Mix Examples - Polynomials Questions in English

(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 $-3$.

(A) According to the division algorithm for polynomials, $p(x) = s(x) \cdot q(x) + r(x)$.
Given $p(x) = x^{3} - 3x^{2} + x + 2$, $q(x) = x - 2$, and $r(x) = -2x + 4$.
Substituting these values into the formula: $x^{3} - 3x^{2} + x + 2 = s(x) \cdot (x - 2) + (-2x + 4)$.
Rearranging to solve for $s(x) \cdot (x - 2)$: $s(x) \cdot (x - 2) = (x^{3} - 3x^{2} + x + 2) - (-2x + 4)$.
$s(x) \cdot (x - 2) = x^{3} - 3x^{2} + x + 2 + 2x - 4$.
$s(x) \cdot (x - 2) = x^{3} - 3x^{2} + 3x - 2$.
Now, divide $x^{3} - 3x^{2} + 3x - 2$ by $(x - 2)$ to find $s(x)$.
Using polynomial division: $(x^{3} - 2x^{2}) - (x^{2} - 2x) + (x - 2) = x^{2}(x - 2) - x(x - 2) + 1(x - 2) = (x^{2} - x + 1)(x - 2)$.
Therefore, $s(x) = x^{2} - x + 1$.

(B) To find what must be added to $p(x)$ to make it divisible by $g(x) = 3x^2 - 2x + 4$,we perform polynomial long division of $p(x)$ by $g(x)$.
Dividing $6x^5 + 5x^4 + 11x^3 - 3x^2 + x + 1$ by $3x^2 - 2x + 4$:
$1$. Divide $6x^5$ by $3x^2$ to get $2x^3$. Multiply $2x^3(3x^2 - 2x + 4) = 6x^5 - 4x^4 + 8x^3$. Subtracting this from $p(x)$ gives $9x^4 + 3x^3 - 3x^2 + x + 1$.
$2$. Divide $9x^4$ by $3x^2$ to get $3x^2$. Multiply $3x^2(3x^2 - 2x + 4) = 9x^4 - 6x^3 + 12x^2$. Subtracting this gives $9x^3 - 15x^2 + x + 1$.
$3$. Divide $9x^3$ by $3x^2$ to get $3x$. Multiply $3x(3x^2 - 2x + 4) = 9x^3 - 6x^2 + 12x$. Subtracting this gives $-9x^2 - 11x + 1$.
$4$. Divide $-9x^2$ by $3x^2$ to get $-3$. Multiply $-3(3x^2 - 2x + 4) = -9x^2 + 6x - 12$. Subtracting this gives $-17x + 13$.
The remainder is $r(x) = -17x + 13$.
To make the polynomial divisible,we must add $-r(x) = -(-17x + 13) = 17x - 13$.

(C) To find what must be subtracted,we divide the polynomial $P(x) = 8x^4 + 14x^3 - 2x^2 + 8x - 12$ by $D(x) = 4x^2 + 3x - 2$.
Step $1$: Divide $8x^4$ by $4x^2$ to get $2x^2$. Multiply $2x^2(4x^2 + 3x - 2) = 8x^4 + 6x^3 - 4x^2$. Subtracting this from $P(x)$ gives $(14x^3 - 6x^3) + (-2x^2 + 4x^2) + 8x - 12 = 8x^3 + 2x^2 + 8x - 12$.
Step $2$: Divide $8x^3$ by $4x^2$ to get $2x$. Multiply $2x(4x^2 + 3x - 2) = 8x^3 + 6x^2 - 4x$. Subtracting this from the remainder gives $(2x^2 - 6x^2) + (8x + 4x) - 12 = -4x^2 + 12x - 12$.
Step $3$: Divide $-4x^2$ by $4x^2$ to get $-1$. Multiply $-1(4x^2 + 3x - 2) = -4x^2 - 3x + 2$. Subtracting this from the current remainder gives $(12x + 3x) + (-12 - 2) = 15x - 14$.
Since the remainder is $15x - 14$,subtracting this from the original polynomial will make it exactly divisible by $4x^2 + 3x - 2$.

(D) To find the other polynomial,we divide the product $x^{5}+3x^{4}+5x^{3}+3x^{2}+9x+15$ by the given polynomial $x^{2}+3$.
Performing polynomial long division:
$1$. Divide $x^{5}$ by $x^{2}$ to get $x^{3}$. Multiply $x^{3}(x^{2}+3) = x^{5}+3x^{3}$. Subtracting this from the original polynomial leaves $3x^{4}+2x^{3}+3x^{2}+9x+15$.
$2$. Divide $3x^{4}$ by $x^{2}$ to get $3x^{2}$. Multiply $3x^{2}(x^{2}+3) = 3x^{4}+9x^{2}$. Subtracting this leaves $2x^{3}-6x^{2}+9x+15$.
$3$. Wait,let us re-examine the division: $(x^{5}+3x^{4}+5x^{3}+3x^{2}+9x+15) / (x^{2}+3)$.
Factor by grouping: $x^{4}(x+3) + 5x^{2}(x+3) + 3(x+3) = (x^{4}+5x^{2}+3)(x+3)$.
Actually,let's divide directly: $(x^{2}+3)(x^{3}+3x^{2}+2x+5)$ is not correct. Let's re-divide: $(x^{5}+3x^{4}+5x^{3}+3x^{2}+9x+15) = x^{3}(x^{2}+3) + 3x^{2}(x^{2}+3) + 5(x^{2}+3) = (x^{3}+3x^{2}+5)(x^{2}+3)$.
Therefore,the other polynomial is $x^{3}+3x^{2}+5$.

(A) According to the division algorithm for polynomials, the dividend $p(x)$ is given by the formula: $p(x) = (\text{divisor} \times \text{quotient}) + \text{remainder}$.
Given:
Divisor $d(x) = x^{2}+4x+2$
Quotient $q(x) = x^{2}-4x+14$
Remainder $r(x) = 9x-13$
Substituting these values into the formula:
$p(x) = (x^{2}+4x+2)(x^{2}-4x+14) + (9x-13)$
First, multiply the divisor and the quotient:
$(x^{2}+4x+2)(x^{2}-4x+14) = x^{2}(x^{2}-4x+14) + 4x(x^{2}-4x+14) + 2(x^{2}-4x+14)$
$= (x^{4}-4x^{3}+14x^{2}) + (4x^{3}-16x^{2}+56x) + (2x^{2}-8x+28)$
Combine like terms:
$= x^{4} + (-4x^{3}+4x^{3}) + (14x^{2}-16x^{2}+2x^{2}) + (56x-8x) + 28$
$= x^{4} + 0x^{3} + 0x^{2} + 48x + 28$
Now, add the remainder $r(x) = 9x-13$:
$p(x) = (x^{4}+48x+28) + (9x-13)$
$p(x) = x^{4} + (48x+9x) + (28-13)$
$p(x) = x^{4} + 57x + 15$.

(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 $= -\frac{7}{1} = -7$.

(C) To find the points where the graph of $p(x) = x^2 - 9$ intersects the $X$-axis,we set $p(x) = 0$.
$x^2 - 9 = 0$
$(x - 3)(x + 3) = 0$
This gives $x = 3$ and $x = -3$.
Since there are two distinct real roots,the graph intersects the $X$-axis at two distinct points: $(3, 0)$ and $(-3, 0)$.
Therefore,the correct option is $C$.

(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.

(C) polynomial is classified based on its degree,which is the highest power of the variable $x$ present in the expression.
In the given polynomial $P(x) = 4x^{3} + 3x^{2} + 2x + 1$,the highest power of $x$ is $3$.
$A$ polynomial of degree $3$ is called a cubic polynomial.
Therefore,$P(x)$ is a cubic 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) 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 $4$.

(B) The degree of a polynomial is defined as the highest power of the variable present in the polynomial expression.
Given the polynomial $p(x) = x^2 - x^3 + x + 1$.
The powers of the variable $x$ are $2, 3, 1,$ and $0$ (for the constant term $1$).
The highest power among these is $3$.
Therefore,the degree of the polynomial $p(x)$ is $3$.

(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 $D$.

(D) To find the factors of the polynomial $p(x) = x^{3} - 3x^{2} + 9x - 27$,we can use the method of grouping.
Step $1$: Group the terms of the polynomial: $p(x) = (x^{3} - 3x^{2}) + (9x - 27)$.
Step $2$: Factor out the common terms from each group: $p(x) = x^{2}(x - 3) + 9(x - 3)$.
Step $3$: Factor out the common binomial $(x - 3)$: $p(x) = (x - 3)(x^{2} + 9)$.
Thus,the factors are $(x - 3)$ and $(x^{2} + 9)$.
Comparing this with the given options,$(x - 3)$ is one of the factors.

(B) To find the factors of $p(x) = x^{3} + 2x^{2} + 3x + 2$,we can use the Factor Theorem.
If $(x + 1)$ is a factor,then $p(-1)$ must be $0$.
$p(-1) = (-1)^{3} + 2(-1)^{2} + 3(-1) + 2$
$p(-1) = -1 + 2(1) - 3 + 2$
$p(-1) = -1 + 2 - 3 + 2 = 0$
Since $p(-1) = 0$,by the Factor Theorem,$(x + 1)$ is a factor of $p(x)$.

(C) linear polynomial is of the form $p(x) = ax + b$,where $a \neq 0$ and $a, b$ are real numbers.
The graph of the equation $y = ax + b$ represents a straight line in the Cartesian plane.
Therefore,the graph of a linear polynomial is always a straight line.

(A) linear polynomial is defined as $p(x) = ax + b$,where $a \neq 0$.
To find the intersection with the $X$-axis,we set $p(x) = 0$.
$ax + b = 0$
$ax = -b$
$x = -b/a$
Since $a \neq 0$,there is exactly one unique value of $x$ for which $p(x) = 0$.
Therefore,the graph of a linear polynomial intersects the $X$-axis at exactly one point,which is $(-b/a, 0)$.

(D) To find the point where the graph of the polynomial $p(x) = 3x - 6$ intersects the $X$-axis,we set $p(x) = 0$.
Setting the polynomial to zero gives: $3x - 6 = 0$.
Adding $6$ to both sides: $3x = 6$.
Dividing by $3$: $x = 2$.
Since the point lies on the $X$-axis,its $y$-coordinate is $0$.
Therefore,the graph intersects the $X$-axis at the point $(2, 0)$.

(D) The given polynomial is $p(x) = x^{2} + 5x + 6$.
This is a quadratic polynomial of the form $ax^{2} + bx + c$,where $a = 1$,$b = 5$,and $c = 6$.
The graph of any quadratic polynomial $p(x) = ax^{2} + bx + c$ (where $a \neq 0$) is always a curve known as a parabola.
Since the coefficient of $x^{2}$ is positive $(a = 1 > 0)$,the parabola opens upwards.
Therefore,the graph of $p(x) = x^{2} + 5x + 6$ is a parabola.

(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) The given polynomial is a quadratic polynomial of the form $p(x) = ax^{2} + bx + c$,where $a = 1$,$b = 6$,and $c = 8$.
Since the coefficient of $x^{2}$ is $a = 1$,which is greater than $0$ $(a > 0)$,the graph of the quadratic polynomial is a parabola that opens upwards.
Therefore,the correct option is $A$.

(B) To determine the number of points where the graph of the quadratic polynomial $p(x) = x^{2} - x - 2$ intersects the $X$-axis,we need to find the number of real roots of the equation $p(x) = 0$.
$1$. Set the polynomial equal to zero: $x^{2} - x - 2 = 0$.
$2$. Calculate the discriminant $D = b^{2} - 4ac$,where $a = 1$,$b = -1$,and $c = -2$.
$3$. $D = (-1)^{2} - 4(1)(-2) = 1 + 8 = 9$.
$4$. Since $D > 0$,the quadratic equation has two distinct real roots.
$5$. Therefore,the graph of the polynomial intersects the $X$-axis at two distinct points.

(D) The given polynomial is a quadratic polynomial of the form $p(x) = ax^2 + bx + c$,where $a = -1$,$b = 1$,and $c = 6$.
Since the degree of the polynomial is $2$,its graph is a parabola.
In a quadratic polynomial $ax^2 + bx + c$,if $a > 0$,the parabola opens upwards,and if $a < 0$,the parabola opens downwards.
Here,$a = -1$,which is less than $0$.
Therefore,the graph of $p(x) = -x^2 + x + 6$ is a parabola opening downwards.

(C) To determine how the graph of $p(x) = x^{2} + 6x + 9$ intersects the $X$-axis,we find the zeros of the polynomial by setting $p(x) = 0$.
$x^{2} + 6x + 9 = 0$
This is a perfect square trinomial,which can be factored as $(x + 3)^{2} = 0$.
Thus,$x = -3$ is the only root (a repeated root).
Since there is only one distinct real root,the parabola represented by the quadratic polynomial touches the $X$-axis at exactly one point,which is $(-3, 0)$.

(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 $D$.