{x}}\Big)-\text{x}^3-\frac{1}{\text{x}^3}=0$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^3+3\Big(\text{x}-\frac{1}{\text{x}}\Big)-14=0$
Let $\Rightarrow\text{x}-\frac{1}{\text{x}}=\text{t}$
$\Rightarrow t^3+3 t-14=0$
$\Rightarrow t^3-2 t^2+2 t^2-4 t+7 t-14=0$
$\Rightarrow t(t-2)+2 t(t-2)+7(t-2)=0$
$\Rightarrow(t-2)(t+2 t+7)=0$
$\Rightarrow t^2+2 t+7=0 \text { has no real roots }$
So, $t = 2$ is a solution
$\Rightarrow\text{x}-\frac{1}{\text{x}}=2$
Hence, correct option is $(d).$

 $a+b+c=0$
$\Rightarrow a+b=-c$
$\Rightarrow(a+b)^3=(-c)^3$
$\Rightarrow a^3+b^3+3 a b(a+b)=-c^3$
$\Rightarrow a^3+b^3+3 a b(-c)=-c^3$
$\Rightarrow a^3+b^3+c^3=3 a b c$

$ 49\text{x}^2-\text{k}=\Big(7\text{x}+\frac{1}{3}\Big)\Big(7\text{x}-\frac{1}{3}\Big),$
$\Rightarrow\text{49}\text{x}^\text{2}-\text{k}=49\text{x}^2-\frac{1}{9} [$Using identity $(a + b) (a - b) = a^2 -b^2]$
On comapring $\text{k}=\frac{1}{9}$

 $p(x)=x^3-3 x^2+4 x+32$
$x+2=0 \Rightarrow x=-2$
By the renainder theorem, we know that when $p(x)$ is divided by
$(\mathrm{x}+2)$, the remainder is $\mathrm{p}(-2)$.
$\text { Now, } p(-2)=x^3-3 x^2+4 x+32$
$=(-2)^3-3(-2)^2+4(-2)+32$
$=-8-12-8+32$
$=4$

 $\frac{\text{x}}{\text{y}}+\frac{\text{y}}{\text{x}}=-1$
$\Rightarrow\frac{\text{x}^2+\text{y}^2}{\text{xy}}=-1$
$\Rightarrow x^2+y^2=-x y$
$\Rightarrow x^2+y^2+x y=0$
Thus, we have:
$\left(x^3-y^3\right)=(x-y)\left(x^2+y^2+x y\right)$
$=(x-y) \times 0$
$=0$