Relationship between Zeroes and Coefficients of a Polynomial Questions in English

(N/A) We have the quadratic polynomial $p(x) = x^{2}+7x+10$.
To find the zeroes,we set $p(x) = 0$:
$x^{2}+7x+10 = 0$
$x^{2}+5x+2x+10 = 0$
$x(x+5)+2(x+5) = 0$
$(x+2)(x+5) = 0$
Thus,the zeroes are $x = -2$ and $x = -5$.
Verification:
Sum of zeroes $= (-2) + (-5) = -7 = \frac{-7}{1} = \frac{-(\text{Coefficient of } x)}{\text{Coefficient of } x^{2}}$.
Product of zeroes $= (-2) \times (-5) = 10 = \frac{10}{1} = \frac{\text{Constant term}}{\text{Coefficient of } x^{2}}$.
Hence,the relationship is verified.

(A) Let the quadratic polynomial be $p(x) = ax^2 + bx + c,$ and its zeroes be $\alpha$ and $\beta$.
The relationship between the zeroes and coefficients of a quadratic polynomial is given by:
Sum of zeroes: $\alpha + \beta = -\frac{b}{a} = -3$
Product of zeroes: $\alpha \beta = \frac{c}{a} = 2$
If we assume $a = 1$,then:
$-b = -3 \implies b = 3$
$c = 2$
Substituting these values into the standard form $ax^2 + bx + c$,we get the polynomial $x^2 + 3x + 2$.
Thus,the required quadratic polynomial is $x^2 + 3x + 2$.

(A) Comparing the given polynomial $p(x) = 3x^3 - 5x^2 - 11x - 3$ with the standard form $ax^3 + bx^2 + cx + d$,we get $a = 3, b = -5, c = -11, d = -3$.
First,we verify the zeroes:
$p(3) = 3(3)^3 - 5(3)^2 - 11(3) - 3 = 81 - 45 - 33 - 3 = 0$
$p(-1) = 3(-1)^3 - 5(-1)^2 - 11(-1) - 3 = -3 - 5 + 11 - 3 = 0$
$p(-\frac{1}{3}) = 3(-\frac{1}{3})^3 - 5(-\frac{1}{3})^2 - 11(-\frac{1}{3}) - 3 = -\frac{1}{9} - \frac{5}{9} + \frac{11}{3} - 3 = -\frac{6}{9} + \frac{33}{9} - \frac{27}{9} = 0$
Thus,$3, -1, -\frac{1}{3}$ are the zeroes.
Let $\alpha = 3, \beta = -1, \gamma = -\frac{1}{3}$.
$1$. Sum of zeroes: $\alpha + \beta + \gamma = 3 - 1 - \frac{1}{3} = 2 - \frac{1}{3} = \frac{5}{3} = -\frac{b}{a}$.
$2$. Sum of product of zeroes taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = (3)(-1) + (-1)(-\frac{1}{3}) + (-\frac{1}{3})(3) = -3 + \frac{1}{3} - 1 = -4 + \frac{1}{3} = -\frac{11}{3} = \frac{c}{a}$.
$3$. Product of zeroes: $\alpha\beta\gamma = (3)(-1)(-\frac{1}{3}) = 1 = -\frac{d}{a} = -(\frac{-3}{3}) = 1$.
All relationships are verified.

(N/A) Given polynomial: $p(x) = x^{2}-2x-8$.
To find the zeroes,we set $p(x) = 0$:
$x^{2}-2x-8 = 0$
$x^{2}-4x+2x-8 = 0$
$x(x-4)+2(x-4) = 0$
$(x-4)(x+2) = 0$
Thus,the zeroes are $x = 4$ and $x = -2$.
Verification:
Comparing $x^{2}-2x-8$ with $ax^{2}+bx+c$,we get $a=1, b=-2, c=-8$.
Sum of zeroes $= 4 + (-2) = 2$.
Formula: $-b/a = -(-2)/1 = 2$.
Since $2 = 2$,the sum of zeroes is verified.
Product of zeroes $= 4 \times (-2) = -8$.
Formula: $c/a = -8/1 = -8$.
Since $-8 = -8$,the product of zeroes is verified.

(N/A) Given polynomial: $p(s) = 4s^{2}-4s+1$.
To find the zeroes,set $p(s) = 0$:
$4s^{2}-4s+1 = 0$
$(2s-1)^{2} = 0$
$2s-1 = 0 \implies s = \frac{1}{2}$.
Thus,the zeroes are $\frac{1}{2}$ and $\frac{1}{2}$.
Verification:
Sum of zeroes $= \frac{1}{2} + \frac{1}{2} = 1 = \frac{-(-4)}{4} = \frac{-(\text{Coefficient of } s)}{\text{Coefficient of } s^{2}}$.
Product of zeroes $= \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \frac{\text{Constant term}}{\text{Coefficient of } s^{2}}$.
Hence,the relationship is verified.

(N/A) First,rewrite the polynomial in standard form: $6x^{2}-7x-3$.
To find the zeroes,factorize the quadratic polynomial:
$6x^{2}-7x-3 = 6x^{2}-9x+2x-3 = 3x(2x-3)+1(2x-3) = (3x+1)(2x-3)$.
The value of $6x^{2}-7x-3$ is zero when $3x+1=0$ or $2x-3=0$,which gives $x = -1/3$ or $x = 3/2$.
Therefore,the zeroes are $\alpha = -1/3$ and $\beta = 3/2$.
Verification of the relationship between zeroes and coefficients:
Sum of zeroes: $\alpha + \beta = -1/3 + 3/2 = (-2+9)/6 = 7/6 = -(-7)/6 = -(\text{Coefficient of } x) / (\text{Coefficient of } x^{2})$.
Product of zeroes: $\alpha \times \beta = (-1/3) \times (3/2) = -3/6 = -1/2 = -3/6 = (\text{Constant term}) / (\text{Coefficient of } x^{2})$.

(N/A) Given polynomial: $p(u) = 4u^{2} + 8u$.
To find the zeroes,set $p(u) = 0$:
$4u^{2} + 8u = 0$
$4u(u + 2) = 0$
This implies $4u = 0$ or $u + 2 = 0$.
So,$u = 0$ or $u = -2$.
The zeroes are $0$ and $-2$.
Verification:
Comparing $4u^{2} + 8u$ with $au^{2} + bu + c$,we get $a = 4, b = 8, c = 0$.
Sum of zeroes $= 0 + (-2) = -2$.
Relationship: $\frac{-b}{a} = \frac{-8}{4} = -2$.
Since Sum of zeroes $= \frac{-b}{a}$,the relationship is verified.
Product of zeroes $= 0 \times (-2) = 0$.
Relationship: $\frac{c}{a} = \frac{0}{4} = 0$.
Since Product of zeroes $= \frac{c}{a}$,the relationship is verified.

(N/A) Given polynomial: $p(t) = t^{2}-15$.
To find the zeroes,set $p(t) = 0$:
$t^{2}-15 = 0$
$t^{2} = 15$
$t = \pm \sqrt{15}$
So,the zeroes are $\alpha = \sqrt{15}$ and $\beta = -\sqrt{15}$.
Comparing $t^{2}-15$ with the standard form $at^{2}+bt+c$,we get $a=1, b=0, c=-15$.
Verification:
Sum of zeroes: $\alpha + \beta = \sqrt{15} + (-\sqrt{15}) = 0$.
From coefficients: $\frac{-b}{a} = \frac{-0}{1} = 0$.
Thus,$\alpha + \beta = \frac{-b}{a}$ is verified.
Product of zeroes: $\alpha \cdot \beta = (\sqrt{15})(-\sqrt{15}) = -15$.
From coefficients: $\frac{c}{a} = \frac{-15}{1} = -15$.
Thus,$\alpha \cdot \beta = \frac{c}{a}$ is verified.

(N/A) Given polynomial: $p(x) = 3x^{2}-x-4$.
To find the zeroes,we set $p(x) = 0$:
$3x^{2}-x-4 = 0$
$3x^{2}-4x+3x-4 = 0$
$x(3x-4)+1(3x-4) = 0$
$(3x-4)(x+1) = 0$
Thus,the zeroes are $x = \frac{4}{3}$ and $x = -1$.
Verification:
Sum of zeroes $= \frac{4}{3} + (-1) = \frac{4-3}{3} = \frac{1}{3}$.
From the polynomial,$-\frac{\text{coefficient of } x}{\text{coefficient of } x^{2}} = -\frac{-1}{3} = \frac{1}{3}$.
Since $\frac{1}{3} = \frac{1}{3}$,the sum is verified.
Product of zeroes $= \frac{4}{3} \times (-1) = -\frac{4}{3}$.
From the polynomial,$\frac{\text{constant term}}{\text{coefficient of } x^{2}} = \frac{-4}{3}$.
Since $-\frac{4}{3} = -\frac{4}{3}$,the product is verified.

(A) Let the quadratic polynomial be $ax^2 + bx + c$ and its zeroes be $\alpha$ and $\beta$.
Given that the sum of zeroes $\alpha + \beta = \frac{1}{4}$ and the product of zeroes $\alpha \beta = -1$.
We know that a quadratic polynomial can be expressed as $k[x^2 - (\alpha + \beta)x + \alpha \beta]$,where $k$ is a non-zero constant.
Substituting the given values,we get $k[x^2 - (\frac{1}{4})x + (-1)]$.
To simplify,let $k = 4$.
Then the polynomial becomes $4[x^2 - \frac{1}{4}x - 1] = 4x^2 - x - 4$.
Thus,the required quadratic polynomial is $4x^2 - x - 4$.

(D) Let the quadratic polynomial be $p(x) = ax^2 + bx + c$,and its zeroes be $\alpha$ and $\beta$.
The sum of the zeroes is given by $\alpha + \beta = \sqrt{2} = -\frac{b}{a}$.
The product of the zeroes is given by $\alpha \beta = \frac{1}{3} = \frac{c}{a}$.
To express these with a common denominator $a$,we write $\alpha + \beta = \frac{3\sqrt{2}}{3} = -\frac{b}{a}$.
Comparing the coefficients,we get $a = 3$,$b = -3\sqrt{2}$,and $c = 1$.
Substituting these values into the general form $ax^2 + bx + c$,we get the quadratic polynomial $3x^2 - 3\sqrt{2}x + 1$.

(D) Let the quadratic polynomial be $p(x) = ax^2 + bx + c$,and its zeroes be $\alpha$ and $\beta$.
The sum of the zeroes is given by $\alpha + \beta = 0 = \frac{0}{1} = -\frac{b}{a}$.
The product of the zeroes is given by $\alpha \times \beta = \sqrt{5} = \frac{\sqrt{5}}{1} = \frac{c}{a}$.
Comparing the coefficients,if we take $a = 1$,then $b = 0$ and $c = \sqrt{5}$.
Substituting these values into the general form $ax^2 + bx + c$,we get the quadratic polynomial $x^2 + 0x + \sqrt{5}$,which simplifies to $x^2 + \sqrt{5}$.

(A) Let the quadratic polynomial be $p(x) = ax^2 + bx + c$,and its zeroes be $\alpha$ and $\beta$.
The sum of zeroes is given by $\alpha + \beta = 1 = \frac{-b}{a}$.
The product of zeroes is given by $\alpha \times \beta = 1 = \frac{c}{a}$.
If we assume $a = 1$,then:
$-b = 1 \implies b = -1$
$c = 1$
Substituting these values into the general form $ax^2 + bx + c$,we get the polynomial $x^2 - x + 1$.

(A) Let the quadratic polynomial be $p(x) = ax^2 + bx + c$,and its zeroes be $\alpha$ and $\beta$.
The sum of zeroes is given by $\alpha + \beta = -\frac{1}{4} = -\frac{b}{a}$.
The product of zeroes is given by $\alpha \times \beta = \frac{1}{4} = \frac{c}{a}$.
Comparing the ratios,if we take $a = 4$,then $b = 1$ and $c = 1$.
Substituting these values into the general form $ax^2 + bx + c$,we get the polynomial $4x^2 + x + 1$.

(A) Let the quadratic polynomial be $p(x) = ax^2 + bx + c$,where $\alpha$ and $\beta$ are its zeroes.
The sum of the zeroes is given by $\alpha + \beta = 4 = \frac{4}{1} = \frac{-b}{a}$.
The product of the zeroes is given by $\alpha \times \beta = 1 = \frac{1}{1} = \frac{c}{a}$.
By comparing the ratios,if we assume $a = 1$,then $b = -4$ and $c = 1$.
Substituting these values into the general form $ax^2 + bx + c$,we get the quadratic polynomial $x^2 - 4x + 1$.

(N/A) Let $p(x) = 3x^{4} + 6x^{3} - 2x^{2} - 10x - 5$.
Since $\sqrt{\frac{5}{3}}$ and $-\sqrt{\frac{5}{3}}$ are two zeroes of $p(x)$,then $(x - \sqrt{\frac{5}{3}})(x + \sqrt{\frac{5}{3}}) = (x^{2} - \frac{5}{3})$ is a factor of $p(x)$.
To find the other zeroes,we divide $p(x)$ by $(x^{2} - \frac{5}{3})$:
$3x^{4} + 6x^{3} - 2x^{2} - 10x - 5 = (x^{2} - \frac{5}{3})(3x^{2} + 6x + 3)$
$= 3(x^{2} - \frac{5}{3})(x^{2} + 2x + 1)$
$= 3(x^{2} - \frac{5}{3})(x + 1)^{2}$
Now,to find the remaining zeroes,we set $(x + 1)^{2} = 0$,which gives $x + 1 = 0$,so $x = -1$.
Since the factor is $(x + 1)^{2}$,the zero $x = -1$ is repeated twice.
Thus,the other two zeroes of the polynomial are $-1$ and $-1$.

(A) Let the polynomial be $p(x) = x^{2} + 3x + k$.
Since $2$ is a zero of the polynomial,$p(2) = 0$.
Substituting $x = 2$ into the polynomial:
$p(2) = (2)^{2} + 3(2) + k = 0$
$4 + 6 + k = 0$
$10 + k = 0$
$k = -10$

(B) Let the zeroes of the cubic polynomial $p(x) = ax^3 + bx^2 + cx + d$ be $\alpha, \beta,$ and $\gamma$.
Given that two zeroes are $0$, let $\alpha = 0$ and $\beta = 0$.
The sum of the zeroes of a cubic polynomial is given by $\alpha + \beta + \gamma = -\frac{b}{a}$.
Substituting $\alpha = 0$ and $\beta = 0$ into the equation, we get $0 + 0 + \gamma = -\frac{b}{a}$.
Therefore, the third zero is $\gamma = -\frac{b}{a}$.

(C) Given that,one of the zeroes of the quadratic polynomial $p(x) = (k-1)x^2 + kx + 1$ is $-3$.
Therefore,$p(-3) = 0$.
Substituting $x = -3$ in the polynomial:
$(k-1)(-3)^2 + k(-3) + 1 = 0$
$9(k-1) - 3k + 1 = 0$
$9k - 9 - 3k + 1 = 0$
$6k - 8 = 0$
$6k = 8$
$k = \frac{8}{6} = \frac{4}{3}$

(D) Let the quadratic polynomial be $p(x) = k(x^2 - (\text{sum of zeroes})x + (\text{product of zeroes}))$, where $k$ is a non-zero constant.
Given zeroes are $\alpha = -3$ and $\beta = 4$.
Sum of zeroes $(\alpha + \beta) = -3 + 4 = 1$.
Product of zeroes $(\alpha \cdot \beta) = -3 \times 4 = -12$.
Substituting these values into the formula, we get $p(x) = k(x^2 - 1x + (-12)) = k(x^2 - x - 12)$.
If we choose $k = \frac{1}{2}$, the polynomial becomes $\frac{1}{2}(x^2 - x - 12) = \frac{x^2}{2} - \frac{x}{2} - 6$.
Since the zeroes of a polynomial remain unchanged when multiplied by a non-zero constant, option $D$ is a valid quadratic polynomial with the given zeroes.

(A) Let $p(x) = x^{2} + (a+1)x + b$.
Given that $2$ and $-3$ are the zeroes of the quadratic polynomial $p(x)$.
Therefore,$p(2) = 0$ and $p(-3) = 0$.
For $x = 2$:
$2^{2} + (a+1)(2) + b = 0$
$4 + 2a + 2 + b = 0$
$2a + b = -6$ ..... $(i)$
For $x = -3$:
$(-3)^{2} + (a+1)(-3) + b = 0$
$9 - 3a - 3 + b = 0$
$-3a + b = -6$ ..... $(ii)$
Subtracting equation $(ii)$ from equation $(i)$:
$(2a + b) - (-3a + b) = -6 - (-6)$
$2a + b + 3a - b = 0$
$5a = 0 \Rightarrow a = 0$.
Substituting $a = 0$ in equation $(i)$:
$2(0) + b = -6 \Rightarrow b = -6$.
Thus,the required values are $a = 0$ and $b = -6$.

(B) Let $p(x) = a(x^2 - (\text{sum of zeroes})x + (\text{product of zeroes}))$ be the general form of the polynomial.
Given zeroes are $\alpha = -2$ and $\beta = 5$.
Sum of zeroes $= \alpha + \beta = -2 + 5 = 3$.
Product of zeroes $= \alpha \cdot \beta = -2 \times 5 = -10$.
Thus, the polynomial is of the form $p(x) = a(x^2 - 3x - 10)$, where $a$ is any non-zero real constant.
Since $a$ can take any non-zero real value (e.g., $1, 2, 3, \dots, 0.5, \dots$), there are infinitely many such polynomials.
Therefore, the number of such polynomials is more than $3$.

(C) Let $p(x) = ax^3 + bx^2 + cx + d$ be the cubic polynomial.
Let the zeroes of the polynomial be $\alpha, \beta,$ and $\gamma$.
Given that one of the zeroes is zero,let $\alpha = 0$.
According to the relationship between zeroes and coefficients of a cubic polynomial,the sum of the product of zeroes taken two at a time is given by $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$.
Substituting $\alpha = 0$ into the equation:
$(0)\beta + \beta\gamma + \gamma(0) = \frac{c}{a}$
$0 + \beta\gamma + 0 = \frac{c}{a}$
$\beta\gamma = \frac{c}{a}$
Thus,the product of the other two zeroes is $\frac{c}{a}$.

(D) Let the cubic polynomial be $p(x) = x^{3} + a x^{2} + b x + c$.
Let the zeroes of the polynomial be $\alpha, \beta,$ and $\gamma$.
Given that one zero is $\alpha = -1$.
Since $\alpha = -1$ is a zero,$p(-1) = 0$.
Substituting $x = -1$ into the polynomial:
$(-1)^{3} + a(-1)^{2} + b(-1) + c = 0$
$-1 + a - b + c = 0$
$c = 1 - a + b$ ... $(i)$
We know that for a cubic polynomial $Ax^{3} + Bx^{2} + Cx + D$,the product of the zeroes is given by $-\frac{D}{A}$.
Here,the product of the zeroes is $\alpha \cdot \beta \cdot \gamma = -\frac{c}{1} = -c$.
Substituting $\alpha = -1$:
$(-1) \cdot \beta \cdot \gamma = -c$
$\beta \cdot \gamma = c$.
Substituting the value of $c$ from equation $(i)$:
$\beta \cdot \gamma = 1 - a + b$.
Thus,the product of the other two zeroes is $b - a + 1$.

(A) Let the given quadratic polynomial be $p(x) = x^{2} + 99x + 127$.
Comparing $p(x)$ with the standard form $ax^{2} + bx + c$,we get $a = 1$,$b = 99$,and $c = 127$.
For a quadratic polynomial $ax^{2} + bx + c$,if $a, b, c > 0$,then the product of zeroes $\alpha \beta = \frac{c}{a} = 127 > 0$ and the sum of zeroes $\alpha + \beta = -\frac{b}{a} = -99 < 0$.
Since the product of the zeroes is positive,both zeroes must have the same sign.
Since the sum of the zeroes is negative,both zeroes must be negative.
Alternatively,using the quadratic formula $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$:
$x = \frac{-99 \pm \sqrt{99^{2} - 4(1)(127)}}{2(1)} = \frac{-99 \pm \sqrt{9801 - 508}}{2} = \frac{-99 \pm \sqrt{9293}}{2}$.
Since $\sqrt{9293} \approx 96.4$,the roots are approximately $\frac{-99 + 96.4}{2} = -1.3$ and $\frac{-99 - 96.4}{2} = -97.7$.
Thus,both zeroes are negative.

(B) Let $p(x) = x^{2} + kx + k, k \neq 0$.
For the zeroes to be real,the discriminant $D$ must be greater than or equal to $0$.
$D = b^{2} - 4ac = k^{2} - 4k \geq 0$.
$k(k - 4) \geq 0$.
This implies $k \in (-\infty, 0] \cup [4, \infty)$. Since $k \neq 0$,$k \in (-\infty, 0) \cup [4, \infty)$.
Let the zeroes be $\alpha$ and $\beta$. Then:
Sum of zeroes $\alpha + \beta = -k/1 = -k$.
Product of zeroes $\alpha \beta = k/1 = k$.
Case $I$: If $k \in (-\infty, 0)$,then $k < 0$. Since the product $\alpha \beta = k < 0$,the zeroes must have opposite signs.
Case $II$: If $k \in [4, \infty)$,then $k > 0$. Since the product $\alpha \beta = k > 0$,both zeroes have the same sign. The sum $\alpha + \beta = -k < 0$,so both zeroes must be negative.
In both cases,the zeroes cannot both be positive.

(C) For a quadratic polynomial $ax^{2} + bx + c$,the zeroes are equal if the discriminant $D = b^{2} - 4ac = 0$.
This implies $b^{2} = 4ac$.
Since $b^{2}$ is always non-negative for any real $b$,$4ac$ must be non-negative.
Given $c \neq 0$,if $a = 0$,it would not be a quadratic polynomial. Thus,$a \neq 0$.
Since $4ac = b^{2} \geq 0$,the product $ac$ must be greater than or equal to $0$.
Since $c \neq 0$ and $a \neq 0$,$ac > 0$,which means $a$ and $c$ must have the same sign.
For example:
$(i)$ $x^{2} + 4x + 4 = 0 \Rightarrow (x+2)^{2} = 0 \Rightarrow x = -2, -2$ (Here $a=1, c=4$,both positive).
$(ii)$ $x^{2} - 4x + 4 = 0 \Rightarrow (x-2)^{2} = 0 \Rightarrow x = 2, 2$ (Here $a=1, c=4$,both positive).

(D) Let the zeroes of the quadratic polynomial $p(x) = x^{2} + ax + b$ be $\alpha$ and $-\alpha$.
Sum of zeroes = $-\frac{\text{coefficient of } x}{\text{coefficient of } x^{2}} = -a$.
Since $\alpha + (-\alpha) = 0$,we have $0 = -a$,which implies $a = 0$.
Product of zeroes = $\frac{\text{constant term}}{\text{coefficient of } x^{2}} = b$.
Since $\alpha \times (-\alpha) = -\alpha^{2}$,we have $-\alpha^{2} = b$.
Since $\alpha^{2}$ is always positive for non-zero $\alpha$,$- \alpha^{2}$ must be negative,so $b < 0$.
Therefore,the polynomial has no linear term $(a = 0)$ and the constant term is negative $(b < 0)$.

(A) The statement is True.
Let the zeroes of the quadratic polynomial $p(x) = ax^2 + bx + c$ be $\alpha$ and $\beta$,where $\alpha < 0$ and $\beta < 0$.
From the relationship between zeroes and coefficients:
$1$. The sum of the zeroes is $\alpha + \beta = -\frac{b}{a}$. Since both $\alpha$ and $\beta$ are negative,their sum $(\alpha + \beta)$ must be negative. Therefore,$-\frac{b}{a} < 0$,which implies $\frac{b}{a} > 0$. This means $a$ and $b$ must have the same sign.
$2$. The product of the zeroes is $\alpha \cdot \beta = \frac{c}{a}$. Since the product of two negative numbers is positive,$\frac{c}{a} > 0$. This means $a$ and $c$ must have the same sign.
Since $a$ and $b$ have the same sign,and $a$ and $c$ have the same sign,it follows that $a, b,$ and $c$ all have the same sign.

(N/A) According to the division algorithm for polynomials,$p(x) = g(x) \cdot q(x) + r(x)$,where $q(x)$ is the quotient and $r(x)$ is the remainder.
Given that the quotient $q(x) = 0$,the equation becomes $p(x) = g(x) \cdot 0 + r(x)$,which simplifies to $p(x) = r(x)$.
In polynomial division,the degree of the remainder $r(x)$ must be strictly less than the degree of the divisor $g(x)$,i.e.,$\text{deg}(r(x)) < \text{deg}(g(x))$.
Since $p(x) = r(x)$,it follows that $\text{deg}(p(x)) < \text{deg}(g(x))$.
Therefore,the degree of $p(x)$ is less than the degree of $g(x)$.

(B) The statement is False.
Let the zeroes of the quadratic polynomial $ax^2 + bx + c$ be $\alpha$ and $\beta$. Since both zeroes are positive,$\alpha > 0$ and $\beta > 0$.
From the relationship between zeroes and coefficients:
$1$. The product of zeroes $\alpha \cdot \beta = \frac{c}{a}$. Since $\alpha > 0$ and $\beta > 0$,their product $\alpha \cdot \beta > 0$. Therefore,$\frac{c}{a} > 0$,which means $a$ and $c$ must have the same sign.
$2$. The sum of zeroes $\alpha + \beta = -\frac{b}{a}$. Since $\alpha > 0$ and $\beta > 0$,their sum $\alpha + \beta > 0$. Therefore,$-\frac{b}{a} > 0$,which implies $\frac{b}{a} < 0$. This means $a$ and $b$ must have opposite signs.
Thus,$a$ and $c$ have the same sign,but $b$ has the opposite sign to $a$ and $c$. Hence,$a, b,$ and $c$ do not all have the same sign.

(A) True. Let the zeroes of the cubic polynomial be $\alpha, \beta,$ and $\gamma$. It is given that two of the zeroes are $0$. Let $\alpha = 0$ and $\beta = 0$.
The cubic polynomial can be expressed as $f(x) = k(x - \alpha)(x - \beta)(x - \gamma)$,where $k$ is a non-zero constant.
Substituting the values: $f(x) = k(x - 0)(x - 0)(x - \gamma) = k(x^2)(x - \gamma) = k(x^3 - \gamma x^2) = kx^3 - k\gamma x^2$.
Comparing this with the general form $ax^3 + bx^2 + cx + d$,we see that the coefficients of the linear term $(c)$ and the constant term $(d)$ are both $0$. Thus,the polynomial does not have linear and constant terms.

(A) The statement is True.
Let the cubic polynomial be $f(x) = a(x - \alpha)(x - \beta)(x - \gamma)$,where $\alpha, \beta, \gamma < 0$ are the negative zeroes.
Let $\alpha = -p, \beta = -q, \gamma = -r$,where $p, q, r > 0$.
Then $f(x) = a(x + p)(x + q)(x + r)$.
Expanding this,we get $f(x) = a[x^3 + (p + q + r)x^2 + (pq + qr + rp)x + pqr]$.
Comparing this with the standard form $f(x) = ax^3 + bx^2 + cx + d$,we get:
$b = a(p + q + r)$
$c = a(pq + qr + rp)$
$d = a(pqr)$
Since $p, q, r > 0$,the terms $(p + q + r)$,$(pq + qr + rp)$,and $(pqr)$ are all positive.
Therefore,$a, b, c,$ and $d$ all have the same sign as $a$.

(B) False. Let $\alpha, \beta$ and $\gamma$ be the three zeroes of the cubic polynomial $p(x) = x^{3}+ax^{2}-bx+c$.
According to the relationship between zeroes and coefficients:
$1$. Product of zeroes: $\alpha \beta \gamma = -\frac{c}{1} = -c$. Since $\alpha, \beta, \gamma > 0$,their product $\alpha \beta \gamma > 0$. Therefore,$-c > 0$,which implies $c < 0$.
$2$. Sum of zeroes: $\alpha + \beta + \gamma = -\frac{a}{1} = -a$. Since $\alpha, \beta, \gamma > 0$,their sum $\alpha + \beta + \gamma > 0$. Therefore,$-a > 0$,which implies $a < 0$.
$3$. Sum of product of zeroes taken two at a time: $\alpha \beta + \beta \gamma + \gamma \alpha = \frac{-b}{1} = -b$. Since $\alpha, \beta, \gamma > 0$,their sum of products $\alpha \beta + \beta \gamma + \gamma \alpha > 0$. Therefore,$-b > 0$,which implies $b < 0$.
Thus,for all three zeroes to be positive,$a, b$ and $c$ must all be negative. Hence,the statement that at least one of them is non-negative is False.

(A) To find the zeroes,set the polynomial to zero: $x^{2}+\frac{1}{6} x-2 = 0$.
Multiply by $6$ to simplify: $6x^{2}+x-12 = 0$.
Factorizing the quadratic: $6x^{2}+9x-8x-12 = 0$.
$3x(2x+3)-4(2x+3) = 0 \implies (3x-4)(2x+3) = 0$.
Thus,the zeroes are $\alpha = \frac{4}{3}$ and $\beta = -\frac{3}{2}$.
Verification:
Sum of zeroes: $\alpha + \beta = \frac{4}{3} - \frac{3}{2} = \frac{8-9}{6} = -\frac{1}{6}$.
From the polynomial $ax^2+bx+c$,sum $= -\frac{b}{a} = -\frac{1/6}{1} = -\frac{1}{6}$.
Product of zeroes: $\alpha \beta = \frac{4}{3} \times (-\frac{3}{2}) = -2$.
From the polynomial,product $= \frac{c}{a} = \frac{-2}{1} = -2$.
Both relations are verified.

(N/A) Let $f(x) = 4x^2 - 3x - 1$.
To find the zeroes,we factorise the polynomial by splitting the middle term:
$f(x) = 4x^2 - 4x + x - 1$
$f(x) = 4x(x - 1) + 1(x - 1)$
$f(x) = (x - 1)(4x + 1)$
The zeroes are found by setting $f(x) = 0$:
$x - 1 = 0 \implies x = 1$
$4x + 1 = 0 \implies x = -\frac{1}{4}$
Thus,the zeroes are $\alpha = 1$ and $\beta = -\frac{1}{4}$.
Verification:
Sum of zeroes $= \alpha + \beta = 1 + (-\frac{1}{4}) = \frac{3}{4} = -\frac{-3}{4} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2}$.
Product of zeroes $= \alpha \cdot \beta = 1 \cdot (-\frac{1}{4}) = -\frac{1}{4} = \frac{-1}{4} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}$.
Hence,the relationship is verified.

(N/A) Let $f(x) = 3x^2 + 4x - 4$.
To find the zeroes,we factorize the polynomial by splitting the middle term:
$3x^2 + 6x - 2x - 4 = 3x(x + 2) - 2(x + 2) = (x + 2)(3x - 2)$.
The zeroes are found by setting $f(x) = 0$:
$(x + 2)(3x - 2) = 0$,which gives $x = -2$ or $x = \frac{2}{3}$.
Verification:
Sum of zeroes $= -2 + \frac{2}{3} = \frac{-6 + 2}{3} = -\frac{4}{3}$.
From the polynomial $ax^2 + bx + c$,where $a=3, b=4, c=-4$,the sum of zeroes is $-\frac{b}{a} = -\frac{4}{3}$.
Product of zeroes $= (-2) \times \frac{2}{3} = -\frac{4}{3}$.
From the polynomial,the product of zeroes is $\frac{c}{a} = \frac{-4}{3}$.
Since the sum and product match the coefficients,the relationship is verified.

(A) Let $f(t) = 5 t^{2}+12 t+7$.
To find the zeroes,we factorise the polynomial by splitting the middle term:
$5 t^{2}+12 t+7 = 5 t^{2}+5 t+7 t+7$
$= 5 t(t+1)+7(t+1)$
$= (5 t+7)(t+1)$
The zeroes are found by setting $f(t) = 0$:
$5 t+7 = 0 \implies t = -7/5$
$t+1 = 0 \implies t = -1$
Thus,the zeroes are $\alpha = -7/5$ and $\beta = -1$.
Verification:
Sum of zeroes $= \alpha + \beta = -7/5 - 1 = -12/5$.
From the polynomial $5 t^{2}+12 t+7$,the coefficient of $t$ is $12$ and the coefficient of $t^{2}$ is $5$.
Sum of zeroes $= -(\text{coefficient of } t) / (\text{coefficient of } t^{2}) = -12/5$.
Product of zeroes $= \alpha \cdot \beta = (-7/5) \cdot (-1) = 7/5$.
From the polynomial,the constant term is $7$ and the coefficient of $t^{2}$ is $5$.
Product of zeroes $= (\text{constant term}) / (\text{coefficient of } t^{2}) = 7/5$.
Hence,the relationship is verified.

(A) Let $f(t) = t^{3}-2 t^{2}-15 t$.
$= t(t^{2}-2 t-15)$
$= t(t^{2}-5 t+3 t-15)$ [by splitting the middle term]
$= t[t(t-5)+3(t-5)]$
$= t(t-5)(t+3)$
So,the value of $f(t)$ is zero when $t=0$,$t-5=0$,or $t+3=0$.
Thus,the zeroes are $t=0, t=5$,and $t=-3$.
Verification:
Sum of zeroes $= 0 + 5 + (-3) = 2 = -(-2)/1 = -(\text{Coefficient of } t^{2}) / (\text{Coefficient of } t^{3})$.
Sum of product of zeroes taken two at a time $= (0)(5) + (5)(-3) + (-3)(0) = 0 - 15 + 0 = -15 = (-15)/1 = (\text{Coefficient of } t) / (\text{Coefficient of } t^{3})$.
Product of zeroes $= (0)(5)(-3) = 0 = -(0)/1 = -(\text{Constant term}) / (\text{Coefficient of } t^{3})$.
Hence,the relationship is verified.

(N/A) Let $f(x) = 2x^2 + \frac{7}{2}x + \frac{3}{4}$.
To find the zeroes,we set $f(x) = 0$,so $2x^2 + \frac{7}{2}x + \frac{3}{4} = 0$.
Multiplying the entire equation by $4$,we get $8x^2 + 14x + 3 = 0$.
Splitting the middle term: $8x^2 + 12x + 2x + 3 = 0$.
$4x(2x + 3) + 1(2x + 3) = 0$.
$(2x + 3)(4x + 1) = 0$.
Thus,the zeroes are $x = -\frac{3}{2}$ and $x = -\frac{1}{4}$.
Verification:
Sum of zeroes $= -\frac{3}{2} + (-\frac{1}{4}) = -\frac{6}{4} - \frac{1}{4} = -\frac{7}{4}$.
Coefficient relation: $-\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} = -\frac{7/2}{2} = -\frac{7}{4}$.
Product of zeroes $= (-\frac{3}{2}) \times (-\frac{1}{4}) = \frac{3}{8}$.
Coefficient relation: $\frac{\text{Constant term}}{\text{Coefficient of } x^2} = \frac{3/4}{2} = \frac{3}{8}$.
Since both relations hold,the relationship is verified.

(N/A) Let $f(x) = 4x^{2} + 5\sqrt{2}x - 3$.
To factorise,we split the middle term: $4x^{2} + 6\sqrt{2}x - \sqrt{2}x - 3$.
$= 2\sqrt{2}x(\sqrt{2}x + 3) - 1(\sqrt{2}x + 3)$.
$= (\sqrt{2}x + 3)(2\sqrt{2}x - 1)$.
Setting $f(x) = 0$,we get $\sqrt{2}x + 3 = 0$ or $2\sqrt{2}x - 1 = 0$.
Thus,the zeroes are $x = -\frac{3}{\sqrt{2}}$ and $x = \frac{1}{2\sqrt{2}}$.
Verification:
Sum of zeroes $= -\frac{3}{\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{-6 + 1}{2\sqrt{2}} = -\frac{5}{2\sqrt{2}} = -\frac{5\sqrt{2}}{4}$.
Coefficient of $x$ is $5\sqrt{2}$ and coefficient of $x^{2}$ is $4$. So,$-\frac{\text{Coefficient of } x}{\text{Coefficient of } x^{2}} = -\frac{5\sqrt{2}}{4}$. (Verified)
Product of zeroes $= (-\frac{3}{\sqrt{2}}) \times (\frac{1}{2\sqrt{2}}) = -\frac{3}{4}$.
Constant term is $-3$ and coefficient of $x^{2}$ is $4$. So,$\frac{\text{Constant term}}{\text{Coefficient of } x^{2}} = -\frac{3}{4}$. (Verified)

(N/A) Let $f(s) = 2s^{2} - (1 + 2\sqrt{2})s + \sqrt{2}$.
$= 2s^{2} - s - 2\sqrt{2}s + \sqrt{2}$
$= s(2s - 1) - \sqrt{2}(2s - 1)$
$= (2s - 1)(s - \sqrt{2})$
The value of $f(s)$ is zero when $2s - 1 = 0$ or $s - \sqrt{2} = 0$.
Thus,$s = \frac{1}{2}$ or $s = \sqrt{2}$.
So,the zeroes of the polynomial are $\frac{1}{2}$ and $\sqrt{2}$.
Verification:
Sum of zeroes $= \frac{1}{2} + \sqrt{2} = \frac{1 + 2\sqrt{2}}{2} = \frac{-[-(1 + 2\sqrt{2})]}{2} = \frac{-(\text{Coefficient of } s)}{\text{Coefficient of } s^{2}}$.
Product of zeroes $= \frac{1}{2} \times \sqrt{2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} = \frac{\text{Constant term}}{\text{Coefficient of } s^{2}}$.
Hence,the relationship between the zeroes and the coefficients is verified.

(A) Let $f(v) = v^{2} + 4 \sqrt{3} v - 15$.
To factorise,we split the middle term: $v^{2} + (5 \sqrt{3} - \sqrt{3}) v - 15 = 0$.
$= v^{2} + 5 \sqrt{3} v - \sqrt{3} v - 15 = 0$.
$= v(v + 5 \sqrt{3}) - \sqrt{3}(v + 5 \sqrt{3}) = 0$.
$= (v + 5 \sqrt{3})(v - \sqrt{3}) = 0$.
Thus,the zeroes are $v = -5 \sqrt{3}$ and $v = \sqrt{3}$.
Verification:
Sum of zeroes $= -5 \sqrt{3} + \sqrt{3} = -4 \sqrt{3} = -\frac{\text{Coefficient of } v}{\text{Coefficient of } v^{2}}$.
Product of zeroes $= (-5 \sqrt{3})(\sqrt{3}) = -5 \times 3 = -15 = \frac{\text{Constant term}}{\text{Coefficient of } v^{2}}$.
Hence,the relationship is verified.

(A) Let $f(y) = y^{2} + \frac{3}{2} \sqrt{5} y - 5$.
To find the zeroes,we set $f(y) = 0$,which implies $y^{2} + \frac{3}{2} \sqrt{5} y - 5 = 0$.
Multiplying by $2$,we get $2y^{2} + 3\sqrt{5}y - 10 = 0$.
Splitting the middle term: $2y^{2} + 4\sqrt{5}y - \sqrt{5}y - 10 = 0$.
$2y(y + 2\sqrt{5}) - \sqrt{5}(y + 2\sqrt{5}) = 0$.
$(y + 2\sqrt{5})(2y - \sqrt{5}) = 0$.
Thus,the zeroes are $y = -2\sqrt{5}$ and $y = \frac{\sqrt{5}}{2}$.
Verification:
Sum of zeroes $= -2\sqrt{5} + \frac{\sqrt{5}}{2} = \frac{-4\sqrt{5} + \sqrt{5}}{2} = -\frac{3\sqrt{5}}{2}$.
From the polynomial $y^{2} + \frac{3}{2} \sqrt{5} y - 5$,the coefficient of $y$ is $\frac{3\sqrt{5}}{2}$ and the coefficient of $y^{2}$ is $1$. Thus,$-\frac{\text{coefficient of } y}{\text{coefficient of } y^{2}} = -\frac{3\sqrt{5}}{2}$.
Product of zeroes $= (-2\sqrt{5}) \times (\frac{\sqrt{5}}{2}) = -5$.
The constant term is $-5$ and the coefficient of $y^{2}$ is $1$. Thus,$\frac{\text{constant term}}{\text{coefficient of } y^{2}} = -5$.
Hence,the relationship is verified.

(A) Let $f(y) = 7y^2 - \frac{11}{3}y - \frac{2}{3}$.
To find the zeroes,set $f(y) = 0$,which implies $\frac{1}{3}(21y^2 - 11y - 2) = 0$,so $21y^2 - 11y - 2 = 0$.
Splitting the middle term: $21y^2 - 14y + 3y - 2 = 0$.
$7y(3y - 2) + 1(3y - 2) = 0$.
$(3y - 2)(7y + 1) = 0$.
Thus,the zeroes are $y = \frac{2}{3}$ and $y = -\frac{1}{7}$.
Verification:
Sum of zeroes = $\frac{2}{3} + (-\frac{1}{7}) = \frac{14 - 3}{21} = \frac{11}{21}$.
From the polynomial $7y^2 - \frac{11}{3}y - \frac{2}{3}$,the coefficient of $y$ is $-\frac{11}{3}$ and the coefficient of $y^2$ is $7$.
Sum of zeroes = $-\frac{\text{coefficient of } y}{\text{coefficient of } y^2} = -\frac{-11/3}{7} = \frac{11}{21}$.
Product of zeroes = $(\frac{2}{3})(-\frac{1}{7}) = -\frac{2}{21}$.
Constant term is $-\frac{2}{3}$,coefficient of $y^2$ is $7$.
Product of zeroes = $\frac{\text{constant term}}{\text{coefficient of } y^2} = \frac{-2/3}{7} = -\frac{2}{21}$.
Hence,the relationship is verified.

(N/A) Let the quadratic polynomial be $p(x) = ax^2 + bx + c$ and its zeroes be $\alpha$ and $\beta$.
Given,sum of zeroes $\alpha + \beta = \sqrt{2}$ and product of zeroes $\alpha \beta = -\frac{3}{2}$.
$A$ quadratic polynomial is given by $k[x^2 - (\alpha + \beta)x + \alpha \beta]$.
Substituting the values,we get $k[x^2 - \sqrt{2}x - \frac{3}{2}]$.
For $k=2$,the polynomial is $2x^2 - 2\sqrt{2}x - 3$.
To find the zeroes,set $2x^2 - 2\sqrt{2}x - 3 = 0$.
$2x^2 - 3\sqrt{2}x + \sqrt{2}x - 3 = 0$.
$\sqrt{2}x(\sqrt{2}x - 3) + 1(\sqrt{2}x - 3) = 0$.
$(\sqrt{2}x + 1)(\sqrt{2}x - 3) = 0$.
Thus,the zeroes are $x = -\frac{1}{\sqrt{2}}$ and $x = \frac{3}{\sqrt{2}}$.

(A) Let $p(x) = 2x^3 + x^2 - 5x + 2$.
The given values are $\alpha = \frac{1}{2}, \beta = 1, \gamma = -2$.
First,we verify that these are zeroes:
$p(\frac{1}{2}) = 2(\frac{1}{8}) + \frac{1}{4} - \frac{5}{2} + 2 = \frac{1}{4} + \frac{1}{4} - \frac{5}{2} + 2 = \frac{1}{2} - \frac{5}{2} + 2 = -2 + 2 = 0$.
$p(1) = 2(1)^3 + (1)^2 - 5(1) + 2 = 2 + 1 - 5 + 2 = 0$.
$p(-2) = 2(-8) + 4 - 5(-2) + 2 = -16 + 4 + 10 + 2 = 0$.
Since $p(\frac{1}{2}) = 0, p(1) = 0,$ and $p(-2) = 0$,these are indeed the zeroes.
Comparing $p(x)$ with $ax^3 + bx^2 + cx + d$,we get $a = 2, b = 1, c = -5, d = 2$.
Verification of relationships:
$1$. Sum of zeroes: $\alpha + \beta + \gamma = \frac{1}{2} + 1 - 2 = -\frac{1}{2} = \frac{-b}{a}$.
$2$. Sum of product of zeroes taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = (\frac{1}{2})(1) + (1)(-2) + (-2)(\frac{1}{2}) = \frac{1}{2} - 2 - 1 = \frac{1-6}{2} = -\frac{5}{2} = \frac{c}{a}$.
$3$. Product of zeroes: $\alpha\beta\gamma = (\frac{1}{2})(1)(-2) = -1 = \frac{-d}{a} = \frac{-2}{2} = -1$.
Thus,the relationship is verified.

(A) Let the polynomial be $p(x) = x^{3} - 4x^{2} + 5x - 2$.
The given numbers are $2, 1, 1$.
For $x = 2$: $p(2) = (2)^{3} - 4(2)^{2} + 5(2) - 2 = 8 - 16 + 10 - 2 = 0$.
For $x = 1$: $p(1) = (1)^{3} - 4(1)^{2} + 5(1) - 2 = 1 - 4 + 5 - 2 = 0$.
Since $p(2) = 0$ and $p(1) = 0$,the numbers $2, 1, 1$ are indeed the zeroes of the polynomial.
Comparing $p(x)$ with the standard form $ax^{3} + bx^{2} + cx + d$,we get $a = 1, b = -4, c = 5, d = -2$.
Verification of relationships:
$1$. Sum of zeroes: $2 + 1 + 1 = 4 = -(-4)/1 = -b/a$.
$2$. Sum of product of zeroes taken two at a time: $(2)(1) + (1)(1) + (2)(1) = 2 + 1 + 2 = 5 = 5/1 = c/a$.
$3$. Product of zeroes: $2 \times 1 \times 1 = 2 = -(-2)/1 = -d/a$.
Thus,the relationship between the zeroes and the coefficients is verified.

(A) Let the cubic polynomial be $p(x) = ax^3 + bx^2 + cx + d$ and its zeroes be $\alpha, \beta,$ and $\gamma$.
The relationships between the coefficients and the zeroes are given by:
$1.$ Sum of zeroes: $\alpha + \beta + \gamma = -b/a = 2/1$
$2.$ Sum of the product of zeroes taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = c/a = -7/1$
$3.$ Product of zeroes: $\alpha\beta\gamma = -d/a = -14/1$
Comparing the values,we get $a = 1, b = -2, c = -7,$ and $d = 14$.
Substituting these values into the general form $ax^3 + bx^2 + cx + d$,we get the polynomial:
$p(x) = x^3 - 2x^2 - 7x + 14$.

(A) Given polynomial is $p(x) = x^{3}-3x^{2}+x+1$.
The zeroes are given as $a-b, a, a+b$.
Comparing the polynomial with the standard form $Ax^{3}+Bx^{2}+Cx+D$,we get $A=1, B=-3, C=1, D=1$.
Using the relationship between zeroes and coefficients:
Sum of zeroes $= (a-b) + a + (a+b) = -B/A$.
$3a = -(-3)/1 = 3$.
$a = 1$.
Product of zeroes $= (a-b)(a)(a+b) = -D/A$.
$(1-b)(1)(1+b) = -1/1$.
$1-b^{2} = -1$.
$b^{2} = 2$.
$b = \pm\sqrt{2}$.
Thus,$a=1$ and $b=\pm\sqrt{2}$.