Mix Examples - Polynomials Questions in English

(C) For a quadratic polynomial of the form $ax^2 + bx + c$,the sum of the zeros is given by the formula $-\frac{b}{a}$.
Comparing $p(x) = 4x^2 + 12x + 5$ with $ax^2 + bx + c$,we get $a = 4$,$b = 12$,and $c = 5$.
Sum of zeros = $-\frac{b}{a} = -\frac{12}{4} = -3$.
Therefore,the correct option is $C$.

(A) For a quadratic polynomial of the form $p(x) = ax^{2} + bx + c$,the product of the zeros is given by the formula $\text{Product} = c/a$.
Comparing $p(x) = 4x^{2} + 12x + 5$ with the standard form,we have $a = 4$,$b = 12$,and $c = 5$.
Therefore,the product of the zeros is $c/a = 5/4$.

(C) quadratic polynomial with zeros $\alpha$ and $\beta$ is given by the formula: $p(x) = k[x^2 - (\alpha + \beta)x + (\alpha\beta)]$,where $k$ is a constant.
Given that the sum of the zeros $(\alpha + \beta) = -7$ and the product of the zeros $(\alpha\beta) = 12$.
Substituting these values into the formula:
$p(x) = x^2 - (-7)x + (12)$
$p(x) = x^2 + 7x + 12$
Thus,the correct option is $C$.

(D) The general form of a quadratic polynomial is given by $p(x) = k[x^{2} - (\text{sum of zeros})x + (\text{product of zeros})],$ where $k$ is a non-zero constant.
Given that the sum of the zeros is $2$ and the product of the zeros is $3.$
Substituting these values into the formula,we get $p(x) = k[x^{2} - 2x + 3].$
Comparing this with the given options,if we take $k = 4,$ the polynomial becomes $p(x) = 4(x^{2} - 2x + 3).$
Therefore,the correct option is $D.$

(A) quadratic polynomial is given by the formula $p(x) = k[x^2 - (\text{sum of zeros})x + (\text{product of zeros})]$.
Given that the sum of the zeros is $-3$ and the product of the zeros is $-4$.
Substituting these values into the formula,we get:
$p(x) = k[x^2 - (-3)x + (-4)]$
$p(x) = k[x^2 + 3x - 4]$
Therefore,the correct quadratic polynomial is $p(x) = k(x^2 + 3x - 4)$.

(D) The general form of a quadratic polynomial is given by $p(x) = ax^2 + bx + c$.
Given the values $a = 6$,$b = 11$,and $c = 4$.
Substituting these values into the general form,we get:
$p(x) = (6)x^2 + (11)x + (4)$
$p(x) = 6x^2 + 11x + 4$.
Therefore,the correct option is $D$.

(A) The general form of a cubic polynomial is given by $p(x) = ax^{3} + bx^{2} + cx + d$.
Given the coefficients:
$a = 3$
$b = -5$
$c = -11$
$d = -3$
Substituting these values into the general form,we get:
$p(x) = (3)x^{3} + (-5)x^{2} + (-11)x + (-3)$
$p(x) = 3x^{3} - 5x^{2} - 11x - 3$
Therefore,the correct option is $A$.

(B) To find the remainder when a polynomial $p(x) = x^{2} + 6x + 10$ is divided by $x + 2$,we use the Remainder Theorem.
According to the Remainder Theorem,if a polynomial $p(x)$ is divided by $(x - a)$,the remainder is $p(a)$.
Here,the divisor is $x + 2$,which can be written as $x - (-2)$.
Therefore,$a = -2$.
Now,we calculate $p(-2)$:
$p(-2) = (-2)^{2} + 6(-2) + 10$
$p(-2) = 4 - 12 + 10$
$p(-2) = 14 - 12$
$p(-2) = 2$
Thus,the remainder is $2$.

(B) Let the two polynomials be $P(x)$ and $Q(x)$.
Given that $P(x) \cdot Q(x) = x^{2}+8x+15$.
We are given one polynomial $P(x) = x+3$.
To find the other polynomial $Q(x)$,we divide the product by the given polynomial:
$Q(x) = \frac{x^{2}+8x+15}{x+3}$.
We can factorize the quadratic polynomial $x^{2}+8x+15$ by splitting the middle term:
$x^{2}+5x+3x+15 = x(x+5)+3(x+5) = (x+3)(x+5)$.
Substituting this back into the division:
$Q(x) = \frac{(x+3)(x+5)}{x+3} = x+5$.
Therefore,the other polynomial is $x+5$.

(D) Let the two polynomials be $P(x)$ and $Q(x)$.
Given that $P(x) \times Q(x) = x^{2}-x-72$.
One polynomial is $P(x) = x+8$.
To find the other polynomial $Q(x)$,we divide the product by the given polynomial:
$Q(x) = \frac{x^{2}-x-72}{x+8}$.
We can factorize the quadratic expression $x^{2}-x-72$ by splitting the middle term:
$x^{2}-9x+8x-72 = x(x-9)+8(x-9) = (x+8)(x-9)$.
Therefore,$Q(x) = \frac{(x+8)(x-9)}{x+8} = x-9$.
Thus,the other polynomial is $x-9$.

(C) Given that $1$ is a zero of the polynomial $p(x) = ax^2 - 11x + 3$.
Therefore,$p(1) = 0$.
Substituting $x = 1$ into the polynomial:
$a(1)^2 - 11(1) + 3 = 0$
$a - 11 + 3 = 0$
$a - 8 = 0$
$a = 8$.

(C) Given that $3$ is a zero of the polynomial $p(x) = 3x^{3} - x^{2} - ax - 45$,it implies that $p(3) = 0$.
Substituting $x = 3$ into the polynomial:
$3(3)^{3} - (3)^{2} - a(3) - 45 = 0$
$3(27) - 9 - 3a - 45 = 0$
$81 - 9 - 3a - 45 = 0$
$27 - 3a = 0$
$3a = 27$
$a = 9$

(A) To find the quotient when $2x^3 - x^2 - 2x - 8$ is divided by $x - 2$,we can use synthetic division:
$\begin{array}{c|rrrr} 2 & 2 & -1 & -2 & -8 \\ & 0 & 4 & 6 & 8 \\ \hline & 2 & 3 & 4 & 0 \end{array}$
The coefficients of the quotient polynomial are $2, 3, 4$.
Therefore,the quotient polynomial is $2x^2 + 3x + 4$ and the remainder is $0$.

(D) polynomial of degree $2$ is called a quadratic polynomial.
In option $A$,$p(x) = 2 - 4x$ is a linear polynomial (degree $1$).
In option $B$,$p(x) = 2x^3 - 1$ is a cubic polynomial (degree $3$).
In option $C$,$p(x) = 2x - 5$ is a linear polynomial (degree $1$).
In option $D$,$p(x) = 1 - x^2$ is a quadratic polynomial because the highest power of the variable $x$ is $2$.

(B) polynomial is defined as an expression consisting of variables and coefficients,where the exponents of the variables must be non-negative integers.
$A$ cubic polynomial is a polynomial of degree $3$,meaning the highest power of the variable $x$ is $3$.
In option $A$,the term $\sqrt{x}$ is $x^{1/2}$,which is not an integer.
In option $B$,$p(x) = 2 - 3x - x^3$. The highest power of $x$ is $3$,and all exponents are non-negative integers. Thus,it is a cubic polynomial.
In option $C$,$\sqrt{x^3} = x^{3/2}$,which is not an integer.
In option $D$,the exponent $1/3$ is not an integer.
Therefore,the correct option is $B$.

(A) According to the Factor Theorem,if $(x-a)$ is a factor of a polynomial $p(x)$,then $p(a) = 0$.
Here,the given factor is $(x-1)$,so we must have $p(1) = 0$.
Let us check each option:
$A) p(1) = (1)^{2} + 2(1) - 3 = 1 + 2 - 3 = 0$. Since $p(1) = 0$,$(x-1)$ is a factor.
$B) p(1) = (1)^{2} + 4(1) + 3 = 1 + 4 + 3 = 8 \neq 0$.
$C) p(1) = (1)^{2} + 5(1) + 6 = 1 + 5 + 6 = 12 \neq 0$.
$D) p(1) = (1)^{2} + (1) - 6 = 1 + 1 - 6 = -4 \neq 0$.
Therefore,the correct option is $A$.

(D) According to the Factor Theorem,$(x+1)$ is a factor of a polynomial $p(x)$ if and only if $p(-1) = 0$.
For option $A$: $p(-1) = (-1)^3 + 2(-1)^2 + 2(-1) + 1 = -1 + 2 - 2 + 1 = 0$. Thus,$(x+1)$ is a factor.
For option $B$: $p(-1) = (-1)^3 - 2(-1) - 1 = -1 + 2 - 1 = 0$. Thus,$(x+1)$ is a factor.
For option $C$: $p(-1) = (-1)^3 + 2(-1)^2 - 1 = -1 + 2 - 1 = 0$. Thus,$(x+1)$ is a factor.
For option $D$: $p(-1) = (-1)^3 - 1 = -1 - 1 = -2$. Since $p(-1) \neq 0$,$(x+1)$ is not a factor of $p(x) = x^3 - 1$.

(D) polynomial of degree $1$ is called a linear polynomial.
The graph of a linear polynomial $p(x) = ax + b$ (where $a \neq 0$) is always a straight line.
In the given options:
$A) p(x) = x^{2} + 8$ is a quadratic polynomial (degree $2$).
$B) p(x) = x^{3} - 2$ is a cubic polynomial (degree $3$).
$C) p(x) = x^{2} + 6x + 9$ is a quadratic polynomial (degree $2$).
$D) p(x) = 5x - 10$ is a linear polynomial (degree $1$).
Therefore,the graph of $p(x) = 5x - 10$ is a line.

(A) The graph of a quadratic polynomial $p(x) = ax^2 + bx + c$ is a parabola.
If $a > 0$,the parabola opens upwards.
If $a < 0$,the parabola opens downwards.
In option $A$,$p(x) = x^2 + 8x + 12$,here $a = 1$,which is greater than $0$.
Therefore,the graph of $p(x) = x^2 + 8x + 12$ is a curve open upwards.

(B) For a quadratic polynomial $p(x) = ax^{2} + bx + c$,the graph is a parabola.
If $a > 0$,the parabola opens upwards.
If $a < 0$,the parabola opens downwards.
Let us examine the coefficient of $x^{2}$ for each option:
$A) p(x) = x^{2} - 4x + 3$,here $a = 1 > 0$ (opens upwards).
$B) p(x) = 6x - x^{2} - 9$,which is $p(x) = -x^{2} + 6x - 9$. Here $a = -1 < 0$ (opens downwards).
$C) p(x) = 3 - 6x + 3x^{2}$,here $a = 3 > 0$ (opens upwards).
$D) p(x) = x^{2} + 6x + 5$,here $a = 1 > 0$ (opens upwards).
Therefore,the correct option is $B$.

(B) The number of real zeros of a polynomial $y = p(x)$ is equal to the number of points where the graph of the polynomial intersects the $x$-axis.
In the given figure,the line representing $y = p(x)$ intersects the $x$-axis at exactly one point.
Therefore,the number of real zeros of $y = p(x)$ is $1$.

(B) The number of real zeros of a polynomial $y=p(x)$ is equal to the number of points where the graph of the polynomial intersects the $x$-axis.
In the given figure,the curve representing $y=p(x)$ intersects the $x$-axis at exactly $2$ distinct points.
Therefore,the number of real zeros of $y=p(x)$ is $2$.

(D) The number of real zeros of a polynomial $y=p(x)$ is equal to the number of points where the graph of the polynomial intersects the $X$-axis.
In the given figure,the parabola $y=p(x)$ lies entirely above the $X$-axis and does not intersect or touch the $X$-axis at any point.
Therefore,the number of real zeros of the polynomial $p(x)$ is $0$.

(A) The number of real zeros of a polynomial $y=p(x)$ is equal to the number of points where the graph of the polynomial intersects the $X$-axis.
In the given figure,the parabola $y=p(x)$ touches the $X$-axis at only one point,which is the origin $(0, 0)$.
Since the graph intersects/touches the $X$-axis at exactly one point,the number of real zeros is $1$.

(B) The number of real zeros of a polynomial $y=p(x)$ is equal to the number of points where the graph of the polynomial intersects the $x$-axis.
In the given figure,the curve $y=p(x)$ intersects the $x$-axis at $3$ distinct points.
Therefore,the number of real zeros of $p(x)$ is $3$.

(B) The number of real zeros of a polynomial $y = p(x)$ is equal to the number of points where the graph of the polynomial intersects the $x$-axis.
In the given figure,the parabola $y = p(x)$ intersects the $x$-axis at exactly $2$ distinct points.
Therefore,the number of real zeros of the polynomial is $2$.

(A) The number of real zeros of a polynomial $y=p(x)$ is equal to the number of points where the graph of the polynomial intersects the $x$-axis.
In the given figure,the curve $y=p(x)$ intersects the $x$-axis at $4$ distinct points.
Therefore,the number of real zeros of $p(x)$ is $4$.

(A) To find the zeros of the polynomial $p(x) = 4x^3 - x$,we set $p(x) = 0$.
$4x^3 - x = 0$
Factor out $x$ from the expression:
$x(4x^2 - 1) = 0$
Using the difference of squares formula $a^2 - b^2 = (a - b)(a + b)$,we can factor $4x^2 - 1$ as $(2x - 1)(2x + 1)$:
$x(2x - 1)(2x + 1) = 0$
Setting each factor to zero,we get:
$x = 0$,$2x - 1 = 0 \implies x = 1/2$,and $2x + 1 = 0 \implies x = -1/2$.
Thus,the zeros are $0, 1/2, -1/2$.
There are $3$ zeros in total.

(C) To find the zeros of the polynomial $p(x) = x^{2} - 2x - 15$,we set $p(x) = 0$.
$x^{2} - 2x - 15 = 0$
We need to factorize the quadratic equation by splitting the middle term. We look for two numbers whose product is $-15$ and whose sum is $-2$.
These numbers are $-5$ and $3$.
$x^{2} - 5x + 3x - 15 = 0$
$x(x - 5) + 3(x - 5) = 0$
$(x - 5)(x + 3) = 0$
Setting each factor to zero:
$x - 5 = 0 \implies x = 5$
$x + 3 = 0 \implies x = -3$
Therefore,the zeros of the polynomial are $5$ and $-3$.

(A) quadratic polynomial with zeros $\alpha$ and $\beta$ is given by the formula: $p(x) = k[x^{2} - (\alpha + \beta)x + (\alpha \cdot \beta)]$,where $k$ is a constant.
Given zeros are $\alpha = -3$ and $\beta = 4$.
Sum of zeros: $\alpha + \beta = -3 + 4 = 1$.
Product of zeros: $\alpha \cdot \beta = (-3) \times 4 = -12$.
Substituting these values into the formula (taking $k=1$): $p(x) = x^{2} - (1)x + (-12) = x^{2} - x - 12$.
Therefore,the correct polynomial is $p(x) = x^{2} - x - 12$.

(B) Given the quadratic polynomial $p(x) = x^{2} - 3x + 2$.
Comparing this with the standard form $ax^{2} + bx + c$,we get $a = 1$,$b = -3$,and $c = 2$.
For a quadratic polynomial,the sum of zeros $\alpha + \beta = -\frac{b}{a} = -(\frac{-3}{1}) = 3$.
The product of zeros $\alpha \beta = \frac{c}{a} = \frac{2}{1} = 2$.
We need to find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$.
Taking the common denominator,we get $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta}$.
Substituting the values,$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{2}$.

(B) For a cubic polynomial $p(x) = ax^3 + bx^2 + cx + d,$ the relationships between zeros and coefficients are:
$1$. Sum of zeros: $\alpha + \beta + \gamma = -\frac{b}{a}$
$2$. Sum of product of zeros taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$
$3$. Product of zeros: $\alpha\beta\gamma = -\frac{d}{a}$
Now,we need to find the value of $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}.$
Taking the least common multiple $(LCM)$ of the denominators:
$\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma}$
Substituting the values from the relationships:
$= \frac{c/a}{-d/a} = \frac{c}{a} \times \left(-\frac{a}{d}\right) = -\frac{c}{d}.$

(B) For a cubic polynomial $p(x) = ax^3 + bx^2 + cx + d$,where $a, b, c, d$ are real numbers and $a \neq 0$.
Let the zeros of the polynomial be $\alpha, \beta,$ and $\gamma$.
According to the relationship between zeros and coefficients of a cubic polynomial:
$1$. The sum of zeros: $\alpha + \beta + \gamma = -\frac{b}{a}$.
$2$. The sum of the product of zeros taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$.
$3$. The product of zeros: $\alpha\beta\gamma = -\frac{d}{a}$.
Here,$b$ is the coefficient of $x^2$ and $a$ is the coefficient of $x^3$.
Therefore,$\alpha + \beta + \gamma = -\frac{\text{coefficient of } x^2}{\text{coefficient of } x^3}$.

(C) For a cubic polynomial $p(x) = ax^3 + bx^2 + cx + d$,the relationship between the zeros $\alpha, \beta, \gamma$ and the coefficients is given by Vieta's formulas:
$1$. Sum of zeros: $\alpha + \beta + \gamma = -\frac{b}{a} = -\frac{\text{coefficient of } x^2}{\text{coefficient of } x^3}$
$2$. Sum of the product of zeros taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{\text{coefficient of } x}{\text{coefficient of } x^3}$
$3$. Product of zeros: $\alpha\beta\gamma = -\frac{d}{a} = -\frac{\text{constant term}}{\text{coefficient of } x^3}$
Therefore,$\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{\text{coefficient of } x}{\text{coefficient of } x^3}$.

(D) For a cubic polynomial $p(x) = ax^3 + bx^2 + cx + d$,where $a \neq 0$,let the zeros be $\alpha, \beta,$ and $\gamma$.
According to the relationship between zeros and coefficients:
$1$. Sum of zeros: $\alpha + \beta + \gamma = -\frac{b}{a} = -\frac{\text{coefficient of } x^2}{\text{coefficient of } x^3}$
$2$. Sum of product of zeros taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{\text{coefficient of } x}{\text{coefficient of } x^3}$
$3$. Product of zeros: $\alpha\beta\gamma = -\frac{d}{a} = -\frac{\text{constant term}}{\text{coefficient of } x^3}$
Since the given options do not include $-\frac{d}{a}$,the correct choice is $D$.

(C) For a cubic polynomial $p(x) = ax^3 + bx^2 + cx + d$,the relationship between the zeros and the coefficients is given by Vieta's formulas.
Let the zeros be $\alpha, \beta$,and $\gamma$.
The sum of the zeros is given by the formula:
$\alpha + \beta + \gamma = -\frac{\text{coefficient of } x^2}{\text{coefficient of } x^3}$
Substituting the coefficients from the given polynomial:
$\alpha + \beta + \gamma = -\frac{b}{a}$
Therefore,the correct option is $C$.

(A) For a cubic polynomial $p(x) = ax^3 + bx^2 + cx + d$,where $a, b, c, d$ are real numbers and $a \neq 0$,let the zeros be $\alpha, \beta$,and $\gamma$.
According to the relationship between the coefficients and the zeros of a polynomial:
$1$. The sum of the zeros: $\alpha + \beta + \gamma = -\frac{b}{a}$
$2$. The sum of the product of the zeros taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$
$3$. The product of the zeros: $\alpha\beta\gamma = -\frac{d}{a}$
Therefore,the value of $\alpha\beta + \beta\gamma + \gamma\alpha$ is $\frac{c}{a}$.

(B) For a cubic polynomial $p(x) = ax^3 + bx^2 + cx + d$ with zeros $\alpha, \beta,$ and $\gamma$:
$1$. The sum of zeros is $\alpha + \beta + \gamma = -\frac{b}{a}$.
$2$. The sum of the product of zeros taken two at a time is $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$.
$3$. The product of the zeros is $\alpha\beta\gamma = -\frac{d}{a}$.
Thus,the correct option is $B$.

(B) For a cubic polynomial of the form $p(x) = ax^{3} + bx^{2} + cx + d$,the sum of the zeros $\alpha + \beta + \gamma$ is given by the formula $-\frac{b}{a}$.
Given the polynomial $p(x) = x^{3} - 6x^{2} - 7x$,we identify the coefficients as $a = 1$,$b = -6$,$c = -7$,and $d = 0$.
Substituting these values into the formula:
$\alpha + \beta + \gamma = -\frac{-6}{1} = 6$.

(C) For a cubic polynomial of the form $p(x) = ax^{3} + bx^{2} + cx + d$,the product of the zeros $\alpha \beta \gamma$ is given by the formula $\alpha \beta \gamma = -\frac{d}{a}$.
Comparing the given polynomial $p(x) = x^{3} - 3x^{2} - 6x + 8$ with the standard form,we have $a = 1$,$b = -3$,$c = -6$,and $d = 8$.
Substituting these values into the formula,we get $\alpha \beta \gamma = -\frac{8}{1} = -8$.

(A) For a cubic polynomial of the form $p(x) = ax^{3} + bx^{2} + cx + d$,the relationship between the coefficients and the zeros $\alpha, \beta, \gamma$ is given by Vieta's formulas.
Specifically,the sum of the product of zeros taken two at a time is given by $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$.
Comparing the given polynomial $p(x) = x^{3} + x^{2} - 17x + 15$ with the standard form,we have $a = 1$,$b = 1$,$c = -17$,and $d = 15$.
Substituting these values into the formula:
$\alpha\beta + \beta\gamma + \gamma\alpha = \frac{-17}{1} = -17$.

(B) According to the Fundamental Theorem of Algebra,a polynomial of degree $n$ has at most $n$ real zeros.
Given that the degree of the polynomial is $k+1$,the maximum number of zeros it can have is equal to its degree.
Therefore,the maximum number of zeros is $k+1$.

(B) The given expression $p(x) = ax^4 + bx^3 + cx^2 + dx + e$ is a polynomial of degree $4$ because the highest power of the variable $x$ is $4$ and $a \neq 0$.
According to the Fundamental Theorem of Algebra,a polynomial of degree $n$ has at most $n$ real zeros.
Since the degree of the given polynomial is $4$,the maximum number of zeros it can have is $4$.

(C) Given the cubic polynomial $p(x) = ax^3 + bx^2 + cx + d$,the relationships between the zeros $\alpha, \beta, \gamma$ and the coefficients are:
$\alpha + \beta + \gamma = -\frac{b}{a}$
$\alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a}$
$\alpha \beta \gamma = -\frac{d}{a}$
We need to evaluate the expression: $\alpha^2 \beta \gamma + \alpha \beta^2 \gamma + \alpha \beta \gamma^2$
Factor out the common term $\alpha \beta \gamma$:
$= \alpha \beta \gamma (\alpha + \beta + \gamma)$
Substitute the known values:
$= (-\frac{d}{a}) \times (-\frac{b}{a})$
$= \frac{bd}{a^2}$

(C) Given the polynomial $p(x) = 3x^2 - x - 4$,comparing it with $ax^2 + bx + c$,we have $a = 3$,$b = -1$,and $c = -4$.
The sum of the zeros is $\alpha + \beta = -\frac{b}{a} = -\frac{-1}{3} = \frac{1}{3}$.
The product of the zeros is $\alpha \beta = \frac{c}{a} = \frac{-4}{3} = -\frac{4}{3}$.
We need to find the value of $\alpha^2 \beta + \alpha \beta^2$.
Factoring out $\alpha \beta$,we get $\alpha^2 \beta + \alpha \beta^2 = \alpha \beta(\alpha + \beta)$.
Substituting the values,we get $\left(-\frac{4}{3}\right) \left(\frac{1}{3}\right) = -\frac{4}{9}$.

(A) polynomial of degree $n$ has at most $n$ zeros.
For $A$: $p(x) = x^3 + 8$. This is a cubic polynomial (degree $3$),so it has $3$ zeros. Thus,this statement is true.
For $B$: $p(x) = x^2 - 36$. This is a quadratic polynomial (degree $2$),so it has at most $2$ zeros. This statement is false.
For $C$: $p(x) = 4x^3 + x = x(4x^2 + 1)$. The zeros are $x = 0$ and $4x^2 = -1$ (no real zeros). It has only $1$ real zero,though $3$ complex zeros. However,in the context of standard school mathematics,we often look for real roots. Even if we consider complex roots,$A$ is a more direct example of a cubic polynomial having $3$ roots.
For $D$: $p(x) = 4x^3 - x = x(4x^2 - 1) = x(2x - 1)(2x + 1)$. The zeros are $x = 0, 1/2, -1/2$. This also has $3$ zeros.
Note: Both $A, C,$ and $D$ have $3$ zeros. However,$p(x) = x^3 + 8$ is a standard cubic form. Given the options,$A$ is a correct statement.

(C) To find the degree of the polynomial $p(x) = x^2(1 + x + x^2) + 5$,we first expand the expression.
By distributing $x^2$ into the parentheses,we get:
$p(x) = x^2(1) + x^2(x) + x^2(x^2) + 5$
$p(x) = x^2 + x^3 + x^4 + 5$
Rearranging the terms in descending order of their powers,we have:
$p(x) = x^4 + x^3 + x^2 + 5$
The degree of a polynomial is defined as the highest power of the variable $x$ present in the expression.
In this polynomial,the highest power of $x$ is $4$.
Therefore,the degree of $p(x)$ is $4$.

(D) To find the cost of one camera,we divide the total cost by the number of cameras.
Cost of one camera = $\frac{x^{3}+3x^{2}+5x+3}{x+1}$.
Using polynomial long division:
$1$. Divide $x^{3}$ by $x$ to get $x^{2}$.
$2$. Multiply $x^{2}(x+1) = x^{3}+x^{2}$.
$3$. Subtract $(x^{3}+3x^{2}+5x+3) - (x^{3}+x^{2}) = 2x^{2}+5x+3$.
$4$. Divide $2x^{2}$ by $x$ to get $2x$.
$5$. Multiply $2x(x+1) = 2x^{2}+2x$.
$6$. Subtract $(2x^{2}+5x+3) - (2x^{2}+2x) = 3x+3$.
$7$. Divide $3x$ by $x$ to get $3$.
$8$. Multiply $3(x+1) = 3x+3$.
$9$. Subtract $(3x+3) - (3x+3) = 0$.
The quotient is $x^{2}+2x+3$. Thus,the cost of one camera is ₹ $(x^{2}+2x+3)$.

(A) To find the remainder when the polynomial $p(x) = x^{3}+2x^{2}+3x+5$ is divided by $x+1$,we use the Remainder Theorem.
According to the Remainder Theorem,if a polynomial $p(x)$ is divided by $(x-a)$,the remainder is $p(a)$.
Here,the divisor is $x+1$,which can be written as $x-(-1)$. Thus,$a = -1$.
Now,we calculate $p(-1)$:
$p(-1) = (-1)^{3} + 2(-1)^{2} + 3(-1) + 5$
$p(-1) = -1 + 2(1) - 3 + 5$
$p(-1) = -1 + 2 - 3 + 5$
$p(-1) = 3$
Therefore,the remainder is $3$.

(C) The degree of a polynomial is the highest power of the variable present in the expression.
For option $A$: $x^{3} + x^{2}$,the degree is $3$.
For option $B$: $5x + x^{2}$,the degree is $2$.
For option $C$: $x^{3}(x^{2} + 1) = x^{5} + x^{3}$,the degree is $5$.
For option $D$: $x(x^{5} - 2) = x^{6} - 2x$,the degree is $6$.
Therefore,the polynomial with degree $5$ is $x^{3}(x^{2} + 1)$.