Textbook - Polynomials Questions in English

(A) The degree of a polynomial is defined as the highest power of the variable present in the polynomial.
In the given polynomial $x^{5}-x^{4}+3$,the powers of the variable $x$ are $5$,$4$,and $0$ (since $3 = 3x^{0}$).
The highest power among these is $5$.
Therefore,the degree of the polynomial is $5$.

(B) The degree of a polynomial is defined as the highest power of the variable present in the expression.
In the given polynomial $2 - y^{2} - y^{3} + 2y^{8}$,the powers of the variable $y$ are $0, 2, 3,$ and $8$.
The highest power among these is $8$.
Therefore,the degree of the polynomial is $8$.

(C) The given polynomial is a constant polynomial $2$.
Any non-zero constant polynomial can be expressed as $a x^{0}$,where $a$ is a non-zero constant.
Thus,$2$ can be written as $2 x^{0}$.
The exponent of the variable $x$ is $0$.
Therefore,the degree of the constant polynomial $2$ is $0$.

(A) The given expression is $4 x^{2}-3 x+7$.
In this expression,the variable is $x$.
The exponents of $x$ in the terms $4 x^{2}$,$-3 x^{1}$,and $7 x^{0}$ are $2, 1,$ and $0$,respectively.
Since all the exponents of the variable $x$ are non-negative integers (whole numbers),the expression $4 x^{2}-3 x+7$ is a polynomial in one variable.

(A) $(i)$ In the expression $2+x^2+x$,the term containing $x^2$ is $1x^2$.
Therefore,the coefficient of $x^2$ is $1$.
$(ii)$ In the expression $2-x^2+x^3$,the term containing $x^2$ is $-1x^2$.
Therefore,the coefficient of $x^2$ is $-1$.

(N/A) The degree of a polynomial is defined as the highest power of the variable present in the polynomial.
$A$ binomial is a polynomial that contains exactly two terms. Therefore,a binomial of degree $35$ can be represented as $x^{35} + 1$.
$A$ monomial is a polynomial that contains exactly one term. Therefore,a monomial of degree $100$ can be represented as $x^{100}$.

(C) The degree of a polynomial is defined as the highest power of the variable present in the polynomial.
$(i)$ For the polynomial $5x^3 + 4x^2 + 7x$:
The variable is $x$. The powers of $x$ are $3, 2,$ and $1$. The highest power is $3$. Therefore,the degree of this polynomial is $3$.
$(ii)$ For the polynomial $4 - y^2$:
The variable is $y$. The power of $y$ is $2$. Therefore,the degree of this polynomial is $2$.

(N/A) $(i)$ $x^{2}+x$
Since the degree of $x^{2}+x$ is $2$,it is a quadratic polynomial.
$(ii)$ $x-x^{3}$
Since the degree of $x-x^{3}$ is $3$,it is a cubic polynomial.
$(iii)$ $y+y^{2}+4$
Since the degree of $y+y^{2}+4$ is $2$,it is a quadratic polynomial.

(9) Given the polynomial $p(x) = 5x^2 - 3x + 7$.
To find the value of the polynomial at $x = 1$,we substitute $1$ for $x$ in the expression:
$p(1) = 5(1)^2 - 3(1) + 7$
$p(1) = 5(1) - 3 + 7$
$p(1) = 5 - 3 + 7$
$p(1) = 2 + 7$
$p(1) = 9$
Therefore,the value of the polynomial at $x = 1$ is $9$.

(A) Let the polynomial be $p(x) = x + 2$.
To check if $x = a$ is a zero of the polynomial,we evaluate $p(a)$. If $p(a) = 0$,then $a$ is a zero.
For $x = 2$:
$p(2) = 2 + 2 = 4$.
Since $p(2) \neq 0$,$2$ is not a zero of the polynomial.
For $x = -2$:
$p(-2) = -2 + 2 = 0$.
Since $p(-2) = 0$,$-2$ is a zero of the polynomial.
Therefore,$-2$ is a zero of the polynomial $x + 2$,but $2$ is not.

(B) To find a zero of the polynomial $p(x)$,we set the polynomial equal to zero:
$p(x) = 0$
$2x + 1 = 0$
Subtract $1$ from both sides:
$2x = -1$
Divide both sides by $2$:
$x = -\frac{1}{2}$
Therefore,$-\frac{1}{2}$ is the zero of the polynomial $p(x) = 2x + 1$.

(A) Let $p(x) = x^{2} - 2x$.
To verify if $2$ is a zero,we calculate $p(2)$:
$p(2) = (2)^{2} - 2(2) = 4 - 4 = 0$.
Since $p(2) = 0$,$2$ is a zero of the polynomial.
To verify if $0$ is a zero,we calculate $p(0)$:
$p(0) = (0)^{2} - 2(0) = 0 - 0 = 0$.
Since $p(0) = 0$,$0$ is a zero of the polynomial.
Hence,both $2$ and $0$ are zeroes of the polynomial $x^{2} - 2x$.

(A) Let the polynomial be $p(x) = 5x - 4x^2 + 3$.
To find the value of the polynomial at $x = 0$,we substitute $0$ for $x$ in the expression.
$p(0) = 5(0) - 4(0)^2 + 3$
$p(0) = 0 - 4(0) + 3$
$p(0) = 0 - 0 + 3$
$p(0) = 3$
Therefore,the value of the polynomial at $x = 0$ is $3$.

(N/A) Given polynomial: $p(y) = y^{2} - y + 1$.
To find $p(0)$,substitute $y = 0$ into the polynomial:
$p(0) = (0)^{2} - (0) + 1 = 0 - 0 + 1 = 1$.
To find $p(1)$,substitute $y = 1$ into the polynomial:
$p(1) = (1)^{2} - (1) + 1 = 1 - 1 + 1 = 1$.
To find $p(2)$,substitute $y = 2$ into the polynomial:
$p(2) = (2)^{2} - (2) + 1 = 4 - 2 + 1 = 3$.

(N/A) Given the polynomial: $p(t) = 2 + t + 2t^2 - t^3$.
To find $p(0)$,substitute $t = 0$ into the polynomial:
$p(0) = 2 + (0) + 2(0)^2 - (0)^3 = 2 + 0 + 0 - 0 = 2$.
To find $p(1)$,substitute $t = 1$ into the polynomial:
$p(1) = 2 + (1) + 2(1)^2 - (1)^3 = 2 + 1 + 2(1) - 1 = 2 + 1 + 2 - 1 = 4$.
To find $p(2)$,substitute $t = 2$ into the polynomial:
$p(2) = 2 + (2) + 2(2)^2 - (2)^3 = 2 + 2 + 2(4) - 8 = 2 + 2 + 8 - 8 = 4$.

Given polynomial is $p(x) = x^{3}$.
To find $p(0)$,substitute $x = 0$ in the polynomial:
$p(0) = (0)^{3} = 0$.
To find $p(1)$,substitute $x = 1$ in the polynomial:
$p(1) = (1)^{3} = 1$.
To find $p(2)$,substitute $x = 2$ in the polynomial:
$p(2) = (2)^{3} = 2 \times 2 \times 2 = 8$.

(N/A) Given the polynomial $p(x) = (x - 1)(x + 1)$.
To find $p(0)$,substitute $x = 0$ into the polynomial:
$p(0) = (0 - 1)(0 + 1) = (-1)(1) = -1$.
To find $p(1)$,substitute $x = 1$ into the polynomial:
$p(1) = (1 - 1)(1 + 1) = (0)(2) = 0$.
To find $p(2)$,substitute $x = 2$ into the polynomial:
$p(2) = (2 - 1)(2 + 1) = (1)(3) = 3$.

(A) To verify if $x = -\frac{1}{3}$ is a zero of the polynomial $p(x) = 3x + 1$,we substitute $x = -\frac{1}{3}$ into the polynomial.
$p\left(-\frac{1}{3}\right) = 3\left(-\frac{1}{3}\right) + 1$
$p\left(-\frac{1}{3}\right) = -1 + 1$
$p\left(-\frac{1}{3}\right) = 0$
Since the value of the polynomial at $x = -\frac{1}{3}$ is $0$,it is confirmed that $x = -\frac{1}{3}$ is a zero of the given polynomial.

(B) To verify if $x = \frac{4}{5}$ is a zero of the polynomial $p(x) = 5x - \pi$,we substitute $x = \frac{4}{5}$ into the polynomial.
$p\left(\frac{4}{5}\right) = 5\left(\frac{4}{5}\right) - \pi$
$p\left(\frac{4}{5}\right) = 4 - \pi$
Since $4 - \pi \neq 0$,the value of the polynomial at $x = \frac{4}{5}$ is not zero.
Therefore,$x = \frac{4}{5}$ is not a zero of the given polynomial $p(x) = 5x - \pi$.

(A) To verify if $x = 1$ and $x = -1$ are zeroes of the polynomial $p(x) = x^{2} - 1$,we evaluate the polynomial at these values.
For $x = 1$:
$p(1) = (1)^{2} - 1 = 1 - 1 = 0$
For $x = -1$:
$p(-1) = (-1)^{2} - 1 = 1 - 1 = 0$
Since $p(1) = 0$ and $p(-1) = 0$,both $x = 1$ and $x = -1$ are zeroes of the given polynomial $p(x) = x^{2} - 1$.

(N/A) If $x = -1$ and $x = 2$ are zeroes of the polynomial $p(x) = (x + 1)(x - 2)$,then $p(-1)$ and $p(2)$ must be equal to $0$.
First,substitute $x = -1$ into the polynomial:
$p(-1) = (-1 + 1)(-1 - 2) = (0)(-3) = 0$.
Next,substitute $x = 2$ into the polynomial:
$p(2) = (2 + 1)(2 - 2) = (3)(0) = 0$.
Since both $p(-1) = 0$ and $p(2) = 0$,it is verified that $x = -1$ and $x = 2$ are indeed the zeroes of the given polynomial.

(A) To verify if $x = 0$ is a zero of the polynomial $p(x) = x^{2}$,we need to calculate the value of $p(0)$.
Substitute $x = 0$ into the polynomial:
$p(0) = (0)^{2} = 0$.
Since $p(0) = 0$,the value $x = 0$ is indeed a zero of the polynomial $p(x) = x^{2}$.

(A) To verify if $x = -\frac{m}{l}$ is a zero of the polynomial $p(x) = lx + m$,we substitute the value of $x$ into the polynomial.
$p\left(-\frac{m}{l}\right) = l\left(-\frac{m}{l}\right) + m$
$p\left(-\frac{m}{l}\right) = -m + m$
$p\left(-\frac{m}{l}\right) = 0$
Since the value of the polynomial at $x = -\frac{m}{l}$ is $0$,it is confirmed that $x = -\frac{m}{l}$ is a zero of the given polynomial.

(N/A) To verify if the given values are zeroes of the polynomial $p(x) = 3x^2 - 1$,we substitute the values of $x$ into the polynomial. If the result is $0$,then the value is a zero of the polynomial.
Step $1$: Check for $x = -\frac{1}{\sqrt{3}}$:
$p\left(-\frac{1}{\sqrt{3}}\right) = 3\left(-\frac{1}{\sqrt{3}}\right)^2 - 1$
$= 3\left(\frac{1}{3}\right) - 1$
$= 1 - 1 = 0$
Since $p\left(-\frac{1}{\sqrt{3}}\right) = 0$,$x = -\frac{1}{\sqrt{3}}$ is a zero of the polynomial.
Step $2$: Check for $x = \frac{2}{\sqrt{3}}$:
$p\left(\frac{2}{\sqrt{3}}\right) = 3\left(\frac{2}{\sqrt{3}}\right)^2 - 1$
$= 3\left(\frac{4}{3}\right) - 1$
$= 4 - 1 = 3$
Since $p\left(\frac{2}{\sqrt{3}}\right) \neq 0$,$x = \frac{2}{\sqrt{3}}$ is not a zero of the polynomial.

(B) If $x = \frac{1}{2}$ is a zero of the polynomial $p(x) = 2x + 1$,then $p(\frac{1}{2})$ must be equal to $0$.
Substituting $x = \frac{1}{2}$ into the polynomial:
$p(\frac{1}{2}) = 2(\frac{1}{2}) + 1$
$p(\frac{1}{2}) = 1 + 1 = 2$
Since $p(\frac{1}{2}) = 2 \neq 0$,the value $x = \frac{1}{2}$ is not a zero of the given polynomial.

(B) To find the zero of the polynomial $p(x)$,we set $p(x) = 0$.
Given $p(x) = x + 5$.
Setting $p(x) = 0$,we get $x + 5 = 0$.
Solving for $x$,we subtract $5$ from both sides: $x = -5$.
Therefore,the zero of the polynomial $p(x) = x + 5$ is $-5$.

(C) To find the zero of the polynomial $p(x) = x - 5$,we set the polynomial equal to zero.
$p(x) = 0$
$x - 5 = 0$
Adding $5$ to both sides,we get:
$x = 5$
Thus,the zero of the polynomial $p(x) = x - 5$ is $5$.

(D) To find the zero of the polynomial $p(x)$,we set $p(x) = 0$.
Given $p(x) = 2x + 5$.
Setting the polynomial equal to zero:
$2x + 5 = 0$
Subtract $5$ from both sides:
$2x = -5$
Divide both sides by $2$:
$x = \frac{-5}{2}$
Thus,the zero of the polynomial $p(x) = 2x + 5$ is $\frac{-5}{2}$.

(A) To find the zero of the polynomial $p(x) = 3x - 2$,we set $p(x) = 0$.
$3x - 2 = 0$
Adding $2$ to both sides,we get $3x = 2$.
Dividing both sides by $3$,we get $x = \frac{2}{3}$.
Thus,the zero of the polynomial $p(x) = 3x - 2$ is $\frac{2}{3}$.

(B) To find the zero of the polynomial $p(x) = 3x$,we set $p(x) = 0$.
$3x = 0$
Dividing both sides by $3$,we get:
$x = \frac{0}{3}$
$x = 0$
Thus,the zero of the polynomial $p(x) = 3x$ is $0$.

(C) To find the zero of the polynomial $p(x)$,we set $p(x) = 0$.
Given $p(x) = ax$ and $a \neq 0$.
Setting the polynomial equal to zero: $ax = 0$.
Since $a \neq 0$,we can divide both sides by $a$: $x = 0/a$.
Therefore,$x = 0$.
Thus,the zero of the polynomial $p(x) = ax$ is $0$.

(A) To find the zero of the polynomial $p(x) = cx + d$,we set $p(x) = 0$.
$cx + d = 0$
Subtracting $d$ from both sides,we get:
$cx = -d$
Dividing both sides by $c$ (since $c \neq 0$):
$x = -\frac{d}{c}$
Thus,the zero of the polynomial $p(x) = cx + d$ is $-\frac{d}{c}$.

(N/A) We carry out the process of division by means of the following steps:
Step $1$: We write the dividend $x + 3x^2 - 1$ and the divisor $1 + x$ in the standard form,i.e.,after arranging the terms in the descending order of their degrees. So,the dividend is $3x^2 + x - 1$ and the divisor is $x + 1$.
Step $2$: We divide the first term of the dividend by the first term of the divisor,i.e.,we divide $3x^2$ by $x$,and get $3x$. This gives us the first term of the quotient.
Step $3$: We multiply the divisor by the first term of the quotient and subtract this product from the dividend,i.e.,we multiply $x + 1$ by $3x$ and subtract the product $3x^2 + 3x$ from the dividend $3x^2 + x - 1$. This gives us the remainder as $-2x - 1$.
Step $4$: We treat the remainder $-2x - 1$ as the new dividend. The divisor remains the same. We repeat Step $2$ to get the next term of the quotient,i.e.,we divide the first term $-2x$ of the new dividend by the first term $x$ of the divisor and obtain $-2$. Thus,$-2$ is the second term in the quotient.
Step $5$: We multiply the divisor by the second term of the quotient and subtract the product from the dividend. That is,we multiply $x + 1$ by $-2$ and subtract the product $-2x - 2$ from the dividend $-2x - 1$. This gives us $1$ as the remainder.
Step $6$: Thus,the quotient is $3x - 2$ and the remainder is $1$.
Verification: $3x^2 + x - 1 = (x + 1)(3x - 2) + 1$.

(D) To divide $p(x) = 3x^{4} - 4x^{3} - 3x - 1$ by $x - 1$,we use the long division method:
$1$. Divide the first term of the dividend $(3x^{4})$ by the first term of the divisor $(x)$ to get $3x^{3}$.
$2$. Multiply $3x^{3}$ by $(x - 1)$ to get $3x^{4} - 3x^{3}$. Subtract this from the dividend to get $-x^{3} - 3x - 1$.
$3$. Divide $-x^{3}$ by $x$ to get $-x^{2}$. Multiply $-x^{2}$ by $(x - 1)$ to get $-x^{3} + x^{2}$. Subtract to get $-x^{2} - 3x - 1$.
$4$. Divide $-x^{2}$ by $x$ to get $-x$. Multiply $-x$ by $(x - 1)$ to get $-x^{2} + x$. Subtract to get $-4x - 1$.
$5$. Divide $-4x$ by $x$ to get $-4$. Multiply $-4$ by $(x - 1)$ to get $-4x + 4$. Subtract to get $-5$.
The quotient is $3x^{3} - x^{2} - x - 4$ and the remainder is $-5$.
Alternatively,using the Remainder Theorem,the zero of $x - 1$ is $1$. Substituting $x = 1$ into $p(x)$:
$p(1) = 3(1)^{4} - 4(1)^{3} - 3(1) - 1$
$p(1) = 3 - 4 - 3 - 1 = -5$.
Thus,the remainder is $-5$.

(A) To find the remainder when $p(x) = x^3 + 1$ is divided by $x + 1$,we can 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 $x + 1$,which can be written as $x - (-1)$. Thus,$a = -1$.
Now,we calculate $p(-1)$:
$p(-1) = (-1)^3 + 1$
$p(-1) = -1 + 1$
$p(-1) = 0$
Therefore,the remainder obtained is $0$.

(A) Let $p(x) = x^4+x^3-2x^2+x+1$.
According to the Remainder Theorem,if a polynomial $p(x)$ is divided by $(x-a)$,the remainder is $p(a)$.
Here,the divisor is $x-1$,so $a = 1$.
Now,calculate $p(1)$:
$p(1) = (1)^4 + (1)^3 - 2(1)^2 + 1 + 1$
$p(1) = 1 + 1 - 2(1) + 1 + 1$
$p(1) = 1 + 1 - 2 + 1 + 1$
$p(1) = 2$.
Therefore,the remainder is $2$.

(A) To check if $q(t)$ is a multiple of $2t + 1$,we need to determine if $2t + 1$ is a factor of $q(t)$.
According to the Factor Theorem,$2t + 1$ is a factor of $q(t)$ if $q(-\frac{1}{2}) = 0$.
Setting $2t + 1 = 0$,we get $t = -\frac{1}{2}$.
Now,substituting $t = -\frac{1}{2}$ into the polynomial $q(t)$:
$q(-\frac{1}{2}) = 4(-\frac{1}{2})^3 + 4(-\frac{1}{2})^2 - (-\frac{1}{2}) - 1$
$q(-\frac{1}{2}) = 4(-\frac{1}{8}) + 4(\frac{1}{4}) + \frac{1}{2} - 1$
$q(-\frac{1}{2}) = -\frac{1}{2} + 1 + \frac{1}{2} - 1 = 0$.
Since the remainder is $0$,$2t + 1$ is a factor of $q(t)$.
Therefore,$q(t)$ is a multiple of $2t + 1$.

(A) Let $p(x) = x^{3}+3x^{2}+3x+1$.
According to the Remainder Theorem,when a polynomial $p(x)$ is divided by $(x - a)$,the remainder is $p(a)$.
Here,the divisor is $x + 1$,which can be written as $x - (-1)$. Thus,$a = -1$.
To find the remainder,we calculate $p(-1)$:
$p(-1) = (-1)^{3} + 3(-1)^{2} + 3(-1) + 1$
$p(-1) = -1 + 3(1) - 3 + 1$
$p(-1) = -1 + 3 - 3 + 1$
$p(-1) = 0$.
Therefore,the required remainder is $0$.

(B) According to the Remainder Theorem,if a polynomial $p(x)$ is divided by $(x - a)$,the remainder is $p(a)$.
Here,$p(x) = x^{3} + 3x^{2} + 3x + 1$ and the divisor is $x - \frac{1}{2}$.
Setting the divisor to zero,we get $x - \frac{1}{2} = 0$,which implies $x = \frac{1}{2}$.
To find the remainder,we calculate $p\left(\frac{1}{2}\right)$:
$p\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^{3} + 3\left(\frac{1}{2}\right)^{2} + 3\left(\frac{1}{2}\right) + 1$
$= \frac{1}{8} + 3\left(\frac{1}{4}\right) + \frac{3}{2} + 1$
$= \frac{1}{8} + \frac{3}{4} + \frac{3}{2} + 1$
Taking the least common multiple $(LCM)$ as $8$:
$= \frac{1 + 6 + 12 + 8}{8} = \frac{27}{8}$.
Thus,the required remainder is $\frac{27}{8}$.

(C) Let $p(x) = x^{3} + 3x^{2} + 3x + 1$.
According to the Remainder Theorem,when a polynomial $p(x)$ is divided by $(x - a)$,the remainder is $p(a)$.
Here,we are dividing by $x$,which is equivalent to $(x - 0)$. Thus,$a = 0$.
To find the remainder,we calculate $p(0)$:
$p(0) = (0)^{3} + 3(0)^{2} + 3(0) + 1$
$p(0) = 0 + 0 + 0 + 1 = 1$.
Therefore,the remainder is $1$.

(A) Let $p(x) = x^{3}+3x^{2}+3x+1$.
According to the Remainder Theorem,when a polynomial $p(x)$ is divided by $(x-a)$,the remainder is $p(a)$.
Here,the divisor is $x+\pi$,which can be written as $x-(-\pi)$.
So,we need to find $p(-\pi)$.
Substituting $x = -\pi$ in $p(x)$:
$p(-\pi) = (-\pi)^{3} + 3(-\pi)^{2} + 3(-\pi) + 1$
$p(-\pi) = -\pi^{3} + 3\pi^{2} - 3\pi + 1$
Thus,the required remainder is $-\pi^{3}+3\pi^{2}-3\pi+1$.

(B) Let $p(x) = x^{3}+3x^{2}+3x+1$.
To find the remainder when $p(x)$ is divided by $5+2x$,we use the Remainder Theorem.
First,find the zero of the divisor $5+2x$:
$5+2x = 0 \implies 2x = -5 \implies x = -\frac{5}{2}$.
Now,evaluate $p(-\frac{5}{2})$:
$p(-\frac{5}{2}) = (-\frac{5}{2})^{3} + 3(-\frac{5}{2})^{2} + 3(-\frac{5}{2}) + 1$
$= -\frac{125}{8} + 3(\frac{25}{4}) - \frac{15}{2} + 1$
$= -\frac{125}{8} + \frac{75}{4} - \frac{15}{2} + 1$
$= \frac{-125 + 150 - 60 + 8}{8}$
$= \frac{-27}{8}$.
Thus,the required remainder is $-\frac{27}{8}$.

(C) Let the polynomial be $p(x) = x^{3} - ax^{2} + 6x - a$.
According to the Remainder Theorem,when a polynomial $p(x)$ is divided by $(x - a)$,the remainder is $p(a)$.
To find the zero of the divisor $(x - a)$,we set $x - a = 0$,which gives $x = a$.
Now,substitute $x = a$ into the polynomial $p(x)$:
$p(a) = (a)^{3} - a(a)^{2} + 6(a) - a$
$p(a) = a^{3} - a^{3} + 6a - a$
$p(a) = 0 + 5a$
$p(a) = 5a$
Therefore,the remainder is $5a$.

(B) Let $p(x) = 3x^3 + 7x$.
To check if $(7+3x)$ is a factor of $p(x)$,we find the zero of $(7+3x)$ by setting it to $0$:
$7 + 3x = 0 \implies 3x = -7 \implies x = -\frac{7}{3}$.
According to the Factor Theorem,$(7+3x)$ is a factor of $p(x)$ if and only if $p(-\frac{7}{3}) = 0$.
Now,calculate $p(-\frac{7}{3})$:
$p(-\frac{7}{3}) = 3(-\frac{7}{3})^3 + 7(-\frac{7}{3})$
$= 3(-\frac{343}{27}) + (-\frac{49}{3})$
$= -\frac{343}{9} - \frac{49}{3}$
$= \frac{-343 - 147}{9} = -\frac{490}{9}$.
Since $p(-\frac{7}{3}) = -\frac{490}{9} \neq 0$,the remainder is not $0$.
Therefore,$(7+3x)$ is not a factor of $3x^3+7x$.

(N/A) The zero of $x+2$ is $-2$. Let $p(x) = x^{3}+3x^{2}+5x+6$ and $s(x) = 2x+4$.
For $p(x)$:
$p(-2) = (-2)^{3} + 3(-2)^{2} + 5(-2) + 6$
$p(-2) = -8 + 3(4) - 10 + 6$
$p(-2) = -8 + 12 - 10 + 6 = 0$.
Since $p(-2) = 0$,by the Factor Theorem,$x+2$ is a factor of $x^{3}+3x^{2}+5x+6$.
For $s(x)$:
$s(-2) = 2(-2) + 4$
$s(-2) = -4 + 4 = 0$.
Since $s(-2) = 0$,$x+2$ is a factor of $2x+4$. Alternatively,$2x+4 = 2(x+2)$,which clearly shows $x+2$ is a factor.

(B) Let $p(x) = 4x^{3} + 3x^{2} - 4x + k$.
Since $(x-1)$ is a factor of $p(x)$,by the Factor Theorem,$p(1) = 0$.
Substituting $x = 1$ into the polynomial:
$p(1) = 4(1)^{3} + 3(1)^{2} - 4(1) + k = 0$
$4(1) + 3(1) - 4 + k = 0$
$4 + 3 - 4 + k = 0$
$3 + k = 0$
Therefore,$k = -3$.

(N/A) Solution $1:$ (By splitting the middle term method)
We need to find two numbers $p$ and $q$ such that $p + q = 17$ and $pq = 6 \times 5 = 30$.
The factors of $30$ are $(1, 30), (2, 15), (3, 10), (5, 6)$.
Among these,$2 + 15 = 17$.
So,$6x^2 + 17x + 5 = 6x^2 + 2x + 15x + 5$
$= 2x(3x + 1) + 5(3x + 1)$
$= (3x + 1)(2x + 5)$.
Solution $2:$ (Using the Factor Theorem)
Let $p(x) = 6x^2 + 17x + 5 = 6(x^2 + \frac{17}{6}x + \frac{5}{6})$.
Let $a$ and $b$ be the zeroes of the quadratic expression. Then $ab = \frac{5}{6}$.
Possible rational roots are $\pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{5}{3}, \pm \frac{5}{2}, \pm 1$.
Testing $p(-\frac{1}{3}) = 6(-\frac{1}{3})^2 + 17(-\frac{1}{3}) + 5 = 6(\frac{1}{9}) - \frac{17}{3} + 5 = \frac{2}{3} - \frac{17}{3} + \frac{15}{3} = 0$.
Thus,$(x + \frac{1}{3})$ is a factor.
Testing $p(-\frac{5}{2}) = 6(-\frac{5}{2})^2 + 17(-\frac{5}{2}) + 5 = 6(\frac{25}{4}) - \frac{85}{2} + 5 = \frac{75}{2} - \frac{85}{2} + \frac{10}{2} = 0$.
Thus,$(x + \frac{5}{2})$ is a factor.
Therefore,$6x^2 + 17x + 5 = 6(x + \frac{1}{3})(x + \frac{5}{2}) = 6(\frac{3x+1}{3})(\frac{2x+5}{2}) = (3x + 1)(2x + 5)$.

(N/A) Let $p(y) = y^2 - 5y + 6$.
According to the Factor Theorem,if $(y - a)$ is a factor of $p(y)$,then $p(a) = 0$.
We look for factors of the constant term $6$,which are $\pm 1, \pm 2, \pm 3, \pm 6$.
Testing $y = 2$: $p(2) = (2)^2 - 5(2) + 6 = 4 - 10 + 6 = 0$.
Since $p(2) = 0$,$(y - 2)$ is a factor of $p(y)$.
Testing $y = 3$: $p(3) = (3)^2 - 5(3) + 6 = 9 - 15 + 6 = 0$.
Since $p(3) = 0$,$(y - 3)$ is a factor of $p(y)$.
Therefore,the factorisation of $y^2 - 5y + 6$ is $(y - 2)(y - 3)$.

(N/A) Let $p(x) = x^{3}-23 x^{2}+142 x-120$.
We look for factors of the constant term $-120$. Some of these are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 8, \pm 10, \pm 12, \pm 15, \pm 20, \pm 24, \pm 30, \pm 60$.
By trial,we find that $p(1) = 1^{3} - 23(1)^{2} + 142(1) - 120 = 1 - 23 + 142 - 120 = 0$. Therefore,$(x-1)$ is a factor of $p(x)$.
Now,we rewrite the polynomial:
$x^{3}-23 x^{2}+142 x-120 = x^{3}-x^{2}-22 x^{2}+22 x+120 x-120$
$= x^{2}(x-1) - 22x(x-1) + 120(x-1)$
$= (x-1)(x^{2}-22x+120)$.
Next,we factorise the quadratic $x^{2}-22x+120$ by splitting the middle term:
$x^{2}-22x+120 = x^{2}-12x-10x+120$
$= x(x-12) - 10(x-12)$
$= (x-12)(x-10)$.
Thus,the complete factorisation is $x^{3}-23 x^{2}+142 x-120 = (x-1)(x-10)(x-12)$.

(A) According to the Factor Theorem,$(x - a)$ is a factor of a polynomial $p(x)$ if $p(a) = 0$.
Here,the divisor is $(x + 1)$,which can be written as $(x - (-1))$. So,we set $a = -1$.
Let $p(x) = x^{3} + x^{2} + x + 1$.
Now,calculate $p(-1)$:
$p(-1) = (-1)^{3} + (-1)^{2} + (-1) + 1$
$p(-1) = -1 + 1 - 1 + 1$
$p(-1) = 0$
Since the remainder $p(-1) = 0$,by the Factor Theorem,$(x + 1)$ is a factor of the polynomial $x^{3} + x^{2} + x + 1$.