Relationship between Zeroes and Coefficients of a Polynomial Questions in English

(7, -5) Given that $2+\sqrt{3}$ and $2-\sqrt{3}$ are zeroes of the given polynomial.
Therefore,$(x-(2+\sqrt{3}))(x-(2-\sqrt{3})) = ((x-2)-\sqrt{3})((x-2)+\sqrt{3}) = (x-2)^{2} - (\sqrt{3})^{2} = x^{2}-4x+4-3 = x^{2}-4x+1$ is a factor of the given polynomial.
To find the remaining zeroes,we divide the polynomial $x^{4}-6 x^{3}-26 x^{2}+138 x-35$ by $x^{2}-4 x+1$:
$x^{4}-6 x^{3}-26 x^{2}+138 x-35 = (x^{2}-4 x+1)(x^{2}-2 x-35)$
Now,we factorize the quadratic polynomial $x^{2}-2 x-35$:
$x^{2}-2 x-35 = x^{2}-7x+5x-35 = x(x-7)+5(x-7) = (x-7)(x+5)$
Setting these factors to zero,we get $x-7=0$ or $x+5=0$,which gives $x=7$ or $x=-5$.
Hence,the other two zeroes are $7$ and $-5$.

(D) Given that $(x+1)$ is a factor of $f(x) = 2x^{3} + ax^{2} + 2bx + 1$,by the Factor Theorem,$f(-1) = 0$.
Substituting $x = -1$ into the polynomial:
$2(-1)^{3} + a(-1)^{2} + 2b(-1) + 1 = 0$
$-2 + a - 2b + 1 = 0$
$a - 2b = 1$ ... $(i)$
We are also given the equation $2a - 3b = 4$ ... $(ii)$.
From equation $(i)$,we have $a = 2b + 1$.
Substituting this into equation $(ii)$:
$2(2b + 1) - 3b = 4$
$4b + 2 - 3b = 4$
$b + 2 = 4$
$b = 2$
Now,substitute $b = 2$ into $a = 2b + 1$:
$a = 2(2) + 1 = 5$.
Thus,the values are $a = 5$ and $b = 2$.

(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 $-\frac{4}{3\sqrt{2}}$.

(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.

(B) The given polynomial is $p(x) = 2012x + 2011$.
The highest power of the variable $x$ in this expression is $1$.
$A$ polynomial with a degree of $1$ is known as a linear polynomial.
Therefore,$p(x) = 2012x + 2011$ is a linear polynomial.

(C) The given polynomial is $p(x) = 9x^{2} + \frac{7}{2}x^{3} + 5$.
Rearranging the terms in descending order of their powers,we get $p(x) = \frac{7}{2}x^{3} + 9x^{2} + 5$.
The degree of a polynomial is the highest power of the variable present in the expression.
Here,the highest power of $x$ is $3$.
Since the degree of the polynomial is $3$,it is classified as a cubic polynomial.

(B) The given polynomial is $p(x) = x^{2} - \sqrt{3}x^{3} + 4x^{7} + 9$.
To find the degree,we first arrange the terms in descending order of their exponents:
$p(x) = 4x^{7} - \sqrt{3}x^{3} + x^{2} + 9$.
The degree of a polynomial is defined as the highest power of the variable present in the polynomial.
In this expression,the powers of $x$ are $7, 3, 2,$ and $0$ (since $9 = 9x^{0}$).
The maximum power is $7$.
Therefore,the degree of the polynomial is $7$.

(C) Given polynomial is $p(x) = 5x^{100} - (x^{10})^{20} + 3(x^2)^{25} + 2x^{25} - 7$.
First,simplify the exponents using the power rule $(x^a)^b = x^{a \cdot b}$:
$(x^{10})^{20} = x^{10 \cdot 20} = x^{200}$
$(x^2)^{25} = x^{2 \cdot 25} = x^{50}$
Substitute these back into the polynomial:
$p(x) = 5x^{100} - x^{200} + 3x^{50} + 2x^{25} - 7$
Rearrange the terms in descending order of their powers:
$p(x) = -x^{200} + 5x^{100} + 3x^{50} + 2x^{25} - 7$
The degree of a polynomial is the highest power of the variable $x$ present in the expression.
Here,the highest power is $200$.
Therefore,the degree of the polynomial is $200$.

(D) The given polynomial is $p(x) = \frac{7}{2} x - 9$.
The degree of a polynomial is defined as the highest power of the variable $x$ present in the expression.
In the expression $\frac{7}{2} x - 9$,the variable $x$ has an exponent of $1$ (since $x = x^1$).
Therefore,the degree of the polynomial is $1$.

(A) The given polynomial is $p(x) = 5x - 2x^2 + 3$.
Rearranging the terms in descending order of their powers,we get $p(x) = -2x^2 + 5x + 3$.
The degree of a polynomial is defined as the highest power of the variable present in the polynomial.
In this polynomial,the highest power of the variable $x$ is $2$ (from the term $-2x^2$).
Therefore,the degree of the polynomial is $2$.

(N/A) According to the Factor Theorem,if $(x-a)$ is a factor of a polynomial $p(x)$,then $p(a)=0$.
Here,we need to check if $(x+2)$ is a factor of $p(x)=2x^{3}-4x^{2}+5x+42$.
Comparing $(x+2)$ with $(x-a)$,we get $a = -2$.
Now,we calculate $p(-2)$:
$p(-2) = 2(-2)^{3} - 4(-2)^{2} + 5(-2) + 42$
$p(-2) = 2(-8) - 4(4) - 10 + 42$
$p(-2) = -16 - 16 - 10 + 42$
$p(-2) = -42 + 42$
$p(-2) = 0$
Since $p(-2) = 0$,by the Factor Theorem,$(x+2)$ is a factor of $p(x)$.

(D) The degree of a polynomial is the highest power of the variable present in the expression.
For the given polynomial $p(x) = 7x^2 - 5x^3 + 3$,the powers of $x$ are $2$,$3$,and $0$ (since $3 = 3x^0$).
The highest power is $3$.
$A$ polynomial with a degree of $3$ is called a cubic polynomial.
Therefore,the correct option is $D$.

(A) The degree of a polynomial is the highest power of the variable present in the expression.
In the given polynomial $p(x) = 7 - 5x$,the variable is $x$ and its highest power is $1$.
$A$ polynomial of degree $1$ is called a linear polynomial.
Therefore,$p(x) = 7 - 5x$ is a linear polynomial.

(B) To identify the type of polynomial,we first simplify the expression $p(x) = (x - 15)(x + 15)$.
Using the algebraic identity $(a - b)(a + b) = a^2 - b^2$,we get:
$p(x) = x^2 - 15^2$
$p(x) = x^2 - 225$
The highest power (degree) of the variable $x$ in this polynomial is $2$.
$A$ polynomial with a degree of $2$ is called a quadratic polynomial.
Therefore,the correct option is $B$.

(C) The given polynomial is $p(x) = \frac{3}{2} - 0.75x - x^{2}$.
To determine the type of polynomial,we look at the highest power of the variable $x$ present in the expression.
Here,the powers of $x$ are $0$ (in $\frac{3}{2}$),$1$ (in $-0.75x$),and $2$ (in $-x^{2}$).
The highest power (degree) of the polynomial is $2$.
$A$ polynomial of degree $2$ is called a quadratic polynomial.
Therefore,the correct option is $C$.

(D) To find the degree of the polynomial $p(x) = 5x^8 - 6(x^2)^5 - 4x^3 - 14$,we first simplify the expression.
Using the power rule $(x^a)^b = x^{a \times b}$,we simplify the term $6(x^2)^5$ as $6x^{10}$.
Now,the polynomial becomes $p(x) = 5x^8 - 6x^{10} - 4x^3 - 14$.
The degree of a polynomial is defined as the highest power of the variable $x$ present in the expression.
Comparing the powers of $x$ $(8, 10, 3)$,the highest power is $10$.
Therefore,the degree of the polynomial is $10$.

(A) To find the degree of the polynomial $p(x) = (x^{1/2})^{50} - 5x^{10} - 9x^2 + 15$,we first simplify the expression.
Using the power of a power rule $(a^m)^n = a^{m \cdot n}$,we simplify the first term:
$(x^{1/2})^{50} = x^{(1/2) \cdot 50} = x^{25}$.
Now,the polynomial becomes $p(x) = x^{25} - 5x^{10} - 9x^2 + 15$.
The degree of a polynomial is defined as the highest power of the variable $x$ present in the expression.
Comparing the exponents of $x$ in the terms $x^{25}$,$x^{10}$,and $x^2$,the highest exponent is $25$.
Therefore,the degree of the polynomial is $25$.

(B) The degree of a polynomial is defined as the highest power of the variable present in the polynomial expression.
Given the polynomial: $p(x) = 6 - 5x^3 + 4x^5 - 5x^2$.
The powers of the variable $x$ in the terms are $0$ (in $6$),$3$ (in $-5x^3$),$5$ (in $4x^5$),and $2$ (in $-5x^2$).
The highest power among these is $5$.
Therefore,the degree of the polynomial $p(x)$ is $5$.

(C) Let the polynomial be $p(x) = x^{3} + 4x^{2} + 8x + a$.
According to the Factor Theorem,if $(x+2)$ is a factor of $p(x)$,then $p(-2) = 0$.
Substituting $x = -2$ into the polynomial:
$p(-2) = (-2)^{3} + 4(-2)^{2} + 8(-2) + a = 0$
$-8 + 4(4) - 16 + a = 0$
$-8 + 16 - 16 + a = 0$
$-8 + a = 0$
$a = 8$
Therefore,the value of $a$ is $8$.

(TRUE) To determine if $(x+3)$ is a factor of $p(x) = x^{2}+10x+21$,we use the Factor Theorem.
According to the Factor Theorem,$(x-a)$ is a factor of $p(x)$ if $p(a) = 0$.
Here,$(x+3) = (x - (-3))$,so $a = -3$.
Now,calculate $p(-3)$:
$p(-3) = (-3)^{2} + 10(-3) + 21$
$p(-3) = 9 - 30 + 21$
$p(-3) = 30 - 30 = 0$.
Since $p(-3) = 0$,the statement is True.

(B) To check if $(x-2)$ is a factor of the polynomial $p(x) = x^{3}+5x^{2}-2x-25$,we use the Factor Theorem.
According to the Factor Theorem,$(x-a)$ is a factor of $p(x)$ if and only if $p(a) = 0$.
Here,$a = 2$.
Substitute $x = 2$ into the polynomial:
$p(2) = (2)^{3} + 5(2)^{2} - 2(2) - 25$
$p(2) = 8 + 5(4) - 4 - 25$
$p(2) = 8 + 20 - 4 - 25$
$p(2) = 28 - 29$
$p(2) = -1$
Since $p(2) \neq 0$,$(x-2)$ is not a factor of the given polynomial.
Therefore,the statement is False.

(A) To check if $(x+3)$ is a factor of $p(x) = x^{3}+6x^{2}+5x-12$,we use the Factor Theorem.
According to the Factor Theorem,$(x-a)$ is a factor of $p(x)$ if $p(a) = 0$.
Here,$x+3 = x-(-3)$,so we need to evaluate $p(-3)$.
$p(-3) = (-3)^{3} + 6(-3)^{2} + 5(-3) - 12$
$p(-3) = -27 + 6(9) - 15 - 12$
$p(-3) = -27 + 54 - 15 - 12$
$p(-3) = 54 - 54 = 0$.
Since $p(-3) = 0$,$(x+3)$ is indeed a factor of the given polynomial.
Therefore,the statement is True.

(C) To factorize the polynomial $p(x) = 25x^2 - 40x + 16$,we observe that it is in the form of the algebraic identity $a^2 - 2ab + b^2 = (a - b)^2$.
Here,$25x^2 = (5x)^2$,so $a = 5x$.
Also,$16 = 4^2$,so $b = 4$.
Checking the middle term: $-2ab = -2(5x)(4) = -40x$,which matches the given polynomial.
Therefore,$25x^2 - 40x + 16 = (5x)^2 - 2(5x)(4) + 4^2 = (5x - 4)^2$.

(A) To find the zeros of the polynomial $p(x) = x^{2} - 2x$,we set $p(x) = 0$.
$x^{2} - 2x = 0$
Factor out $x$ from the expression:
$x(x - 2) = 0$
This gives two possible values for $x$:
$x = 0$ or $x - 2 = 0$,which implies $x = 2$.
Thus,the zeros of the polynomial are $0$ and $2$.
Since there are two distinct real values,the number of real zeros is $2$.

(B) To find the number of real zeros of the quadratic polynomial $p(x) = ax^2 + bx + c$,we calculate the discriminant $D = b^2 - 4ac$.
For the given polynomial $p(x) = 4x^2 + 11x + 10$,we have $a = 4$,$b = 11$,and $c = 10$.
Substituting these values into the discriminant formula:
$D = (11)^2 - 4(4)(10)$
$D = 121 - 160$
$D = -39$
Since the discriminant $D < 0$,the quadratic equation $p(x) = 0$ has no real roots.
Therefore,the polynomial has $0$ real zeros.

(D) To find the zeros of the polynomial $p(x) = x^{3} - 9x$,we set $p(x) = 0$.
$x^{3} - 9x = 0$
Factor out $x$ from the expression:
$x(x^{2} - 9) = 0$
Using the difference of squares formula $a^{2} - b^{2} = (a - b)(a + b)$,we can factor $x^{2} - 9$ as $(x - 3)(x + 3)$:
$x(x - 3)(x + 3) = 0$
Setting each factor to zero,we get:
$x = 0$,$x - 3 = 0 \implies x = 3$,and $x + 3 = 0 \implies x = -3$.
Thus,the zeros of the polynomial are $0, 3, -3$.
Since there are $3$ distinct real values,the number of real zeros is $3$.

(A) To verify if $\frac{-2}{3}$ is a zero of the polynomial $p(x)=3x+2$,we substitute $x = \frac{-2}{3}$ into the polynomial.
$p\left(\frac{-2}{3}\right) = 3\left(\frac{-2}{3}\right) + 2$
$p\left(\frac{-2}{3}\right) = -2 + 2$
$p\left(\frac{-2}{3}\right) = 0$
Since the value of the polynomial at $x = \frac{-2}{3}$ is $0$,it is verified that $\frac{-2}{3}$ is a zero of the linear polynomial $p(x)=3x+2$.

(N/A) Given polynomial is $p(x) = 10x^2 - 14x - 12$.
To find the zeros,we factorize the polynomial:
$10x^2 - 14x - 12 = 10x^2 - 20x + 6x - 12$
$= 10x(x - 2) + 6(x - 2)$
$= (10x + 6)(x - 2)$
Setting $p(x) = 0$,we get $(10x + 6)(x - 2) = 0$.
This gives $x = -6/10 = -3/5$ or $x = 2$.
Thus,the zeros are $-3/5$ and $2$.
Using the relationship between zeros and coefficients for $ax^2 + bx + c$:
Sum of zeros $= -b/a = -(-14)/10 = 14/10 = 7/5$.
Product of zeros $= c/a = -12/10 = -6/5$.

(N/A) To find the zeros of $p(x),$ set $p(x)=0$.
$4 x^{3}+10 x^{2}+6 x=0$
$2 x(2 x^{2}+5 x+3)=0$
$2 x(2 x^{2}+2 x+3 x+3)=0$
$2 x(2 x(x+1)+3(x+1))=0$
$2 x(2 x+3)(x+1)=0$
Thus,the zeros are $x=0, x=-\frac{3}{2}, x=-1$.
For $p(x)=4 x^{3}+10 x^{2}+6 x+0$,we have $a=4, b=10, c=6, d=0$.
Sum of zeros: $0 + (-\frac{3}{2}) + (-1) = -\frac{5}{2} = -\frac{10}{4} = -\frac{b}{a}$.
Sum of product of zeros taken two at a time: $(0)(-\frac{3}{2}) + (-\frac{3}{2})(-1) + (-1)(0) = 0 + \frac{3}{2} + 0 = \frac{3}{2} = \frac{6}{4} = \frac{c}{a}$.
Product of zeros: $(0)(-\frac{3}{2})(-1) = 0 = -\frac{0}{4} = -\frac{d}{a}$.

(A) To find the zeros of $p(x)$,we set $p(x) = 0$.
$\sqrt{3} x^{2} - 8 x + 4 \sqrt{3} = 0$
We split the middle term $-8x$ into $-6x - 2x$ because $(\sqrt{3}) \times (4\sqrt{3}) = 12$,and we need two numbers whose product is $12$ and sum is $-8$.
$\sqrt{3} x^{2} - 6x - 2x + 4 \sqrt{3} = 0$
$\sqrt{3} x(x - 2\sqrt{3}) - 2(x - 2\sqrt{3}) = 0$
$(x - 2\sqrt{3})(\sqrt{3} x - 2) = 0$
Setting each factor to zero:
$x - 2\sqrt{3} = 0 \implies x = 2\sqrt{3}$
$\sqrt{3} x - 2 = 0 \implies x = \frac{2}{\sqrt{3}}$
Thus,the zeros are $\frac{2}{\sqrt{3}}$ and $2\sqrt{3}$.

(B) To find the zeros of $p(x)$,we set $p(x) = 0$.
$4x^2 - 256 = 0$
Divide the entire equation by $4$:
$x^2 - 64 = 0$
This can be written as a difference of squares:
$x^2 - 8^2 = 0$
$(x - 8)(x + 8) = 0$
Setting each factor to zero:
$x - 8 = 0 \implies x = 8$
$x + 8 = 0 \implies x = -8$
Thus,the zeros of the quadratic polynomial are $-8$ and $8$.

(C) To find the zeros of $p(x)$,we set $p(x) = 0$.
$x^{2} + x - 12 = 0$
We factorize the quadratic equation by splitting the middle term:
$x^{2} + 4x - 3x - 12 = 0$
$x(x + 4) - 3(x + 4) = 0$
$(x + 4)(x - 3) = 0$
Setting each factor to zero:
$x + 4 = 0 \implies x = -4$
$x - 3 = 0 \implies x = 3$
Therefore,the zeros of the quadratic polynomial $p(x)$ are $-4$ and $3$.

(D) To find the zeros of the polynomial $p(x)$,we set $p(x) = 0$.
$3x^2 + 15x = 0$
Factor out the common term $3x$:
$3x(x + 5) = 0$
Since $3 \neq 0$,we have:
$x(x + 5) = 0$
This implies either $x = 0$ or $x + 5 = 0$.
Therefore,$x = 0$ or $x = -5$.
Thus,the zeros of the quadratic polynomial $p(x)$ are $0$ and $-5$.

(N/A) To verify that $3$ is a zero of the polynomial $p(x) = 4x - 12$,we need to evaluate the polynomial at $x = 3$.
Substituting $x = 3$ into the given polynomial:
$p(3) = 4(3) - 12$
$p(3) = 12 - 12$
$p(3) = 0$
Since $p(3) = 0$,it is verified that $3$ is a zero of the polynomial $p(x) = 4x - 12$.

(A) For a quadratic polynomial of the form $ax^{2} + bx + c$,the sum of the zeros is given by $-\frac{b}{a}$ and the product of the zeros is given by $\frac{c}{a}$.
Given the polynomial $4x^{2} - 4x + 1$,we have $a = 4$,$b = -4$,and $c = 1$.
Sum of the zeros $= -\frac{b}{a} = -\frac{-4}{4} = 1$.
Product of the zeros $= \frac{c}{a} = \frac{1}{4}$.
Thus,the sum is $1$ and the product is $\frac{1}{4}$.

(A) For a quadratic polynomial of the form $ax^{2} + bx + c$,the sum of the zeros is given by $-\frac{b}{a}$ and the product of the zeros is given by $\frac{c}{a}$.
Given polynomial: $6x^{2} - 7x - 3$.
Here,$a = 6$,$b = -7$,and $c = -3$.
Sum of zeros = $-\frac{b}{a} = -(\frac{-7}{6}) = \frac{7}{6}$.
Product of zeros = $\frac{c}{a} = \frac{-3}{6} = -\frac{1}{2}$.
Thus,the sum and product are $\frac{7}{6}$ and $-\frac{1}{2}$ respectively.

(D) For a quadratic polynomial of the form $ax^2 + bx + c$,the sum of the zeros is given by $-\frac{b}{a}$ and the product of the zeros is given by $\frac{c}{a}$.
Given the polynomial $3x^2 + x - 4$,we have $a = 3$,$b = 1$,and $c = -4$.
Sum of zeros $= -\frac{b}{a} = -\frac{1}{3}$.
Product of zeros $= \frac{c}{a} = \frac{-4}{3} = -\frac{4}{3}$.
Thus,the sum and product are $-\frac{1}{3}$ and $-\frac{4}{3}$ respectively.

(A) For a quadratic polynomial of the form $ax^{2} + bx + c$,the sum of the zeros is given by $-\frac{b}{a}$ and the product of the zeros is given by $\frac{c}{a}$.
Given the polynomial $x^{2} - 7x + 10$,we have $a = 1$,$b = -7$,and $c = 10$.
Sum of the zeros $= -\frac{b}{a} = -\frac{-7}{1} = 7$.
Product of the zeros $= \frac{c}{a} = \frac{10}{1} = 10$.
Therefore,the sum and the product of the zeros are $7$ and $10$ respectively.

(B) To find the zeros of $p(x) = 6x^2 - x - 2$,we set $p(x) = 0$.
$6x^2 - x - 2 = 0$
$6x^2 - 4x + 3x - 2 = 0$
$2x(3x - 2) + 1(3x - 2) = 0$
$(2x + 1)(3x - 2) = 0$
Thus,the zeros are $x = -1/2$ and $x = 2/3$.
Let $\alpha = 2/3$ and $\beta = -1/2$.
For a quadratic polynomial $ax^2 + bx + c$,the sum of zeros is $-b/a$ and the product of zeros is $c/a$.
Here,$a = 6, b = -1, c = -2$.
Sum of zeros: $\alpha + \beta = 2/3 + (-1/2) = (4 - 3)/6 = 1/6$. Also,$-b/a = -(-1)/6 = 1/6$. Thus,$\alpha + \beta = -b/a$.
Product of zeros: $\alpha \cdot \beta = (2/3) \cdot (-1/2) = -2/6 = -1/3$. Also,$c/a = -2/6 = -1/3$. Thus,$\alpha \cdot \beta = c/a$.
Hence,the relationship is verified.

(C) quadratic polynomial is given by the formula $p(x) = k[x^2 - (\text{sum of zeros})x + (\text{product of zeros})]$,where $k$ is a non-zero constant.
Given that the sum of zeros is $\sqrt{2}$ and the product of zeros is $\frac{1}{3}$,we substitute these values into the formula:
$p(x) = k[x^2 - (\sqrt{2})x + \frac{1}{3}]$
Since the condition $a < 0$ is specified,we multiply the expression by $-1$ (setting $k$ to a negative value or simply distributing the negative sign):
$p(x) = k[-x^2 + \sqrt{2}x - \frac{1}{3}]$
Thus,the required polynomial is $k(-x^2 + \sqrt{2}x - \frac{1}{3})$.

(D) To find the zeros of the quadratic polynomial $p(x) = 5x^2 + 8x + 3$,we set $p(x) = 0$.
$5x^2 + 8x + 3 = 0$
We split the middle term $8x$ into $5x + 3x$:
$5x^2 + 5x + 3x + 3 = 0$
Factor by grouping:
$5x(x + 1) + 3(x + 1) = 0$
$(5x + 3)(x + 1) = 0$
Setting each factor to zero:
$5x + 3 = 0 \implies x = -\frac{3}{5}$
$x + 1 = 0 \implies x = -1$
Thus,the zeros are $-1$ and $-\frac{3}{5}$.

(A) To find the zeros of the quadratic polynomial $p(x) = -21x^2 + 16x + 5$,we set $p(x) = 0$.
$-21x^2 + 16x + 5 = 0$
Multiply the entire equation by $-1$ to simplify:
$21x^2 - 16x - 5 = 0$
We need to find two numbers that multiply to $(21 \times -5) = -105$ and add to $-16$. These numbers are $-21$ and $5$.
$21x^2 - 21x + 5x - 5 = 0$
Factor by grouping:
$21x(x - 1) + 5(x - 1) = 0$
$(21x + 5)(x - 1) = 0$
Setting each factor to zero:
$21x + 5 = 0 \implies x = -\frac{5}{21}$
$x - 1 = 0 \implies x = 1$
Thus,the zeros are $1$ and $-\frac{5}{21}$.

(B) To find the zeros of the quadratic polynomial $p(x) = 15x^2 + 16x + 4$,we set $p(x) = 0$.
$15x^2 + 16x + 4 = 0$
We split the middle term $16x$ into two parts such that their sum is $16x$ and their product is $15 \times 4 = 60x^2$. The factors are $10x$ and $6x$.
$15x^2 + 10x + 6x + 4 = 0$
$5x(3x + 2) + 2(3x + 2) = 0$
$(5x + 2)(3x + 2) = 0$
Setting each factor to zero:
$5x + 2 = 0 \implies x = -\frac{2}{5}$
$3x + 2 = 0 \implies x = -\frac{2}{3}$
Thus,the zeros are $-\frac{2}{3}$ and $-\frac{2}{5}$.

(A) quadratic polynomial is given by the formula: $p(x) = k[x^2 - (\text{sum of zeros})x + (\text{product of zeros})]$.
Given:
Sum of zeros $= \frac{1}{4}$
Product of zeros $= -1$
Substituting these values into the formula:
$p(x) = k[x^2 - (\frac{1}{4})x + (-1)]$
$p(x) = k[x^2 - \frac{1}{4}x - 1]$
To simplify,we can take $k = 4$:
$p(x) = 4[x^2 - \frac{1}{4}x - 1] = 4x^2 - x - 4$.
Thus,the polynomial is $k(4x^2 - x - 4)$.

(D) quadratic polynomial is given by the formula $p(x) = k[x^{2} - (\text{sum of zeros})x + (\text{product of zeros})]$,where $k$ is a non-zero real constant.
Given that the sum of the zeros $= -\frac{1}{4}$ and the product of the zeros $= \frac{1}{4}$.
Substituting these values into the formula:
$p(x) = k[x^{2} - (-\frac{1}{4})x + \frac{1}{4}]$
$p(x) = k[x^{2} + \frac{1}{4}x + \frac{1}{4}]$
To simplify,we can take $k = 4$ (or any multiple of $4$):
$p(x) = 4[x^{2} + \frac{1}{4}x + \frac{1}{4}]$
$p(x) = 4x^{2} + x + 1$
Thus,the general form is $k(4x^{2} + x + 1)$.