Mix Examples - Polynomials Questions in English

(7) To find the degree of the polynomial,first simplify the expression:
$y^{3}(1-y^{4}) = y^{3} - y^{3+4} = y^{3} - y^{7}$.
The degree of a polynomial is defined as the highest power of the variable present in the expression.
In the expression $y^{3} - y^{7}$,the powers of $y$ are $3$ and $7$.
The highest power is $7$.
Therefore,the degree of the polynomial is $7$.

(N/A) $(i)$ The degree of a polynomial is the highest power of the variable present in the expression. In the given polynomial $\frac{1}{5}x^{3} + \frac{2}{5}x + \frac{1}{5} - \frac{7}{2}x^{2} - x^{6}$,the highest power of $x$ is $6$. Thus,the degree is $6$.
$(ii)$ The term containing $x^{3}$ is $\frac{x^{3}}{5}$,which can be written as $\frac{1}{5}x^{3}$. Therefore,the coefficient of $x^{3}$ is $\frac{1}{5}$.
$(iii)$ The term containing $x^{6}$ is $-x^{6}$,which is $-1 \cdot x^{6}$. Therefore,the coefficient of $x^{6}$ is $-1$.
$(iv)$ The constant term is the term independent of $x$. In the expanded form $\frac{1}{5}x^{3} - \frac{7}{2}x^{2} + \frac{2}{5}x + \frac{1}{5}$,the constant term is $\frac{1}{5}$.

(N/A) $(i)$ The given polynomial is $\frac{\pi}{6} x + 1 \cdot x^{2} - 1$. Comparing this with the standard form,the coefficient of $x^{2}$ is $1$.
$(ii)$ The given polynomial $3x - 5$ can be written as $0 \cdot x^{2} + 3x - 5$. Therefore,the coefficient of $x^{2}$ in this polynomial is $0$.

(N/A) $(i)$ The given polynomial can be expanded as:
$(x-1)(3 x-4) = 3 x^{2} - 4 x - 3 x + 4$
$= 3 x^{2} - 7 x + 4$
Comparing this with the standard form $ax^{2} + bx + c$,the coefficient of $x^{2}$ is $3$.
$(ii)$ The given polynomial can be expanded as:
$(2 x-5)(2 x^{2}-3 x+1) = 2 x(2 x^{2}-3 x+1) - 5(2 x^{2}-3 x+1)$
$= 4 x^{3} - 6 x^{2} + 2 x - 10 x^{2} + 15 x - 5$
$= 4 x^{3} - 16 x^{2} + 17 x - 5$
Comparing this with the standard form,the coefficient of $x^{2}$ is $-16$.

(D) polynomial of degree $n$ is classified based on the highest power of the variable present in the expression.
For the given polynomial $p(x) = 2 - x^{2} + x^{3}$,the highest power of the variable $x$ is $3$.
$A$ polynomial of degree $3$ is called a cubic polynomial.
Therefore,$2 - x^{2} + x^{3}$ is a cubic polynomial.

(D) polynomial of degree $3$ is called a cubic polynomial.
Since the degree of the polynomial $3x^3$ is $3$,it is a cubic polynomial.

(B) polynomial of degree $1$ is called a linear polynomial.
In the expression $5t - \sqrt{7}$,the variable $t$ has a power of $1$.
Since the highest degree of the variable is $1$,$5t - \sqrt{7}$ is a linear polynomial.

(C) polynomial of degree $1$ is called a linear polynomial.
$A$ polynomial of degree $2$ is called a quadratic polynomial.
$A$ polynomial of degree $3$ is called a cubic polynomial.
In the expression $4-5 y^{2}$,the highest power of the variable $y$ is $2$.
Since the degree of the polynomial is $2$,it is a quadratic polynomial.

(A) The expression $3$ is a constant polynomial.
This is because $3$ can be written as $3x^0$,where the exponent of the variable $x$ is $0$.
$A$ polynomial of degree $0$ is defined as a constant polynomial.

(B) polynomial of degree $1$ is called a linear polynomial.
The given polynomial is $p(x) = 2+x$.
The highest power of the variable $x$ in this expression is $1$.
Therefore,$2+x$ is a linear polynomial.

(D) polynomial is classified based on its degree,which is the highest power of the variable present in the expression.
For the given polynomial $p(y) = y^{3}-y$,the highest power of the variable $y$ is $3$.
$A$ polynomial of degree $3$ is called a cubic polynomial.
Therefore,$y^{3}-y$ is a cubic polynomial.

(C) polynomial of degree $1$ is called a linear polynomial,a polynomial of degree $2$ is called a quadratic polynomial,and a polynomial of degree $3$ is called a cubic polynomial.
In the given expression $1+x+x^{2}$,the highest power of the variable $x$ is $2$.
Since the degree of the polynomial is $2$,it is classified as a quadratic polynomial.

(QUADRATIC) polynomial of degree $2$ is called a quadratic polynomial.
Since the degree of the polynomial $t^{2}$ is $2$,it is a quadratic polynomial.

(B) polynomial of degree $1$ is called a linear polynomial.
The given expression is $\sqrt{2} x - 1$.
Since the highest power of the variable $x$ is $1$,the degree of the polynomial is $1$.
Therefore,$\sqrt{2} x - 1$ is a linear polynomial.

(N/A) polynomial having only one term is called a monomial,a polynomial having only two terms is called a binomial,and a polynomial having only three terms is called a trinomial.
$(i)$ $3x$ is a monomial of degree $1$.
$(ii)$ $x^{20} - 7$ is a binomial of degree $20$.
$(iii)$ $5x^2 + 3x - 1$ is a trinomial of degree $2$.

(B) Let the polynomial be $p(x) = 3x^{3} - 4x^{2} + 7x - 5$.
To find the value of the polynomial at $x = 3$,we substitute $3$ for $x$ in the expression:
$p(3) = 3(3)^{3} - 4(3)^{2} + 7(3) - 5$
Calculate the powers:
$p(3) = 3(27) - 4(9) + 21 - 5$
Perform the multiplications:
$p(3) = 81 - 36 + 21 - 5$
Perform the addition and subtraction:
$p(3) = 45 + 21 - 5 = 66 - 5 = 61$.
Thus,the value of the polynomial at $x = 3$ is $61$.

(C) Let the polynomial be $p(x) = 3x^{3} - 4x^{2} + 7x - 5$.
To find the value at $x = -3$,we substitute $-3$ for $x$ in the polynomial:
$p(-3) = 3(-3)^{3} - 4(-3)^{2} + 7(-3) - 5$
Calculate the powers:
$p(-3) = 3(-27) - 4(9) - 21 - 5$
Perform the multiplications:
$p(-3) = -81 - 36 - 21 - 5$
Sum the negative values:
$p(-3) = -143$.

(D) Given the polynomial $p(x) = x^2 - 4x + 3$.
First,calculate $p(2)$:
$p(2) = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1$.
Next,calculate $p(-1)$:
$p(-1) = (-1)^2 - 4(-1) + 3 = 1 + 4 + 3 = 8$.
Then,calculate $p\left(\frac{1}{2}\right)$:
$p\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - 4\left(\frac{1}{2}\right) + 3 = \frac{1}{4} - 2 + 3 = \frac{1}{4} + 1 = \frac{5}{4}$.
Finally,evaluate the expression $p(2) - p(-1) + p\left(\frac{1}{2}\right)$:
$= -1 - 8 + \frac{5}{4} = -9 + \frac{5}{4} = \frac{-36 + 5}{4} = \frac{-31}{4}$.

Given the polynomial $p(x) = 10x - 4x^{2} - 3$.
$(i)$ To find $p(0)$,substitute $x = 0$ into the polynomial:
$p(0) = 10(0) - 4(0)^{2} - 3 = 0 - 0 - 3 = -3$.
$(ii)$ To find $p(1)$,substitute $x = 1$ into the polynomial:
$p(1) = 10(1) - 4(1)^{2} - 3 = 10 - 4 - 3 = 3$.
$(iii)$ To find $p(-2)$,substitute $x = -2$ into the polynomial:
$p(-2) = 10(-2) - 4(-2)^{2} - 3 = -20 - 4(4) - 3 = -20 - 16 - 3 = -39$.

(N/A) Given the polynomial $p(y) = (y+2)(y-2)$.
Using the algebraic identity $(a+b)(a-b) = a^2 - b^2$,we can simplify the expression as $p(y) = y^2 - 4$.
Step $1$: Find $p(0)$.
$p(0) = (0)^2 - 4 = 0 - 4 = -4$.
Step $2$: Find $p(1)$.
$p(1) = (1)^2 - 4 = 1 - 4 = -3$.
Step $3$: Find $p(-2)$.
$p(-2) = (-2)^2 - 4 = 4 - 4 = 0$.

(B) zero of a polynomial $p(x)$ is a number $c$ such that $p(c) = 0$.
Let $p(x) = x - 3$.
To check if $-3$ is a zero,we calculate $p(-3)$:
$p(-3) = -3 - 3 = -6$.
Since $p(-3) = -6 \neq 0$,$-3$ is not a zero of the polynomial $x - 3$.
Therefore,the statement is False.

(A) zero of a polynomial $p(x)$ is a value $c$ such that $p(c) = 0$.
Let $p(x) = 3x + 1$.
To verify,substitute $x = -\frac{1}{3}$ into the polynomial:
$p\left(-\frac{1}{3}\right) = 3\left(-\frac{1}{3}\right) + 1$
$p\left(-\frac{1}{3}\right) = -1 + 1 = 0$.
Since $p\left(-\frac{1}{3}\right) = 0$,it is confirmed that $-\frac{1}{3}$ is a zero of the polynomial $3x + 1$. Therefore,the statement is True.

(FALSE) zero of a polynomial $p(y)$ is a value $c$ such that $p(c) = 0$.
Let $p(y) = 4 - 5y$.
Substitute $y = -\frac{4}{5}$ into the polynomial:
$p\left(-\frac{4}{5}\right) = 4 - 5\left(-\frac{4}{5}\right)$
$p\left(-\frac{4}{5}\right) = 4 + 4 = 8$.
Since $p\left(-\frac{4}{5}\right) \neq 0$,the statement is False.
Therefore,$-\frac{4}{5}$ is not a zero of $4 - 5y$.

(TRUE) zero of a polynomial $p(t)$ is a value $c$ such that $p(c) = 0$.
Let $p(t) = t^{2} - 2t$.
For $t = 0$:
$p(0) = (0)^{2} - 2(0) = 0 - 0 = 0$.
For $t = 2$:
$p(2) = (2)^{2} - 2(2) = 4 - 4 = 0$.
Since $p(0) = 0$ and $p(2) = 0$,the statement is True. Thus,$0$ and $2$ are the zeroes of the polynomial $t^{2} - 2t$.

(TRUE) zero of a polynomial $p(y)$ is a value $c$ such that $p(c) = 0$.
Let $p(y) = y^{2} + y - 6$.
Substitute $y = -3$ into the polynomial:
$p(-3) = (-3)^{2} + (-3) - 6$
$p(-3) = 9 - 3 - 6$
$p(-3) = 0$.
Since $p(-3) = 0$,it is True that $-3$ is a zero of the polynomial $y^{2} + y - 6$.

(D) To find the zero of the polynomial $p(x)$,we set $p(x) = 0$.
Given $p(x) = x - 4$.
Setting the polynomial equal to zero,we get:
$x - 4 = 0$
Adding $4$ to both sides of the equation:
$x = 4$
Therefore,the zero of the polynomial $p(x) = x - 4$ is $4$.

(A) To find the zero of the polynomial $g(x) = 3 - 6x$,we set $g(x) = 0$.
$3 - 6x = 0$
Subtract $3$ from both sides:
$-6x = -3$
Divide both sides by $-6$:
$x = \frac{-3}{-6}$
Simplifying the fraction,we get:
$x = \frac{1}{2}$
Therefore,the zero of the polynomial $g(x) = 3 - 6x$ is $\frac{1}{2}$.

(B) To find the zero of the polynomial $q(x) = 2x - 7$,we set the polynomial equal to zero:
$q(x) = 0$
$2x - 7 = 0$
Adding $7$ to both sides,we get:
$2x = 7$
Dividing both sides by $2$,we get:
$x = \frac{7}{2}$
Therefore,the zero of the polynomial $q(x) = 2x - 7$ is $\frac{7}{2}$.

(C) To find the zero of the polynomial $h(y) = 2y$,we set the polynomial equal to zero.
$h(y) = 0$
$2y = 0$
Dividing both sides by $2$,we get:
$y = 0$
Therefore,$0$ is the zero of the polynomial $h(y) = 2y$.

(D) To find the zeroes of the polynomial $p(x)$,we set $p(x) = 0$.
Given $p(x) = (x-2)^{2} - (x+2)^{2}$.
Using the algebraic identity $a^2 - b^2 = (a-b)(a+b)$,where $a = (x-2)$ and $b = (x+2)$:
$p(x) = [(x-2) - (x+2)] \cdot [(x-2) + (x+2)] = 0$
$p(x) = (x - 2 - x - 2) \cdot (x - 2 + x + 2) = 0$
$p(x) = (-4) \cdot (2x) = 0$
$-8x = 0$
$x = 0$
Thus,the zero of the polynomial is $0$.

(N/A) To divide $x^{4}+1$ by $x+1$,we perform long division:
$x+1$ is the divisor and $x^{4}+0x^{3}+0x^{2}+0x+1$ is the dividend.
$1$. Divide the first term of the dividend $(x^{4})$ by the first term of the divisor $(x)$ to get $x^{3}$.
$2$. Multiply $x^{3}(x+1) = x^{4}+x^{3}$ and subtract from the dividend: $(x^{4}+0x^{3}) - (x^{4}+x^{3}) = -x^{3}$.
$3$. Bring down the next term $(0x^{2})$ to get $-x^{3}+0x^{2}$.
$4$. Divide $-x^{3}$ by $x$ to get $-x^{2}$. Multiply $-x^{2}(x+1) = -x^{3}-x^{2}$ and subtract: $(-x^{3}+0x^{2}) - (-x^{3}-x^{2}) = x^{2}$.
$5$. Bring down $0x$ to get $x^{2}+0x$. Divide $x^{2}$ by $x$ to get $x$. Multiply $x(x+1) = x^{2}+x$ and subtract: $(x^{2}+0x) - (x^{2}+x) = -x$.
$6$. Bring down $1$ to get $-x+1$. Divide $-x$ by $x$ to get $-1$. Multiply $-1(x+1) = -x-1$ and subtract: $(-x+1) - (-x-1) = 2$.
Thus,the quotient is $x^{3}-x^{2}+x-1$ and the remainder is $2$.

(B) According to the Remainder Theorem,if a polynomial $p(x)$ is divided by $(x - a)$,the remainder is $p(a)$.
Given $g(x) = x + 1$,we set $g(x) = 0$ to find the value of $x$:
$x + 1 = 0 \Rightarrow x = -1$.
Now,substitute $x = -1$ into $p(x)$:
$p(-1) = (-1)^{3} - 2(-1)^{2} - 4(-1) - 1$
$p(-1) = -1 - 2(1) + 4 - 1$
$p(-1) = -1 - 2 + 4 - 1$
$p(-1) = -4 + 4 = 0$.
Therefore,the remainder is $0$.

(C) According to the Remainder Theorem,if a polynomial $p(x)$ is divided by $(x - a)$,the remainder is $p(a)$.
Given $g(x) = x - 3$,we set $g(x) = 0$ to find the value of $x$:
$x - 3 = 0 \Rightarrow x = 3$.
Now,we substitute $x = 3$ into the polynomial $p(x)$:
$p(3) = (3)^{3} - 3(3)^{2} + 4(3) + 50$
$p(3) = 27 - 3(9) + 12 + 50$
$p(3) = 27 - 27 + 12 + 50$
$p(3) = 0 + 62 = 62$.
Therefore,the remainder is $62$.

(A) According to the Remainder Theorem,if a polynomial $p(x)$ is divided by $(ax - b)$,the remainder is $p\left(\frac{b}{a}\right)$.
Given $g(x) = 2x - 1$,we set $g(x) = 0$ to find the value of $x$:
$2x - 1 = 0 \Rightarrow 2x = 1 \Rightarrow x = \frac{1}{2}$.
Now,substitute $x = \frac{1}{2}$ into $p(x)$:
$p\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right)^3 - 12\left(\frac{1}{2}\right)^2 + 14\left(\frac{1}{2}\right) - 3$.
$p\left(\frac{1}{2}\right) = 4\left(\frac{1}{8}\right) - 12\left(\frac{1}{4}\right) + 7 - 3$.
$p\left(\frac{1}{2}\right) = \frac{1}{2} - 3 + 7 - 3$.
$p\left(\frac{1}{2}\right) = \frac{1}{2} + 1 = \frac{3}{2}$.
Thus,the remainder is $\frac{3}{2}$.

(A) According to the Remainder Theorem,if a polynomial $p(x)$ is divided by $g(x) = ax + b$,the remainder is $p(-b/a)$.
Set $g(x) = 0$ to find the value of $x$:
$1 - \frac{3}{2}x = 0$
$\frac{3}{2}x = 1$
$x = \frac{2}{3}$
Now,substitute $x = \frac{2}{3}$ into $p(x)$:
$p\left(\frac{2}{3}\right) = \left(\frac{2}{3}\right)^{3} - 6\left(\frac{2}{3}\right)^{2} + 2\left(\frac{2}{3}\right) - 4$
$p\left(\frac{2}{3}\right) = \frac{8}{27} - 6\left(\frac{4}{9}\right) + \frac{4}{3} - 4$
$p\left(\frac{2}{3}\right) = \frac{8}{27} - \frac{24}{9} + \frac{4}{3} - 4$
To add these,find the common denominator,which is $27$:
$p\left(\frac{2}{3}\right) = \frac{8}{27} - \frac{72}{27} + \frac{36}{27} - \frac{108}{27}$
$p\left(\frac{2}{3}\right) = \frac{8 - 72 + 36 - 108}{27}$
$p\left(\frac{2}{3}\right) = \frac{-136}{27}$

(N/A) polynomial $p(x)$ is a multiple of $g(x)$ if and only if $g(x)$ divides $p(x)$ completely,which means the remainder must be $0$.
According to the Remainder Theorem,if $g(x) = x - 2$,we set $x - 2 = 0$,which gives $x = 2$.
Now,we calculate the remainder by evaluating $p(2)$:
$p(2) = (2)^{3} - 5(2)^{2} + 4(2) - 3$
$p(2) = 8 - 5(4) + 8 - 3$
$p(2) = 8 - 20 + 8 - 3$
$p(2) = -7$
Since the remainder is $-7$,which is not equal to $0$,$p(x)$ is not a multiple of $g(x)$.

(B) polynomial $p(x)$ is a multiple of $g(x)$ if and only if $g(x)$ divides $p(x)$ completely,which means the remainder must be $0$.
According to the Remainder Theorem,if we divide $p(x)$ by $g(x) = 2x + 1$,we set $g(x) = 0$ to find the value of $x$:
$2x + 1 = 0 \implies x = -\frac{1}{2}$.
Now,calculate $p(-\frac{1}{2})$:
$p(-\frac{1}{2}) = 2(-\frac{1}{2})^3 - 11(-\frac{1}{2})^2 - 4(-\frac{1}{2}) + 5$
$= 2(-\frac{1}{8}) - 11(\frac{1}{4}) + 2 + 5$
$= -\frac{1}{4} - \frac{11}{4} + 7$
$= \frac{-1 - 11 + 28}{4} = \frac{16}{4} = 4$.
Since the remainder is $4$,which is not equal to $0$,$p(x)$ is not a multiple of $g(x)$.

(N/A) Let $p(x) = 69 + 11x - x^2 + x^3$ and $g(x) = x + 3$.
According to the Factor Theorem,$g(x)$ is a factor of $p(x)$ if $p(a) = 0$ where $g(a) = 0$.
Set $g(x) = x + 3 = 0$,which gives $x = -3$.
Now,calculate $p(-3)$:
$p(-3) = 69 + 11(-3) - (-3)^2 + (-3)^3$
$p(-3) = 69 - 33 - 9 - 27$
$p(-3) = 69 - 69 = 0$.
Since $p(-3) = 0$,by the Factor Theorem,$x + 3$ is a factor of $69 + 11x - x^2 + x^3$.

(N/A) Let $p(x) = 2x^3 - 9x^2 + x + 12$ and $g(x) = 2x - 3$.
According to the Factor Theorem,$g(x)$ is a factor of $p(x)$ if $p(a) = 0$ where $x = a$ is the zero of $g(x)$.
Setting $g(x) = 0$,we get $2x - 3 = 0$,which implies $x = \frac{3}{2}$.
Now,we evaluate $p\left(\frac{3}{2}\right)$:
$p\left(\frac{3}{2}\right) = 2\left(\frac{3}{2}\right)^3 - 9\left(\frac{3}{2}\right)^2 + \left(\frac{3}{2}\right) + 12$
$= 2\left(\frac{27}{8}\right) - 9\left(\frac{9}{4}\right) + \frac{3}{2} + 12$
$= \frac{27}{4} - \frac{81}{4} + \frac{6}{4} + \frac{48}{4}$
$= \frac{27 - 81 + 6 + 48}{4} = \frac{81 - 81}{4} = \frac{0}{4} = 0$.
Since $p\left(\frac{3}{2}\right) = 0$,by the Factor Theorem,$2x - 3$ is a factor of $p(x)$.

(A) We know that if $(x-a)$ is a factor of $p(x)$,then $p(a) = 0$.
$(i)$ Let $p(x) = 3x^2 + 6x - 24$.
If $(x-2)$ is a factor of $p(x)$,then $p(2)$ must be equal to $0$.
$p(2) = 3(2)^2 + 6(2) - 24 = 3(4) + 12 - 24 = 12 + 12 - 24 = 0$.
Since $p(2) = 0$,by the factor theorem,$(x-2)$ is a factor of $3x^2 + 6x - 24$.
$(ii)$ Let $p(x) = 4x^2 + x - 2$.
If $(x-2)$ is a factor of $p(x)$,then $p(2)$ must be equal to $0$.
$p(2) = 4(2)^2 + 2 - 2 = 4(4) + 0 = 16$.
Since $p(2)
eq 0$,$(x-2)$ is not a factor of $4x^2 + x - 2$.

(N/A) According to the Factor Theorem,if $(x-a)$ is a factor of a polynomial $f(x)$,then $f(a) = 0$.
For the polynomial $f(p) = p^{10}-1$,we check the factor $(p-1)$ by substituting $p=1$:
$f(1) = (1)^{10} - 1 = 1 - 1 = 0$.
Since the remainder is $0$,$(p-1)$ is a factor of $p^{10}-1$.
For the polynomial $g(p) = p^{11}-1$,we check the factor $(p-1)$ by substituting $p=1$:
$g(1) = (1)^{11} - 1 = 1 - 1 = 0$.
Since the remainder is $0$,$(p-1)$ is a factor of $p^{11}-1$.
Thus,it is proved that $(p-1)$ is a factor of both $p^{10}-1$ and $p^{11}-1$.

(A) If $x^{3}-2 m x^{2}+16$ is divisible by $x+2$,then by the Factor Theorem,$x+2$ is a factor of $p(x) = x^{3}-2 m x^{2}+16$.
For $x+2$ to be a factor,$p(-2)$ must be equal to $0$.
Substituting $x = -2$ into the polynomial:
$p(-2) = (-2)^{3} - 2m(-2)^{2} + 16$
$p(-2) = -8 - 2m(4) + 16$
$p(-2) = -8 - 8m + 16$
$p(-2) = 8 - 8m$
Setting $p(-2) = 0$:
$8 - 8m = 0$
$8m = 8$
$m = 1$
Therefore,for $m = 1$,the polynomial is divisible by $x+2$.

(A) Let $p(x) = x^{5}-4a^{2}x^{3}+2x+2a+3$.
According to the Factor Theorem,if $(x+2a)$ is a factor of $p(x)$,then $p(-2a) = 0$.
Substituting $x = -2a$ into the polynomial:
$p(-2a) = (-2a)^{5} - 4a^{2}(-2a)^{3} + 2(-2a) + 2a + 3$
$p(-2a) = -32a^{5} - 4a^{2}(-8a^{3}) - 4a + 2a + 3$
$p(-2a) = -32a^{5} + 32a^{5} - 2a + 3$
$p(-2a) = -2a + 3$
Since $p(-2a) = 0$,we have:
$-2a + 3 = 0$
$2a = 3$
$a = \frac{3}{2}$

(B) Let $p(x) = 8x^4 + 4x^3 - 16x^2 + 10x + m$.
Since $(2x - 1)$ is a factor of $p(x)$,by the factor theorem,$p(\frac{1}{2}) = 0$.
Substituting $x = \frac{1}{2}$ into the polynomial:
$8(\frac{1}{2})^4 + 4(\frac{1}{2})^3 - 16(\frac{1}{2})^2 + 10(\frac{1}{2}) + m = 0$
$8(\frac{1}{16}) + 4(\frac{1}{8}) - 16(\frac{1}{4}) + 5 + m = 0$
$\frac{1}{2} + \frac{1}{2} - 4 + 5 + m = 0$
$1 - 4 + 5 + m = 0$
$2 + m = 0$
Therefore,$m = -2$.

(C) Let $p(x) = ax^3 + x^2 - 2x + 4a - 9$.
According to the factor theorem,if $(x+1)$ is a factor of $p(x)$,then $p(-1) = 0$.
Substituting $x = -1$ into the polynomial:
$a(-1)^3 + (-1)^2 - 2(-1) + 4a - 9 = 0$
Simplifying the expression:
$-a + 1 + 2 + 4a - 9 = 0$
Combining like terms:
$3a - 6 = 0$
Solving for $a$:
$3a = 6$
$a = 2$

(A) To factorise the quadratic expression $x^{2}+9x+18$,we need to find two numbers $p$ and $q$ such that their sum $p+q = 9$ and their product $pq = 18$.
By observing the factors of $18$,we find that $6+3 = 9$ and $6 \times 3 = 18$.
Now,we split the middle term $9x$ as $6x + 3x$:
$x^{2}+9x+18 = x^{2}+6x+3x+18$
Next,we group the terms and factor out the common terms:
$= x(x+6) + 3(x+6)$
Finally,taking $(x+6)$ as a common factor:
$= (x+6)(x+3)$

(N/A) To factorise the quadratic expression $6x^{2} + 7x - 3$,we use the splitting the middle term method.
We need to find two numbers $p$ and $q$ such that $p + q = 7$ (coefficient of $x$) and $p \times q = 6 \times (-3) = -18$ (product of coefficient of $x^{2}$ and constant term).
Clearly,$9 + (-2) = 7$ and $9 \times (-2) = -18$.
Now,rewrite the middle term $7x$ as $9x - 2x$:
$6x^{2} + 9x - 2x - 3$
Group the terms to factor out common factors:
$= 3x(2x + 3) - 1(2x + 3)$
Finally,factor out the common binomial $(2x + 3)$:
$= (2x + 3)(3x - 1)$

(A) To factorise the quadratic polynomial $2 x^{2}-7 x-15$,we look for two numbers $p$ and $q$ such that their sum is the coefficient of $x$,which is $-7$,and their product is the product of the coefficient of $x^{2}$ and the constant term,which is $2 \times (-15) = -30$.
We need $p+q = -7$ and $p q = -30$.
By testing factors of $-30$,we find that $-10$ and $3$ satisfy these conditions: $(-10) + 3 = -7$ and $(-10) \times 3 = -30$.
Now,split the middle term $-7 x$ as $-10 x + 3 x$:
$2 x^{2}-7 x-15 = 2 x^{2}-10 x+3 x-15$
Factor by grouping:
$= 2 x(x-5) + 3(x-5)$
Taking $(x-5)$ as a common factor:
$= (x-5)(2 x+3)$

(B) To factorise $84 - 2r - 2r^2$,first rearrange the terms in descending order of powers of $r$:
$-2r^2 - 2r + 84$
Factor out the common term $-2$:
$-2(r^2 + r - 42)$
Now,factor the quadratic expression $r^2 + r - 42$. We need two numbers whose product is $-42$ and whose sum is $1$:
These numbers are $7$ and $-6$.
So,$r^2 + 7r - 6r - 42 = r(r + 7) - 6(r + 7) = (r - 6)(r + 7)$.
Combining these,the final factorised form is:
$-2(r - 6)(r + 7)$

(D) Let $f(x) = 2x^{3}-3x^{2}-17x+30$ be the given polynomial.
By testing values,we find $f(2) = 2(8) - 3(4) - 17(2) + 30 = 16 - 12 - 34 + 30 = 0$. Thus,$(x-2)$ is a factor.
By testing $f(-3) = 2(-27) - 3(9) - 17(-3) + 30 = -54 - 27 + 51 + 30 = 0$. Thus,$(x+3)$ is a factor.
Since $(x-2)$ and $(x+3)$ are factors,their product $(x-2)(x+3) = x^{2}+x-6$ is also a factor.
Dividing $2x^{3}-3x^{2}-17x+30$ by $(x^{2}+x-6)$:
$2x^{3}-3x^{2}-17x+30 = (x^{2}+x-6)(2x-5)$.
Further factoring $(x^{2}+x-6)$ into $(x-2)(x+3)$,we get:
$2x^{3}-3x^{2}-17x+30 = (x-2)(x+3)(2x-5)$.