Mix Examples - Polynomials Questions in English

(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 $\beta = 0$.
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 the values: $(0)(0) + (0)(\gamma) + (\gamma)(0) = 0 = \frac{c}{a}$. This implies $c = 0$.
The product of the zeroes is given by $\alpha\beta\gamma = -\frac{d}{a}$.
Substituting the known values: $(0)(0)(\gamma) = 0 = -\frac{d}{a}$. This implies $d = 0$.
The sum of the zeroes is given by $\alpha + \beta + \gamma = -\frac{b}{a}$.
Substituting $\alpha = 0$ and $\beta = 0$: $0 + 0 + \gamma = -\frac{b}{a}$.
Therefore,the third zero $\gamma = -\frac{b}{a}$.

(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) For any quadratic polynomial $ax^{2} + bx + c$ $(a \neq 0)$,the graph of the corresponding equation $y = ax^{2} + bx + c$ is a parabola,which is either open upwards (if $a > 0$) or open downwards (if $a < 0$).
$1$. $A$ quadratic polynomial graph can intersect the $X$-axis at most at two points.
$2$. The graph in option $(A)$ shows a curve that intersects the $X$-axis at three distinct points,which is characteristic of a cubic polynomial,not a quadratic polynomial.
$3$. Therefore,the graph in option $(A)$ is not the graph of a quadratic polynomial.

(B) No,$x-1$ cannot be the remainder.
According to the Division Algorithm for polynomials,if a polynomial $p(x)$ is divided by a divisor $g(x)$,the remainder $r(x)$ must satisfy the condition that either $r(x) = 0$ or the degree of $r(x)$ is strictly less than the degree of $g(x)$.
Here,the divisor is $g(x) = 2x+3$,which has a degree of $1$.
The proposed remainder is $r(x) = x-1$,which also has a degree of $1$.
Since the degree of the remainder is not less than the degree of the divisor,$x-1$ cannot be the remainder.

(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.

(B) No. According to the division algorithm for polynomials,if $p(x)$ is the dividend and $g(x)$ is the divisor,then $p(x) = g(x) \cdot q(x) + r(x)$,where $\text{deg}(r(x)) < \text{deg}(g(x))$.
Given,$\text{deg}(p(x)) = 6$ and $\text{deg}(g(x)) = 5$.
If the quotient $q(x) = x^{2}-1$,then $\text{deg}(q(x)) = 2$.
The degree of the product $g(x) \cdot q(x)$ would be $\text{deg}(g(x)) + \text{deg}(q(x)) = 5 + 2 = 7$.
Since the degree of the dividend $p(x)$ is $6$,and the degree of the product $g(x) \cdot q(x)$ is $7$,it is impossible for $x^{2}-1$ to be the quotient because the degree of the product cannot exceed the degree of the dividend (assuming the remainder is zero or has a lower degree).
Therefore,$x^{2}-1$ cannot be the quotient.

(N/A) Given that,the divisor is $px^{3} + qx^{2} + nx + s$ (where $p \neq 0$) and the dividend is $ax^{2} + bx + c$.
We observe that the degree of the divisor is $3$ and the degree of the dividend is $2$.
According to the division algorithm for polynomials,if the degree of the divisor is greater than the degree of the dividend,the division cannot be performed in the standard sense of obtaining a non-zero quotient.
Therefore,the quotient is $0$ and the remainder is equal to the dividend itself,which is $ax^{2} + bx + c$.

(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 $g(x)$.

(N/A) If a non-zero polynomial $p(x)$ is divided by a polynomial $g(x)$ and the remainder is zero,it implies that $g(x)$ is a factor of $p(x)$. According to the division algorithm for polynomials,$p(x) = g(x) \cdot q(x) + r(x)$,where $r(x) = 0$. Since $p(x)$ and $g(x)$ are non-zero polynomials,the degree of the product $g(x) \cdot q(x)$ must equal the degree of $p(x)$. Therefore,the degree of $g(x)$ must be less than or equal to the degree of $p(x)$,i.e.,$\text{deg}(g(x)) \le \text{deg}(p(x))$.

(A) No. Let $p(x) = x^{2} + kx + k$. If $p(x)$ has equal zeroes,its discriminant $D$ must be equal to $0$.
The discriminant formula is $D = B^{2} - 4AC = 0$ ... $(i)$.
Comparing $p(x)$ with the standard form $Ax^{2} + Bx + C$,we get $A = 1$,$B = k$,and $C = k$.
Substituting these values into Eq. $(i)$:
$k^{2} - 4(1)(k) = 0$
$k^{2} - 4k = 0$
$k(k - 4) = 0$
This gives $k = 0$ or $k = 4$.
Since neither $0$ nor $4$ is an odd integer greater than $1$,the quadratic polynomial $x^{2} + kx + k$ cannot have equal zeroes for any odd integer $k > 1$.

(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.

(B) The statement is 'False'.
$A$ quadratic polynomial $p(x) = ax^2 + bx + c$ can intersect or touch the $x$-axis at only one point if the discriminant $D = b^2 - 4ac$ is equal to $0$.
In this case,the quadratic polynomial has two equal real roots,and the graph touches the $x$-axis at exactly one point (the vertex).
Therefore,a quadratic polynomial can indeed intersect the $x$-axis at only one point.

(A) The statement is 'True'.
If the graph of a polynomial intersects the $x$-axis at exactly two points,it does not necessarily have to be a quadratic polynomial.
$A$ polynomial of degree $n > 2$ can also intersect the $x$-axis at exactly two points if it has two real roots and the remaining $(n-2)$ roots are imaginary (non-real complex roots).
For example,a polynomial of degree $4$ can have two real roots and two imaginary roots,resulting in a graph that crosses the $x$-axis at only two points.

(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.

(B) False.
Let the quadratic polynomial be $f(x) = k x^{2} + x + k$.
For the polynomial to have equal zeros,its discriminant $D$ must be equal to $0$.
The discriminant is given by $D = b^{2} - 4ac$.
Here,$a = k$,$b = 1$,and $c = k$.
Substituting these values into the discriminant formula:
$D = (1)^{2} - 4(k)(k) = 0$
$1 - 4k^{2} = 0$
$4k^{2} = 1$
$k^{2} = \frac{1}{4}$
$k = \pm \frac{1}{2}$
Therefore,there are two values of $k$,which are $\frac{1}{2}$ and $-\frac{1}{2}$,for which the quadratic polynomial has equal zeros. Thus,the statement 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 $p(x) = x^{3}+2 x^{2}+k x+3$.
By the Remainder Theorem,since $p(x)$ is divided by $x-3$,the remainder is $p(3)$.
Given $p(3) = 21$,we have:
$3^{3}+2(3)^{2}+k(3)+3 = 21$
$27 + 18 + 3k + 3 = 21$
$48 + 3k = 21$
$3k = 21 - 48 = -27$
$k = -9$.
Now,dividing $x^{3}+2 x^{2}-9 x+3$ by $x-3$ using long division:
$x^{3}+2 x^{2}-9 x+3 = (x-3)(x^{2}+5x+6) + 21$.
The quotient is $x^{2}+5x+6$.
To find the zeroes of $x^{3}+2 x^{2}-9 x-18$,we factorize it:
$x^{3}+2 x^{2}-9 x-18 = x^{2}(x+2) - 9(x+2)$
$= (x^{2}-9)(x+2)$
$= (x-3)(x+3)(x+2)$.
Setting the polynomial to $0$,the zeroes are $x = 3, -3, -2$.

(N/A) Given that,sum of zeroes $(S) = -\frac{8}{3}$ and product of zeroes $(P) = \frac{4}{3}$.
The general form of a quadratic polynomial is $f(x) = k(x^2 - Sx + P)$,where $k$ is a non-zero constant.
Substituting the values,we get $f(x) = x^2 - (-\frac{8}{3})x + \frac{4}{3} = x^2 + \frac{8}{3}x + \frac{4}{3}$.
To simplify,we can write the polynomial as $3x^2 + 8x + 4$.
To find the zeroes,we factorise $3x^2 + 8x + 4$:
$3x^2 + 6x + 2x + 4 = 3x(x + 2) + 2(x + 2) = (3x + 2)(x + 2)$.
Setting $f(x) = 0$,we get $(3x + 2)(x + 2) = 0$.
Therefore,$3x + 2 = 0 \implies x = -\frac{2}{3}$ and $x + 2 = 0 \implies x = -2$.
Hence,the zeroes of the polynomial are $-2$ and $-\frac{2}{3}$.

(A) Given that,the sum of zeroes $S = \frac{21}{8}$ and the product of zeroes $P = \frac{5}{16}$.
The required quadratic polynomial is given by $f(x) = k(x^2 - Sx + P)$,where $k$ is a constant. Taking $k = 16$ to clear the denominators:
$f(x) = 16(x^2 - \frac{21}{8}x + \frac{5}{16}) = 16x^2 - 42x + 5$.
To find the zeroes,set $f(x) = 0$:
$16x^2 - 42x + 5 = 0$.
Splitting the middle term: $16x^2 - 40x - 2x + 5 = 0$.
$8x(2x - 5) - 1(2x - 5) = 0$.
$(8x - 1)(2x - 5) = 0$.
Thus,the zeroes are $x = \frac{1}{8}$ and $x = \frac{5}{2}$.

(N/A) Given that,sum of zeroes $(S)$ = $-2 \sqrt{3}$ and product of zeroes $(P)$ = $-9$.
The general form of a quadratic polynomial is $f(x) = x^{2} - Sx + P$.
Substituting the values,we get $f(x) = x^{2} - (-2 \sqrt{3})x + (-9) = x^{2} + 2 \sqrt{3}x - 9$.
To find the zeroes by factorization,we split the middle term $2 \sqrt{3}x$ into $3 \sqrt{3}x - \sqrt{3}x$:
$f(x) = x^{2} + 3 \sqrt{3}x - \sqrt{3}x - 9$
$f(x) = x(x + 3 \sqrt{3}) - \sqrt{3}(x + 3 \sqrt{3})$
$f(x) = (x + 3 \sqrt{3})(x - \sqrt{3})$
Setting $f(x) = 0$,we get $x + 3 \sqrt{3} = 0$ or $x - \sqrt{3} = 0$.
Therefore,the zeroes are $-3 \sqrt{3}$ and $\sqrt{3}$.

(N/A) Given that,Sum of zeroes $(S)$ = $-\frac{3}{2 \sqrt{5}}$ and Product of zeroes $(P)$ = $-\frac{1}{2}$.
The general form of a quadratic polynomial is $f(x) = k(x^2 - Sx + P)$,where $k$ is a constant.
Substituting the values,$f(x) = x^2 - (-\frac{3}{2 \sqrt{5}})x + (-\frac{1}{2}) = x^2 + \frac{3}{2 \sqrt{5}}x - \frac{1}{2}$.
To simplify,we can multiply by $2\sqrt{5}$ to get the polynomial $2\sqrt{5}x^2 + 3x - \sqrt{5}$.
Now,factorising $2\sqrt{5}x^2 + 3x - \sqrt{5}$:
$= 2\sqrt{5}x^2 + 5x - 2x - \sqrt{5}$
$= \sqrt{5}x(2x + \sqrt{5}) - 1(2x + \sqrt{5})$
$= (2x + \sqrt{5})(\sqrt{5}x - 1)$
Setting $f(x) = 0$,we get $2x + \sqrt{5} = 0$ or $\sqrt{5}x - 1 = 0$.
Therefore,the zeroes are $x = -\frac{\sqrt{5}}{2}$ and $x = \frac{1}{\sqrt{5}}$.

(A) Let $f(x) = x^{3}-6x^{2}+3x+10$.
Given that $a, (a+b),$ and $(a+2b)$ are the zeroes of $f(x)$.
Sum of the zeroes $= -\frac{\text{Coefficient of } x^{2}}{\text{Coefficient of } x^{3}} = -\frac{-6}{1} = 6$.
So,$a + (a+b) + (a+2b) = 6 \Rightarrow 3a + 3b = 6 \Rightarrow a+b = 2 \Rightarrow b = 2-a \dots (i)$.
Product of the zeroes $= -\frac{\text{Constant term}}{\text{Coefficient of } x^{3}} = -\frac{10}{1} = -10$.
So,$a(a+b)(a+2b) = -10$.
Substituting $a+b=2$ and $b=2-a$,we get $a(2)(a+2(2-a)) = -10$.
$2a(a+4-2a) = -10 \Rightarrow 2a(4-a) = -10 \Rightarrow 8a - 2a^{2} = -10$.
$2a^{2} - 8a - 10 = 0 \Rightarrow a^{2} - 4a - 5 = 0$.
$(a-5)(a+1) = 0$.
Thus,$a = 5$ or $a = -1$.
If $a = 5$,then $b = 2-5 = -3$. The zeroes are $5, 5-3, 5-6$,i.e.,$5, 2, -1$.
If $a = -1$,then $b = 2-(-1) = 3$. The zeroes are $-1, -1+3, -1+6$,i.e.,$-1, 2, 5$.
Thus,the values are $(a=5, b=-3)$ or $(a=-1, b=3)$,and the zeroes are $-1, 2, 5$.

(B) Let $f(x) = 6x^{3}+\sqrt{2}x^{2}-10x-4\sqrt{2}$.
Since $\sqrt{2}$ is a zero of $f(x)$,$(x-\sqrt{2})$ is a factor of $f(x)$.
Dividing $f(x)$ by $(x-\sqrt{2})$:
$6x^{3}+\sqrt{2}x^{2}-10x-4\sqrt{2} = (x-\sqrt{2})(6x^{2}+7\sqrt{2}x+4)$.
Now,factorize the quadratic polynomial $6x^{2}+7\sqrt{2}x+4$:
$6x^{2}+4\sqrt{2}x+3\sqrt{2}x+4 = 2\sqrt{2}x(\sqrt{3}x+2) + ...$ (Wait,let's simplify correctly).
$6x^{2}+7\sqrt{2}x+4 = 6x^{2}+3\sqrt{2}x+4\sqrt{2}x+4 = 3\sqrt{2}x(\sqrt{2}x+1) + 4(\sqrt{2}x+1) = (3\sqrt{2}x+4)(\sqrt{2}x+1)$.
Setting these factors to zero:
$\sqrt{2}x+1 = 0 \implies x = -\frac{1}{\sqrt{2}}$.
$3\sqrt{2}x+4 = 0 \implies x = -\frac{4}{3\sqrt{2}}$.
Thus,the other two zeroes are $-\frac{1}{\sqrt{2}}$ and $-\frac{4}{3\sqrt{2}}$.

(C) Given that $(x^{2}+2x+k)$ is a factor of $2x^{4}+x^{3}-14x^{2}+5x+6$,the remainder must be zero when we perform polynomial division.
Performing long division of $2x^{4}+x^{3}-14x^{2}+5x+6$ by $(x^{2}+2x+k)$:
$1$. Divide $2x^{4}$ by $x^{2}$ to get $2x^{2}$. Multiply $(x^{2}+2x+k)$ by $2x^{2}$ to get $2x^{4}+4x^{3}+2kx^{2}$. Subtracting this from the dividend gives $-3x^{3}-(2k+14)x^{2}+5x+6$.
$2$. Divide $-3x^{3}$ by $x^{2}$ to get $-3x$. Multiply $(x^{2}+2x+k)$ by $-3x$ to get $-3x^{3}-6x^{2}-3kx$. Subtracting this gives $(6-2k-14)x^{2}+(3k+5)x+6 = (-8-2k)x^{2}+(3k+5)x+6$.
$3$. Divide $(-8-2k)x^{2}$ by $x^{2}$ to get $(-8-2k)$. Multiply $(x^{2}+2x+k)$ by $(-8-2k)$ to get $(-8-2k)x^{2} + 2(-8-2k)x + k(-8-2k)$.
Subtracting this from the previous remainder gives the final remainder: $(3k+5+16+4k)x + (6+8k+2k^{2}) = (7k+21)x + (2k^{2}+8k+6)$.
For the remainder to be zero,both coefficients must be zero:
$7k+21=0 \Rightarrow k=-3$.
$2k^{2}+8k+6=0 \Rightarrow 2(k^{2}+4k+3)=0 \Rightarrow 2(k+1)(k+3)=0 \Rightarrow k=-1$ or $k=-3$.
Since $k$ must satisfy both,$k=-3$.
With $k=-3$,the divisor is $x^{2}+2x-3 = (x+3)(x-1)$,so its zeroes are $x=-3, 1$.
The quotient is $2x^{2}-3x-2 = 2x^{2}-4x+x-2 = 2x(x-2)+1(x-2) = (2x+1)(x-2)$,so its zeroes are $x=-1/2, 2$.
Thus,the zeroes of $x^{2}+2x-3$ are $1, -3$ and the zeroes of $2x^{4}+x^{3}-14x^{2}+5x+6$ are $1, -3, 2, -1/2$.

(D) Let $f(x) = x^{3}-3 \sqrt{5} x^{2}+13 x-3 \sqrt{5}$.
Since $(x-\sqrt{5})$ is a factor,we divide $f(x)$ by $(x-\sqrt{5})$ to find the quotient.
Performing polynomial long division:
$x^{3}-3 \sqrt{5} x^{2}+13 x-3 \sqrt{5} = (x-\sqrt{5})(x^{2}-2 \sqrt{5} x+3)$.
Now,we find the roots of the quadratic factor $x^{2}-2 \sqrt{5} x+3 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$:
$x = \frac{2 \sqrt{5} \pm \sqrt{(-2 \sqrt{5})^{2}-4(1)(3)}}{2(1)}$
$x = \frac{2 \sqrt{5} \pm \sqrt{20-12}}{2} = \frac{2 \sqrt{5} \pm \sqrt{8}}{2} = \frac{2 \sqrt{5} \pm 2 \sqrt{2}}{2} = \sqrt{5} \pm \sqrt{2}$.
Thus,the zeroes are $\sqrt{5}, (\sqrt{5}+\sqrt{2})$,and $(\sqrt{5}-\sqrt{2})$.

(A) Given that the zeroes of $q(x) = x^{3} + 2x^{2} + a$ are also the zeroes of $p(x) = x^{5} - x^{4} - 4x^{3} + 3x^{2} + 3x + b$,it implies that $q(x)$ is a factor of $p(x)$.
Performing polynomial long division of $p(x)$ by $q(x)$:
Dividing $x^{5} - x^{4} - 4x^{3} + 3x^{2} + 3x + b$ by $x^{3} + 2x^{2} + a$ yields a quotient of $x^{2} - 3x + 2$ and a remainder of $-(1+a)x^{2} + (3+3a)x + (b-2a)$.
Since $q(x)$ is a factor,the remainder must be zero for all $x$,so the coefficients must be zero:
$-(1+a) = 0 \Rightarrow a = -1$
$3+3a = 0 \Rightarrow 3+3(-1) = 0$ (Consistent)
$b-2a = 0 \Rightarrow b = 2(-1) = -2$.
Thus,$a = -1$ and $b = -2$.
Substituting these values,$p(x) = (x^{3} + 2x^{2} - 1)(x^{2} - 3x + 2)$.
Factoring the quotient: $x^{2} - 3x + 2 = (x-1)(x-2)$.
The zeroes of $p(x)$ are the zeroes of $q(x)$ along with the zeroes of the quotient,which are $x=1$ and $x=2$. Therefore,the zeroes of $p(x)$ that are not zeroes of $q(x)$ are $1$ and $2$.

(B) The given polynomial is $p(x) = 3.14x^2 + 1.57x + 1$.
The degree of a polynomial is the highest power of the variable present in the expression.
In this expression,the highest power of $x$ is $2$.
$A$ polynomial of degree $2$ is known as a quadratic polynomial.
Therefore,$p(x)$ is a quadratic polynomial.