Mix Examples - Polynomials Questions in English

(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) Given the polynomial $p(x) = (x - 1)(2 - x)$.
Expanding the expression: $p(x) = x(2) - x(x) - 1(2) + 1(x) = 2x - x^2 - 2 + x$.
Combining like terms: $p(x) = -x^2 + 3x - 2$.
The highest power (degree) of the variable $x$ in this polynomial is $2$.
$A$ polynomial of degree $2$ is known as a quadratic 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) Given polynomial: $p(x) = 2x^3 + x^2 - x - 1$
Step $1$: Find the value at $x = 0$.
$p(0) = 2(0)^3 + (0)^2 - (0) - 1$
$p(0) = 0 + 0 - 0 - 1$
$p(0) = -1$
Step $2$: Find the value at $x = -2$.
$p(-2) = 2(-2)^3 + (-2)^2 - (-2) - 1$
$p(-2) = 2(-8) + 4 + 2 - 1$
$p(-2) = -16 + 4 + 2 - 1$
$p(-2) = -11$
Thus,the values are $p(0) = -1$ and $p(-2) = -11$.

(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) To find the degree of the polynomial $p(x) = x^{15} - (x^5)^6 + x^3 - 7$,we first simplify the expression.
Using the power rule of exponents $(a^m)^n = a^{m \times n}$,we simplify the term $(x^5)^6$:
$(x^5)^6 = x^{5 \times 6} = x^{30}$.
Now,substitute this back into the polynomial:
$p(x) = x^{15} - x^{30} + x^3 - 7$.
The degree of a polynomial is defined as the highest power of the variable $x$ present in the expression.
Comparing the powers of $x$ in the terms $x^{15}$,$x^{30}$,and $x^3$,the highest power is $30$.
Therefore,the degree of the polynomial is $30$.

(D) To find the value of the polynomial $p(x) = x^{4} + 2x^{3} - x + 2$ at $x = 2$,substitute $2$ for every $x$ in the expression:
$p(2) = (2)^{4} + 2(2)^{3} - (2) + 2$
Calculate each term:
$(2)^{4} = 16$
$2(2)^{3} = 2(8) = 16$
Substitute these values back into the expression:
$p(2) = 16 + 16 - 2 + 2$
$p(2) = 32 - 2 + 2$
$p(2) = 32$
Therefore,the value of the polynomial at $x = 2$ is $32$.

(A) To find the value of the polynomial $p(x) = 2x^4 - x^2 + 5x - 8$ at $x = -3$,we substitute $-3$ for $x$ in the expression:
$p(-3) = 2(-3)^4 - (-3)^2 + 5(-3) - 8$
Calculate each term:
$(-3)^4 = 81$,so $2(81) = 162$
$(-3)^2 = 9$,so $-(9) = -9$
$5(-3) = -15$
Now,combine these values:
$p(-3) = 162 - 9 - 15 - 8$
$p(-3) = 153 - 15 - 8$
$p(-3) = 138 - 8$
$p(-3) = 130$
Therefore,the value of the polynomial at $x = -3$ is $130$.

(B) To find the value of the polynomial $p(x) = 10 - 5x + 3x^2 - x^3$ at $x = 2$:
$p(2) = 10 - 5(2) + 3(2)^2 - (2)^3$
$p(2) = 10 - 10 + 3(4) - 8$
$p(2) = 0 + 12 - 8 = 4$
To find the value of the polynomial at $x = -2$:
$p(-2) = 10 - 5(-2) + 3(-2)^2 - (-2)^3$
$p(-2) = 10 + 10 + 3(4) - (-8)$
$p(-2) = 20 + 12 + 8 = 40$
Thus,the values are $4$ and $40$.

(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,$25x^2 - 40x + 16 = (5x)^2 - 2(5x)(4) + 4^2 = (5x - 4)^2$.

(D) To factorize the quadratic polynomial $p(x) = x^{2} + 5x - 24$,we need to find two numbers whose product is $-24$ and whose sum is $5$.
These two numbers are $8$ and $-3$,because $8 \times (-3) = -24$ and $8 + (-3) = 5$.
Now,split the middle term $5x$ as $8x - 3x$:
$p(x) = x^{2} + 8x - 3x - 24$
Group the terms:
$p(x) = (x^{2} + 8x) - (3x + 24)$
Factor out the common terms:
$p(x) = x(x + 8) - 3(x + 8)$
Finally,factor out the common binomial $(x + 8)$:
$p(x) = (x + 8)(x - 3)$

(A) To factorize the polynomial $p(x) = x^{3} - 4x^{2} + 5x - 20$,we group the terms:
$p(x) = (x^{3} - 4x^{2}) + (5x - 20)$
Now,factor out the common terms from each group:
$p(x) = x^{2}(x - 4) + 5(x - 4)$
Finally,factor out the common binomial $(x - 4)$:
$p(x) = (x - 4)(x^{2} + 5)$

(N/A) Here,$p(x)=x^{2}-3x-4=(x-4)(x+1)$.
To find the zeros of $p(x)$,consider $p(x)=0$.
$\therefore (x-4)(x+1)=0$.
$\therefore x=4$ or $x=-1$.
$\therefore 4$ and $-1$ are the zeros of $p(x)$.
To draw the graph of this polynomial,we take some different values of $x$ and prepare the following table:

$x$	$-2$	$-1$	$0$	$3$	$4$	$5$
$p(x)=x^2-3x-4$	$6$	$0$	$-4$	$-4$	$0$	$6$

Plot these points on a graph paper. Joining all these points $(-2, 6), (-1, 0), (0, -4), (3, -4), (4, 0)$ and $(5, 6)$,we get the shape of the graph as a parabola opening upwards. We can see that this graph intersects the $X$-axis at two points $(-1, 0)$ and $(4, 0)$. Their $X$-coordinates are the zeros of this polynomial. Thus,$-1$ and $4$ are the zeros of $p(x)$.

(N/A) Here,$p(x)=x^{3}-4 x$
$=x(x^{2}-4)$
$=x(x-2)(x+2)$
To find the zeros of $p(x)$,consider $p(x)=0$.
$\therefore x(x-2)(x+2)=0$
$\therefore x=0, x=2$ or $x=-2$.
$\therefore 0, 2$ and $-2$ are the zeros of $p(x)$.
To draw the graph of this polynomial,we take some different values of $x$ and prepare the following table:

$x$	$-2$	$-1$	$0$	$1$	$2$
$p(x)=x^3-4x$	$0$	$3$	$0$	$-3$	$0$

Plot these points as shown in the graph. We can see that this graph intersects the $X$-axis at three distinct points $(-2, 0)$,$(0, 0)$ and $(2, 0)$. Their $X$-coordinates are the zeros of this polynomial. So,the zeros of $p(x)$ are $-2, 0$ and $2$.

(0) The number of zeros of a polynomial $y=p(x)$ is equal to the number of points where the graph of the polynomial intersects the $X-$axis.
In the given figure,the graph of $y=p(x)$ is a line parallel to the $X-$axis,which does not intersect the $X-$axis at any point.
Therefore,the number of real zeros of $p(x)$ is $0$.

(0) The number of zeros of a polynomial $y=p(x)$ is equal to the number of points where the graph of the polynomial intersects the $X-$axis.
In the given figure,the graph of $y=p(x)$ lies entirely above the $X-$axis and does not intersect or touch the $X-$axis at any point.
Therefore,the number of real zeros of the polynomial $p(x)$ is $0$.

(2) The number of zeros of a polynomial $y=p(x)$ is equal to the number of points where its graph intersects the $X$-axis.
In the given figure,the curve $y=p(x)$ intersects the $X$-axis at two distinct points (the origin $O$ and one point on the positive $X$-axis).
Therefore,the number of real zeros of $p(x)$ is $2$.

(A) The number of zeros of a polynomial $y=p(x)$ is equal to the number of points where the graph of the polynomial intersects the $X-$axis.
In the given figure,the graph of $y=p(x)$ intersects the $X-$axis at exactly $1$ point.
Therefore,the number of real zeros of $p(x)$ is $1$.

(4) The number of zeros of a polynomial $p(x)$ is equal to the number of points where the graph of $y=p(x)$ intersects the $X-$axis.
By observing the given graph,we can see that the curve intersects the $X-$axis at $4$ distinct points.
Therefore,the number of real zeros of $p(x)$ is $4$.

(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.

(C) To find the zeros of the polynomial $p(x) = 5x - 4$,we set $p(x) = 0$.
Thus,$5x - 4 = 0$.
Adding $4$ to both sides,we get $5x = 4$.
Dividing by $5$,we get $x = \frac{4}{5}$.
Since this is a linear polynomial of the form $ax + b$ (where $a \neq 0$),it has exactly one real zero.
Therefore,the number of real zeros is $1$.

(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 find the zero of the polynomial $p(x) = 2x - 5$,we set $p(x) = 0$.
$2x - 5 = 0$
$2x = 5$
$x = 5/2$
Thus,the zero of the polynomial is $5/2$ or $2.5$.
To draw the graph,we find two points on the line:
$1$. If $x = 0$,$p(0) = 2(0) - 5 = -5$. Point: $(0, -5)$.
$2$. If $x = 2.5$,$p(2.5) = 2(2.5) - 5 = 0$. Point: $(2.5, 0)$.
Plotting these points and drawing a straight line through them gives the graph of the linear polynomial.

(B) To find the zeros of the polynomial $p(x) = 3 - 2x - x^2$,we set $p(x) = 0$.
$3 - 2x - x^2 = 0$
Multiply by $-1$ to simplify: $x^2 + 2x - 3 = 0$.
Factor the quadratic equation: $x^2 + 3x - x - 3 = 0$.
$x(x + 3) - 1(x + 3) = 0$.
$(x - 1)(x + 3) = 0$.
Setting each factor to zero,we get $x - 1 = 0$ or $x + 3 = 0$.
Thus,the zeros are $x = 1$ and $x = -3$.

(C) To find the zeros of the polynomial $p(x) = x^{2} + x - 12$,we set $p(x) = 0$.
Thus,$x^{2} + x - 12 = 0$.
Factoring the quadratic equation,we look for two numbers that multiply to $-12$ and add to $1$.
These numbers are $4$ and $-3$.
So,$x^{2} + 4x - 3x - 12 = 0$.
$x(x + 4) - 3(x + 4) = 0$.
$(x - 3)(x + 4) = 0$.
Setting each factor to zero,we get $x - 3 = 0$ or $x + 4 = 0$.
Therefore,$x = 3$ or $x = -4$.
The zeros of the polynomial are $-4$ and $3$.

(D) To find the zeros of the polynomial $p(x) = x^3 - 2x^2$,we set $p(x) = 0$.
$x^3 - 2x^2 = 0$
Factor out $x^2$ from the expression:
$x^2(x - 2) = 0$
This gives us two cases:
$x^2 = 0 \implies x = 0$
$x - 2 = 0 \implies x = 2$
Thus,the zeros of the polynomial are $0$ and $2$.

(A) To find the zeros of the polynomial $p(x) = x^3$,we set $p(x) = 0$.
This gives the equation $x^3 = 0$.
Taking the cube root on both sides,we get $x = 0$.
Graphically,the curve $y = x^3$ passes through the origin $(0, 0)$,which is the only point where it intersects the $x$-axis.
Therefore,the zero of the polynomial $p(x) = x^3$ is $0$.

(B) The number of real zeros of a polynomial $y = p(x)$ is equal to the number of points where the graph of the polynomial intersects the $x$-axis.
In the given figure,the curve $y = p(x)$ touches the $x$-axis at exactly one point.
Therefore,the number of real zeros of the polynomial is $1$.

(C) The number of real zeros of a polynomial $y=p(x)$ is equal to the number of points where the graph of the polynomial intersects the $x$-axis.
In the given figure,the line representing $y=p(x)$ intersects the $x$-axis at exactly one point.
Therefore,the number of real zeros of $p(x)$ is $1$.
Thus,the correct option is $C$.

(D) The number of real zeros of a polynomial $y=p(x)$ is equal to the number of points where the graph of the polynomial intersects the $x$-axis.
In the given figure,the curve $y=p(x)$ intersects the $x$-axis at $3$ distinct points.
Therefore,the number of real zeros of the polynomial is $3$.

(D) The number of real zeros of a polynomial $y=p(x)$ is equal to the number of points where the graph of the polynomial intersects the $x$-axis.
In the given figure,the curve $y=p(x)$ intersects the $x$-axis at $3$ distinct points.
Therefore,the number of real zeros of $p(x)$ 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: $(0)(-\frac{3}{2})(-1) = 0 = -\frac{0}{4} = -\frac{d}{a}$.