Relationship between Zeroes and Coefficients of a Polynomial Questions in English

(A) Given polynomial: $p(x) = 2x^3 + x^2 - 5x + 2$.
Comparing with $ax^3 + bx^2 + cx + d$,we get $a = 2, b = 1, c = -5, d = 2$.
To prove these are zeros,we substitute the values into $p(x)$:
$p(1/2) = 2(1/8) + (1/4) - 5(1/2) + 2 = 1/4 + 1/4 - 5/2 + 2 = 1/2 - 2.5 + 2 = 0$.
$p(1) = 2(1)^3 + (1)^2 - 5(1) + 2 = 2 + 1 - 5 + 2 = 0$.
$p(-2) = 2(-8) + 4 - 5(-2) + 2 = -16 + 4 + 10 + 2 = 0$.
Since $p(1/2) = 0, p(1) = 0, p(-2) = 0$,these are zeros.
Verification:
Sum of zeros: $1/2 + 1 + (-2) = -1/2$. Formula: $-b/a = -1/2$. (Verified)
Sum of product of zeros taken two at a time: $(1/2)(1) + (1)(-2) + (-2)(1/2) = 1/2 - 2 - 1 = -2.5 = -5/2$. Formula: $c/a = -5/2$. (Verified)
Product of zeros: $(1/2)(1)(-2) = -1$. Formula: $-d/a = -2/2 = -1$. (Verified)

(A) Given polynomial: $p(x) = 2x^3 - 5x^2 - 14x + 8$.
Comparing with $ax^3 + bx^2 + cx + d$,we get $a = 2, b = -5, c = -14, d = 8$.
Step $1$: Check zeros.
$p(-2) = 2(-2)^3 - 5(-2)^2 - 14(-2) + 8 = 2(-8) - 5(4) + 28 + 8 = -16 - 20 + 28 + 8 = 0$.
$p(4) = 2(4)^3 - 5(4)^2 - 14(4) + 8 = 2(64) - 5(16) - 56 + 8 = 128 - 80 - 56 + 8 = 0$.
$p(1/2) = 2(1/8) - 5(1/4) - 14(1/2) + 8 = 1/4 - 5/4 - 7 + 8 = -4/4 + 1 = -1 + 1 = 0$.
Thus,$-2, 4, 1/2$ are zeros.
Step $2$: Verify relationships.
Sum of zeros: $\alpha + \beta + \gamma = -2 + 4 + 0.5 = 2.5 = 5/2 = -b/a$.
Sum of product of zeros taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = (-2)(4) + (4)(0.5) + (0.5)(-2) = -8 + 2 - 1 = -7 = -14/2 = c/a$.
Product of zeros: $\alpha\beta\gamma = (-2)(4)(0.5) = -4 = -8/2 = -d/a$.
All relationships are verified.

(A) quadratic polynomial in standard form is given by the expression $p(x) = ax^{2} + bx + c$.
Given the coefficients $a=6$,$b=-7$,and $c=-3$,we substitute these values into the standard form.
Thus,the polynomial is $6x^{2} + (-7)x + (-3)$,which simplifies to $6x^{2} - 7x - 3$.

(A) The standard form of a quadratic polynomial is given by $p(x) = ax^{2} + bx + c$.
Given the coefficients $a=1$,$b=-10$,and $c=25$.
Substituting these values into the standard form:
$p(x) = (1)x^{2} + (-10)x + 25$
$p(x) = x^{2} - 10x + 25$.

(N/A) The standard form of a cubic polynomial is given by $P(x) = ax^3 + bx^2 + cx + d$.
Given the coefficients $a=2, b=3, c=-5$,and $d=0$.
Substituting these values into the standard form,we get:
$P(x) = 2x^3 + 3x^2 + (-5)x + 0$.
Therefore,the required polynomial is $2x^3 + 3x^2 - 5x$.

(C) Given that $1$ and $3$ are zeros of $p(x)$,$(x - 1)$ and $(x - 3)$ are factors of $p(x)$.
Therefore,$(x - 1)(x - 3) = x^2 - 4x + 3$ is a factor of $p(x)$.
Dividing $p(x) = 2x^4 - 7x^3 - 13x^2 + 63x - 45$ by $(x^2 - 4x + 3)$:
$2x^4 - 7x^3 - 13x^2 + 63x - 45 = (x^2 - 4x + 3)(2x^2 + x - 15)$.
Now,factorize the quadratic $2x^2 + x - 15$:
$2x^2 + 6x - 5x - 15 = 2x(x + 3) - 5(x + 3) = (2x - 5)(x + 3)$.
Setting these factors to zero,we get $x = -3$ and $x = \frac{5}{2}$.
Thus,the remaining zeros are $-3$ and $\frac{5}{2}$.

(N/A) To divide $p(x) = x^{4} - 1$ by $x - 1$ using synthetic division:
$1$. Write the dividend $p(x)$ in standard form including all missing terms with zero coefficients: $p(x) = 1x^{4} + 0x^{3} + 0x^{2} + 0x - 1$.
$2$. The divisor is $x - 1$,so set $x - 1 = 0$,which gives $x = 1$. Use $1$ for the synthetic division.
$3$. Set up the synthetic division table:
$\begin{array}{c|ccccc} 1 & 1 & 0 & 0 & 0 & -1 \\ & & 1 & 1 & 1 & 1 \\ \hline & 1 & 1 & 1 & 1 & 0 \end{array}$
$4$. The bottom row represents the coefficients of the quotient and the remainder. The last value is the remainder.
Thus,the quotient is $q(x) = x^{3} + x^{2} + x + 1$ and the remainder is $r(x) = 0$.

(N/A) To divide $p(x) = x^{3} - 4x^{2} + 5x + 3$ by $x - 2$ using synthetic division:
$1$. Identify the coefficients of $p(x)$,which are $1, -4, 5, 3$.
$2$. The divisor is $x - 2$,so we use $x = 2$ for synthetic division.
$3$. Set up the synthetic division table:
$2 | 1, -4, 5, 3$
| $2$,-$4$,$2$
-----------------
| $1$,-$2$,$1$,$5$
$4$. The coefficients of the quotient are $1, -2, 1$,which corresponds to the polynomial $x^{2} - 2x + 1$.
$5$. The remainder is the last value,which is $5$.
Therefore,the quotient is $x^{2} - 2x + 1$ and the remainder is $5$.

(A) To divide $p(t) = 2t^2 + 3t + 1$ by $t + 2$ using synthetic division:
$1$. Identify the root of the divisor $t + 2 = 0$,which is $t = -2$.
$2$. Write the coefficients of the dividend $p(t) = 2t^2 + 3t + 1$,which are $2, 3, 1$.
$3$. Perform synthetic division:
- Bring down the first coefficient: $2$.
- Multiply by the root: $2 \times (-2) = -4$.
- Add to the next coefficient: $3 + (-4) = -1$.
- Multiply by the root: $(-1) \times (-2) = 2$.
- Add to the next coefficient: $1 + 2 = 3$.
$4$. The resulting coefficients are $2$ and $-1$,representing the quotient $2t - 1$.
$5$. The final value $3$ is the remainder.
Quotient polynomial: $2t - 1$;
Remainder polynomial: $3$.

(A) Given polynomial: $p(x) = 2x^3 + 11x^2 - 7x - 6$.
Step $1$: Check if $-6, -\frac{1}{2}, 1$ are zeros.
$p(-6) = 2(-6)^3 + 11(-6)^2 - 7(-6) - 6 = 2(-216) + 11(36) + 42 - 6 = -432 + 396 + 42 - 6 = 0$.
$p(-\frac{1}{2}) = 2(-\frac{1}{8}) + 11(\frac{1}{4}) - 7(-\frac{1}{2}) - 6 = -\frac{1}{4} + \frac{11}{4} + \frac{7}{2} - 6 = \frac{10}{4} + \frac{14}{4} - 6 = 6 - 6 = 0$.
$p(1) = 2(1)^3 + 11(1)^2 - 7(1) - 6 = 2 + 11 - 7 - 6 = 0$.
Thus,$-6, -\frac{1}{2}, 1$ are zeros.
Step $2$: Verify relationship.
Sum of zeros: $\alpha + \beta + \gamma = -6 - \frac{1}{2} + 1 = -5 - 0.5 = -5.5 = -\frac{11}{2} = -\frac{\text{coefficient of } x^2}{\text{coefficient of } x^3}$.
Sum of product of zeros taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = (-6)(-\frac{1}{2}) + (-\frac{1}{2})(1) + (1)(-6) = 3 - 0.5 - 6 = -3.5 = -\frac{7}{2} = \frac{\text{coefficient of } x}{\text{coefficient of } x^3}$.
Product of zeros: $\alpha\beta\gamma = (-6)(-\frac{1}{2})(1) = 3 = -\frac{-6}{2} = -\frac{\text{constant term}}{\text{coefficient of } x^3}$.

(D) Given that $-2$ is a zero of $p(x) = x^3 + 6x^2 + 11x + 6$,it implies that $(x + 2)$ is a factor of $p(x)$.
To find the remaining zeros,divide $p(x)$ by $(x + 2)$ using polynomial long division or synthetic division:
$(x^3 + 6x^2 + 11x + 6) \div (x + 2) = x^2 + 4x + 3$.
Now,factorize the quadratic polynomial $x^2 + 4x + 3$:
$x^2 + 3x + x + 3 = x(x + 3) + 1(x + 3) = (x + 1)(x + 3)$.
Setting these factors to zero gives $x + 1 = 0$ and $x + 3 = 0$,which results in $x = -1$ and $x = -3$.
Thus,the remaining zeros are $-1$ and $-3$.

(B) Since $\sqrt{2}$ and $-\sqrt{2}$ are zeros of $p(x)$,$(x - \sqrt{2})(x + \sqrt{2}) = x^{2} - 2$ is a factor of $p(x)$.
Dividing $p(x) = 2x^{4} + 7x^{3} - 19x^{2} - 14x + 30$ by $(x^{2} - 2)$,we get the quotient $2x^{2} + 7x - 15$.
To find the other zeros,we solve $2x^{2} + 7x - 15 = 0$.
$2x^{2} + 10x - 3x - 15 = 0$
$2x(x + 5) - 3(x + 5) = 0$
$(2x - 3)(x + 5) = 0$
Thus,$x = \frac{3}{2}$ and $x = -5$ are the other zeros.

(C) For a quadratic polynomial $p(x) = ax^{2} + bx + c$,the product of the zeros is given by the formula $\frac{c}{a}$.
Given the polynomial $p(x) = ax^{2} - 6x - 6$,we identify the coefficients as $A = a$,$B = -6$,and $C = -6$.
The product of the zeros is given as $4$.
Therefore,$\frac{C}{A} = 4$.
Substituting the values,we get $\frac{-6}{a} = 4$.
Solving for $a$: $4a = -6$,which implies $a = -\frac{6}{4} = -\frac{3}{2}$.
Thus,the correct value of $a$ is $-\frac{3}{2}$.

(D) For a quadratic polynomial $p(x) = ax^2 + bx + c$,the product of the zeros is given by $\alpha \cdot \beta = \frac{c}{a}$.
Given the polynomial $p(x) = 3x^2 - 10x + 3$,we have $a = 3$,$b = -10$,and $c = 3$.
Let the zeros be $\alpha = \frac{1}{3}$ and $\beta$ be the other zero.
Using the product of zeros formula: $\frac{1}{3} \cdot \beta = \frac{3}{3} = 1$.
Solving for $\beta$: $\beta = 1 \cdot 3 = 3$.
Therefore,the other zero is $3$.

(A) To find the zero of the polynomial $p(x) = \sqrt{3}x - 3$,we set $p(x) = 0$.
$\sqrt{3}x - 3 = 0$
$\sqrt{3}x = 3$
$x = \frac{3}{\sqrt{3}}$
Since $3 = \sqrt{3} \times \sqrt{3}$,we have:
$x = \frac{\sqrt{3} \times \sqrt{3}}{\sqrt{3}}$
$x = \sqrt{3}$
Therefore,the zero of the polynomial is $\sqrt{3}$.

(B) For a quadratic polynomial of the form $ax^{2}+bx+c$,the sum of the zeros is given by the formula $-\frac{b}{a}$.
Here,$a = 1$,$b = 7$,and $c = 12$.
Therefore,the sum of the zeros $= -\frac{7}{1} = -7$.

(D) For a cubic polynomial of the form $ax^{3}+bx^{2}+cx+d$,the product of the zeros is given by the formula $-\frac{d}{a}$.
Comparing the given polynomial $x^{3}+4x^{2}+x-6$ with the standard form,we have $a=1$,$b=4$,$c=1$,and $d=-6$.
Substituting these values into the formula:
Product of zeros $= -\frac{-6}{1} = 6$.
Therefore,the correct option is $D$.

(A) For a cubic polynomial of the form $ax^{3}+bx^{2}+cx+d$,the product of its zeros is given by the formula $-\frac{d}{a}$.
Comparing the given polynomial $kx^{3}-6x^{2}+11x-6$ with the standard form,we have $a=k$,$b=-6$,$c=11$,and $d=-6$.
The product of the zeros is given as $4$.
Therefore,$-\frac{-6}{k} = 4$.
$\frac{6}{k} = 4$.
$4k = 6$.
$k = \frac{6}{4} = \frac{3}{2}$.

(B) The standard form of a quadratic polynomial is given by $p(x) = ax^{2} + bx + c$.
Given the values $a = 1$,$b = -2\sqrt{3}$,and $c = 2$.
Substituting these values into the standard form:
$p(x) = (1)x^{2} + (-2\sqrt{3})x + 2$
$p(x) = x^{2} - 2\sqrt{3}x + 2$
Therefore,the correct option is $B$.

(C) According to the Remainder Theorem,if a polynomial $p(x)$ is divided by $(x - a)$,the remainder is $p(a)$.
Here,$p(x) = x^{3} - 3x^{2} - x + 3$ and the divisor is $(x - 1)$,so $a = 1$.
To find the remainder,calculate $p(1)$:
$p(1) = (1)^{3} - 3(1)^{2} - (1) + 3$
$p(1) = 1 - 3(1) - 1 + 3$
$p(1) = 1 - 3 - 1 + 3$
$p(1) = 0$.
Therefore,the remainder is $0$.

(D) To find the zeros of the polynomial $p(x) = x^{3} - 4x$,we set $p(x) = 0$.
$x^{3} - 4x = 0$
Factor out $x$ from the expression:
$x(x^{2} - 4) = 0$
Using the difference of squares formula $a^{2} - b^{2} = (a - b)(a + b)$,we can write $x^{2} - 4$ as $(x - 2)(x + 2)$:
$x(x - 2)(x + 2) = 0$
Setting each factor to zero,we get:
$x = 0$,$x - 2 = 0 \Rightarrow x = 2$,and $x + 2 = 0 \Rightarrow x = -2$.
Thus,the zeros are $0, 2, -2$,which can be written as $0, \pm 2$.

(B) polynomial is classified based on its degree,which is the highest power of the variable $x$ present in the expression.
In the given polynomial $p(x) = 3x^2 + 7x + 4$,the highest power of $x$ is $2$.
$A$ polynomial of degree $2$ is called a quadratic polynomial.
Therefore,$p(x) = 3x^2 + 7x + 4$ is a quadratic polynomial.

(A) To find the value of the polynomial $p(x) = 3x^2 + 7x + 4$ at $x = 1$,we substitute $1$ for $x$ in the expression.
$p(1) = 3(1)^2 + 7(1) + 4$
$p(1) = 3(1) + 7 + 4$
$p(1) = 3 + 7 + 4$
$p(1) = 14$
Therefore,the value of the polynomial at $x = 1$ is $14$.

(C) quadratic polynomial is a polynomial of degree $2$. According to the Fundamental Theorem of Algebra,a polynomial of degree $n$ has at most $n$ roots or zeros. Therefore,a quadratic polynomial $(n=2)$ has at most $2$ zeros.

(D) quadratic polynomial in one variable $x$ is an algebraic expression of the form $ax^2 + bx + c$,where $a, b,$ and $c$ are real numbers $(a, b, c \in R)$ and the coefficient of the squared term must be non-zero $(a \neq 0)$.
If we set the polynomial equal to zero,it becomes a quadratic equation $(ax^2 + bx + c = 0)$.
Since the question asks for the standard form of a quadratic polynomial,the correct expression is $ax^2 + bx + c$ with the condition $a \neq 0$.

(C) To find the value of the polynomial $p(x) = 4x^3 + 3x^2 + 2x + 1$ at $x = 1$,we substitute $1$ for $x$ in the expression.
$p(1) = 4(1)^3 + 3(1)^2 + 2(1) + 1$
$p(1) = 4(1) + 3(1) + 2 + 1$
$p(1) = 4 + 3 + 2 + 1 = 10$
Therefore,the value of the polynomial at $x = 1$ is $10$.

(D) The degree of a polynomial is defined as the highest power of the variable present in the polynomial.
Given the polynomial $p(x) = 3x - x^4 + x^2 + 2x^3 + 7$.
Rearranging the terms in descending order of their powers,we get $p(x) = -x^4 + 2x^3 + x^2 + 3x + 7$.
The powers of $x$ in the terms are $4, 3, 2, 1,$ and $0$ (for the constant term).
The highest power among these is $4$.
Therefore,the degree of the polynomial is $4$.

(D) The given polynomial is $p(x) = 7 - 5x^4 - 2x^3 - x^2$.
In the term $-5x^4$,the variable part is $x^4$ and the numerical coefficient is $-5$.
Therefore,the coefficient of the underlined term $-5x^4$ is $-5$.

(C) To find the value of the polynomial $p(x) = 2x^{4} - 3x^{3} + 7x + 5$ at $x = -2$,we substitute $-2$ for $x$ in the expression:
$p(-2) = 2(-2)^{4} - 3(-2)^{3} + 7(-2) + 5$
Calculate each term:
$2(-2)^{4} = 2(16) = 32$
$-3(-2)^{3} = -3(-8) = 24$
$7(-2) = -14$
Adding these values together:
$p(-2) = 32 + 24 - 14 + 5 = 47$
Therefore,the value of the polynomial at $x = -2$ is $47$.

(B) To find $p(-1)$,substitute $x = -1$ into the given polynomial $p(x) = 3x^3 + 2x^2 + 7x + 8$.
$p(-1) = 3(-1)^3 + 2(-1)^2 + 7(-1) + 8$
$p(-1) = 3(-1) + 2(1) - 7 + 8$
$p(-1) = -3 + 2 - 7 + 8$
$p(-1) = -1 + 1 = 0$
Therefore,the value of $p(-1)$ is $0$.

(D) Given the polynomial $p(x) = x^{2} - 2x - 3$.
To find $p(3)$,substitute $x = 3$ into the polynomial expression:
$p(3) = (3)^{2} - 2(3) - 3$
$p(3) = 9 - 6 - 3$
$p(3) = 3 - 3$
$p(3) = 0$
Therefore,the correct option is $D$.

(C) To find the zero of the polynomial $p(x) = 3x - 6$,we set $p(x) = 0$.
Setting the expression equal to zero gives: $3x - 6 = 0$.
Adding $6$ to both sides,we get: $3x = 6$.
Dividing both sides by $3$,we get: $x = 6 / 3 = 2$.
Therefore,the zero of the polynomial is $2$.

(D) To find the zeros of the polynomial $p(x) = x^{2} + 4x + 3$,we set $p(x) = 0$.
$x^{2} + 4x + 3 = 0$
We factorize the quadratic equation by splitting the middle term:
$x^{2} + 3x + x + 3 = 0$
$x(x + 3) + 1(x + 3) = 0$
$(x + 3)(x + 1) = 0$
Therefore,$x + 3 = 0$ or $x + 1 = 0$.
This gives $x = -3$ or $x = -1$.
Thus,the zeros are $-1$ and $-3$.

(A) To find the zeros of the polynomial $p(x) = 2 + x - x^{2}$,we set $p(x) = 0$.
$2 + x - x^{2} = 0$
Rearranging the terms in standard form $ax^{2} + bx + c = 0$:
$-x^{2} + x + 2 = 0$
Multiplying by $-1$ to simplify:
$x^{2} - x - 2 = 0$
Now,factor the quadratic equation by splitting the middle term:
$x^{2} - 2x + x - 2 = 0$
$x(x - 2) + 1(x - 2) = 0$
$(x - 2)(x + 1) = 0$
Setting each factor to zero:
$x - 2 = 0 \implies x = 2$
$x + 1 = 0 \implies x = -1$
Thus,the zeros are $-1$ and $2$.

(B) To find the zero of the polynomial $p(x) = x^{2} + 6x + 9$,we set $p(x) = 0$.
$x^{2} + 6x + 9 = 0$
This is a quadratic expression of the form $a^{2} + 2ab + b^{2} = (a + b)^{2}$.
Here,$a = x$ and $b = 3$,so $x^{2} + 2(x)(3) + 3^{2} = (x + 3)^{2}$.
Thus,$(x + 3)^{2} = 0$.
Taking the square root on both sides,we get $x + 3 = 0$.
Therefore,$x = -3$.

(A) To find the zero of the polynomial $p(x) = -x^2 + 2x - 1$,we set $p(x) = 0$.
$-x^2 + 2x - 1 = 0$
Multiply the entire equation by $-1$ to simplify:
$x^2 - 2x + 1 = 0$
This is a perfect square trinomial of the form $(a - b)^2 = a^2 - 2ab + b^2$,where $a = x$ and $b = 1$.
$(x - 1)^2 = 0$
Taking the square root of both sides:
$x - 1 = 0$
$x = 1$
Therefore,the zero of the polynomial is $1$.

(A) To find the zeros of the quadratic polynomial $p(x) = 3x^2 + 5x - 8$,we set $p(x) = 0$.
$3x^2 + 5x - 8 = 0$
We factorize the quadratic expression by splitting the middle term. We need two numbers whose product is $3 \times (-8) = -24$ and whose sum is $5$. These numbers are $8$ and $-3$.
$3x^2 - 3x + 8x - 8 = 0$
$3x(x - 1) + 8(x - 1) = 0$
$(3x + 8)(x - 1) = 0$
Setting each factor to zero:
$3x + 8 = 0 \implies x = -\frac{8}{3}$
$x - 1 = 0 \implies x = 1$
Thus,the zeros are $-\frac{8}{3}$ and $1$.

(A) To find the zeros of the polynomial $p(x) = x^{2} - 9$,we set $p(x) = 0$.
$x^{2} - 9 = 0$
$x^{2} = 9$
$x = \pm \sqrt{9}$
$x = 3$ or $x = -3$.
Since there are two distinct values of $x$ for which the polynomial equals zero,the number of zeros is $2$.

(C) To find the zero of a polynomial $p(x)$,we set $p(x) = 0$.
Given the linear polynomial $p(x) = ax + b$,we set $ax + b = 0$.
Subtracting $b$ from both sides,we get $ax = -b$.
Dividing both sides by $a$ (where $a \neq 0$),we get $x = -\frac{b}{a}$.
Therefore,the zero of the polynomial is $-\frac{b}{a}$.

(D) For a quadratic polynomial of the form $p(x) = ax^{2} + bx + c$,the sum of the zeros is given by the formula $-\frac{b}{a}$.
Comparing $p(x) = x^{2} + 3x + 2$ with the standard form,we get $a = 1$,$b = 3$,and $c = 2$.
Therefore,the sum of the zeros $= -\frac{b}{a} = -\frac{3}{1} = -3$.

(B) For a quadratic polynomial of the form $p(x) = ax^{2} + bx + c$,the product of the zeros is given by the formula $\frac{c}{a}$.
Comparing $p(x) = x^{2} + 5x + 6$ with the standard form,we have $a = 1$,$b = 5$,and $c = 6$.
Therefore,the product of the zeros $= \frac{c}{a} = \frac{6}{1} = 6$.

(C) For a quadratic polynomial of the form $ax^2 + bx + c$,the sum of the zeros is given by the formula $-\frac{b}{a}$.
Given the polynomial $p(x) = 3x^2 + 7x + 4$,we identify the coefficients as $a = 3$,$b = 7$,and $c = 4$.
Substituting these values into the formula,we get:
Sum of zeros = $-\frac{b}{a} = -\frac{7}{3}$.
Therefore,the correct option is $C$.

(D) For a quadratic polynomial of the form $ax^2 + bx + c$,the product of the zeros is given by the formula $\frac{c}{a}$.
Given the polynomial $p(x) = 3x^2 + 7x + 4$,we identify the coefficients as $a = 3$,$b = 7$,and $c = 4$.
Substituting these values into the formula,we get:
Product of zeros = $\frac{c}{a} = \frac{4}{3}$.
Therefore,the correct option is $D$.

(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 $D$.

(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 $2$.