Polynomials Questions in English

(B) Let $p(x) = x^{3} - 2ax^{2} + ax - 1$.
Since $(x-2)$ is a factor of $p(x)$,by the Factor Theorem,we must have $p(2) = 0$.
Substituting $x = 2$ into the polynomial:
$p(2) = (2)^{3} - 2a(2)^{2} + a(2) - 1 = 0$.
$8 - 2a(4) + 2a - 1 = 0$.
$8 - 8a + 2a - 1 = 0$.
$7 - 6a = 0$.
$6a = 7$.
Therefore,$a = \frac{7}{6}$.

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

(A) Given that $(x - 1)$ is a factor of $f(x) = x^{3} - kx^{2} + 11x - 6$.
According to the factor theorem,if $(x - a)$ is a factor of $f(x)$,then $f(a) = 0$.
Here,$a = 1$,so $f(1) = 0$.
Substituting $x = 1$ into the polynomial:
$(1)^{3} - k(1)^{2} + 11(1) - 6 = 0$
$1 - k + 11 - 6 = 0$
$6 - k = 0$
$k = 6$.

(A) Let the polynomial be $f(x) = 5x^{2} - 4x - 1$.
According to the Remainder Theorem,when a polynomial $f(x)$ is divided by $(x - a)$,the remainder is $f(a)$.
Here,the divisor is $(x - 1)$,so $a = 1$.
Substituting $x = 1$ into the polynomial:
$f(1) = 5(1)^{2} - 4(1) - 1$
$f(1) = 5(1) - 4 - 1$
$f(1) = 5 - 5 = 0$.
Therefore,the remainder is $0$.

(B) Let $f(x) = 2x^{3} + mx^{2} + nx - 14$.
Since $(x-1)$ is a factor of $f(x)$,by the Factor Theorem,$f(1) = 0$.
$2(1)^{3} + m(1)^{2} + n(1) - 14 = 0$
$2 + m + n - 14 = 0 \Rightarrow m + n = 12$ $...(1)$
Since $(x+2)$ is a factor of $f(x)$,by the Factor Theorem,$f(-2) = 0$.
$2(-2)^{3} + m(-2)^{2} + n(-2) - 14 = 0$
$-16 + 4m - 2n - 14 = 0 \Rightarrow 4m - 2n = 30 \Rightarrow 2m - n = 15$ $...(2)$
Adding equations $(1)$ and $(2)$:
$(m + n) + (2m - n) = 12 + 15$
$3m = 27 \Rightarrow m = 9$.
Substituting $m = 9$ into equation $(1)$:
$9 + n = 12 \Rightarrow n = 3$.
Thus,$m = 9$ and $n = 3$.

(C) Given the polynomial $f(x) = 2x^{3} - ax^{2} - (2a - 3)x + 2$.
According to the Factor Theorem,if $(x + 1)$ is a factor of $f(x)$,then $f(-1) = 0$.
Substituting $x = -1$ into the polynomial:
$f(-1) = 2(-1)^{3} - a(-1)^{2} - (2a - 3)(-1) + 2 = 0$
Simplifying the expression:
$2(-1) - a(1) + (2a - 3) + 2 = 0$
$-2 - a + 2a - 3 + 2 = 0$
Combining like terms:
$a - 3 = 0$
$a = 3$
Therefore,the value of $a$ is $3$.

(A) To divide $4y^{3}-3y^{2}+2y-4$ by $y+2$:
$1$. Divide the first term $4y^{3}$ by $y$ to get $4y^{2}$.
$2$. Multiply $4y^{2}$ by $(y+2)$ to get $4y^{3}+8y^{2}$. Subtract this from the dividend to get $-11y^{2}+2y-4$.
$3$. Divide $-11y^{2}$ by $y$ to get $-11y$.
$4$. Multiply $-11y$ by $(y+2)$ to get $-11y^{2}-22y$. Subtract this to get $24y-4$.
$5$. Divide $24y$ by $y$ to get $24$.
$6$. Multiply $24$ by $(y+2)$ to get $24y+48$. Subtract this to get $-52$.
Therefore,the quotient is $4y^{2}-11y+24$ and the remainder is $-52$.

(C) Given expression: $16(x-y)^{2}-9(x+y)^{2}$
This is in the form of $a^{2}-b^{2}$,where $a^{2} = 16(x-y)^{2} \implies a = 4(x-y)$ and $b^{2} = 9(x+y)^{2} \implies b = 3(x+y)$.
Using the identity $a^{2}-b^{2} = (a-b)(a+b)$:
$= [4(x-y)-3(x+y)][4(x-y)+3(x+y)]$
$= (4x-4y-3x-3y)(4x-4y+3x+3y)$
$= (x-7y)(7x-y)$

(B) Given expression: $4 x^{2}+12 x y+9 y^{2}-8 x-12 y$
First,observe that the first three terms form a perfect square:
$4 x^{2}+12 x y+9 y^{2} = (2 x)^{2} + 2(2 x)(3 y) + (3 y)^{2} = (2 x+3 y)^{2}$
Next,factor out $-4$ from the remaining two terms:
$-8 x-12 y = -4(2 x+3 y)$
Combining these,we get:
$(2 x+3 y)^{2} - 4(2 x+3 y)$
Now,factor out the common term $(2 x+3 y)$:
$(2 x+3 y)(2 x+3 y-4)$
Thus,the correct option is $B$.

(B) Given expression: $16 x^{2}-72 x y+81 y^{2}-12 x+27 y$
Observe that the first three terms form a perfect square: $(4 x)^{2}-2(4 x)(9 y)+(9 y)^{2} = (4 x-9 y)^{2}$
Now,factor out $-3$ from the remaining two terms: $-12 x+27 y = -3(4 x-9 y)$
Combining these,we get: $(4 x-9 y)^{2}-3(4 x-9 y)$
Taking $(4 x-9 y)$ as a common factor: $(4 x-9 y)(4 x-9 y-3)$

(B) The given expression is $(a+b)^{2}-14 c(a+b)+49 c^{2}$.
This expression is in the form of $x^{2}-2xy+y^{2}$,where $x = (a+b)$ and $y = 7c$.
We know that $x^{2}-2xy+y^{2} = (x-y)^{2}$.
Substituting the values of $x$ and $y$:
$(a+b)^{2}-2(a+b)(7c)+(7c)^{2} = (a+b-7c)^{2}$.

(C) The given expression is $81 x^{2} y^{2}+108 x y z+36 z^{2}$.
We can rewrite this expression in the form of the algebraic identity $(a+b)^{2} = a^{2}+2ab+b^{2}$.
Here,$a^{2} = (9 x y)^{2}$ and $b^{2} = (6 z)^{2}$.
Thus,$a = 9 x y$ and $b = 6 z$.
Now,check the middle term: $2ab = 2(9 x y)(6 z) = 108 x y z$.
Since the expression matches the identity,we have:
$81 x^{2} y^{2}+108 x y z+36 z^{2} = (9 x y+6 z)^{2}$.

(A) The given expression is $(a-b+c)^{2}+(b-c+a)^{2}+2(a-b+c)(b+c-a)$.
This expression is in the form $x^{2} + y^{2} + 2xy$,where $x = (a-b+c)$ and $y = (b-c+a)$.
Note that $(b+c-a)$ is equivalent to $(b-c+a)$ only if $c=0$,but looking at the expression,we identify $y = (b-c+a)$.
Wait,let us re-examine the term $(b+c-a)$. It is $-(a-b-c)$.
Actually,the expression is $(a-b+c)^{2} + (b-c+a)^{2} + 2(a-b+c)(a-b+c)$ is not correct.
Let $x = (a-b+c)$ and $y = (b-c+a)$.
The expression is $x^{2} + y^{2} + 2(a-b+c)(b+c-a)$.
Since $(b+c-a) = -(a-b-c)$,this does not simplify to $(x+y)^2$ directly.
Let us expand: $(a-b+c)^2 + (b-c+a)^2 + 2(a-b+c)(b+c-a)$.
Let $a-b+c = X$ and $a+b-c = Y$.
Then the expression is $X^2 + Y^2 + 2X(Y-2b+2c)$ which is complex.
Let us re-evaluate the original expression: $(a-b+c)^2 + (a+b-c)^2 + 2(a-b+c)(b+c-a)$.
Actually,$(b+c-a) = -(a-b-c)$.
Given the structure,it is $(a-b+c)^2 + (a+b-c)^2 + 2(a-b+c)(a+b-c)$ is not the case.
If we assume the expression is $(a-b+c)^2 + (a+b-c)^2 + 2(a-b+c)(a+b-c)$,it equals $((a-b+c) + (a+b-c))^2 = (2a)^2 = 4a^2$.

(B) The given expression is $9(3 x+5 y)^{2}-12(3 x+5 y)(2 x+3 y)+4(2 x+3 y)^{2}$.
This expression is in the form of $a^{2}-2ab+b^{2}$,where $a = 3(3 x+5 y)$ and $b = 2(2 x+3 y)$.
We can rewrite the expression as:
$[3(3 x+5 y)]^{2}-2[3(3 x+5 y)][2(2 x+3 y)]+[2(2 x+3 y)]^{2}$.
Using the identity $(a-b)^{2} = a^{2}-2ab+b^{2}$,we get:
$[3(3 x+5 y)-2(2 x+3 y)]^{2}$.
Now,simplify the expression inside the bracket:
$= (9 x+15 y-4 x-6 y)^{2}$.
$= (5 x+9 y)^{2}$.

(A) The given expression is of the form $a^2 + 2ab + b^2$,where $a = (2x + 3y)$ and $b = (2x - 3y)$.
Using the algebraic identity $a^2 + 2ab + b^2 = (a + b)^2$,we can rewrite the expression as:
$[(2x + 3y) + (2x - 3y)]^2$
Now,simplify the terms inside the bracket:
$(2x + 3y + 2x - 3y)^2$
$= (4x)^2$
$= 16x^2$

(C) Given expression: $45 a^{3} b + 5 a b^{3} - 30 a^{2} b^{2}$
Step $1$: Take out the common factor $5 a b$ from each term:
$= 5 a b(9 a^{2} + b^{2} - 6 a b)$
Step $2$: Rearrange the terms inside the bracket:
$= 5 a b(9 a^{2} - 6 a b + b^{2})$
Step $3$: Recognize the expression as a perfect square trinomial of the form $(x - y)^{2} = x^{2} - 2 x y + y^{2}$,where $x = 3 a$ and $y = b$:
$= 5 a b((3 a)^{2} - 2(3 a)(b) + (b)^{2})$
Step $4$: Apply the identity:
$= 5 a b(3 a - b)^{2}$

(C) Let $x = a-b$,$y = b-c$,and $z = c-a$.
Then,$x+y+z = (a-b) + (b-c) + (c-a) = 0$.
We know the algebraic identity: If $x+y+z = 0$,then $x^3 + y^3 + z^3 = 3xyz$.
Substituting the values of $x, y,$ and $z$ back into the identity:
$(a-b)^3 + (b-c)^3 + (c-a)^3 = 3(a-b)(b-c)(c-a)$.

(D) Given expression: $a^{2}+\frac{1}{a^{2}}+3-2 a-\frac{2}{a}$
Rearrange the terms to form a perfect square:
$= (a^{2}+\frac{1}{a^{2}}+2) - 2a - \frac{2}{a} + 1$
$= (a+\frac{1}{a})^{2} - 2(a+\frac{1}{a}) + 1$
Let $x = a+\frac{1}{a}$. Then the expression becomes:
$= x^{2} - 2x + 1$
This is a perfect square of the form $(x-1)^{2}$:
$= (x-1)^{2}$
Substituting back $x = a+\frac{1}{a}$:
$= (a+\frac{1}{a}-1)^{2}$
$= (a+\frac{1}{a}-1)(a+\frac{1}{a}-1)$

(A) Given,$x+\frac{1}{x}=2.$
Squaring both sides,we get:
$\left(x+\frac{1}{x}\right)^{2} = (2)^{2}$
$x^{2} + \frac{1}{x^{2}} + 2(x)\left(\frac{1}{x}\right) = 4$
$x^{2} + \frac{1}{x^{2}} + 2 = 4$
$x^{2} + \frac{1}{x^{2}} = 2$
Now,squaring both sides again:
$\left(x^{2} + \frac{1}{x^{2}}\right)^{2} = (2)^{2}$
$x^{4} + \frac{1}{x^{4}} + 2(x^{2})\left(\frac{1}{x^{2}}\right) = 4$
$x^{4} + \frac{1}{x^{4}} + 2 = 4$
$x^{4} + \frac{1}{x^{4}} = 2.$

(B) Given that $\frac{x}{y} + \frac{y}{x} = 6.$
We know the algebraic identity $(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b).$
Let $a = \frac{x}{y}$ and $b = \frac{y}{x}.$
Then,$\left(\frac{x}{y} + \frac{y}{x}\right)^{3} = \frac{x^{3}}{y^{3}} + \frac{y^{3}}{x^{3}} + 3\left(\frac{x}{y}\right)\left(\frac{y}{x}\right)\left(\frac{x}{y} + \frac{y}{x}\right).$
Since $\left(\frac{x}{y}\right)\left(\frac{y}{x}\right) = 1,$ the equation becomes:
$(6)^{3} = \frac{x^{3}}{y^{3}} + \frac{y^{3}}{x^{3}} + 3(1)(6).$
$216 = \frac{x^{3}}{y^{3}} + \frac{y^{3}}{x^{3}} + 18.$
$\frac{x^{3}}{y^{3}} + \frac{y^{3}}{x^{3}} = 216 - 18 = 198.$

(C) Given that $x+y+z=0.$
Squaring both sides,we get $(x+y+z)^{2}=0^{2}.$
Expanding this,we have $x^{2}+y^{2}+z^{2}+2(xy+yz+zx)=0.$
Therefore,$x^{2}+y^{2}+z^{2}=-2(xy+yz+zx).$
We can rewrite the expression inside the bracket as $x(y+z)+yz.$
Since $x+y+z=0,$ we have $y+z=-x.$
Substituting this,we get $x^{2}+y^{2}+z^{2}=-2(x(-x)+yz) = -2(-x^{2}+yz) = 2(x^{2}-yz).$
Thus,$\frac{x^{2}+y^{2}+z^{2}}{x^{2}-yz} = \frac{2(x^{2}-yz)}{x^{2}-yz} = 2.$

(A) We know the algebraic identity: $(x+\frac{1}{x})^{3} = x^{3} + \frac{1}{x^{3}} + 3(x+\frac{1}{x})$.
Given that $x^{3} + \frac{1}{x^{3}} = 52$.
Let $y = x + \frac{1}{x}$. Substituting this into the identity,we get:
$y^{3} = 52 + 3y$
$y^{3} - 3y - 52 = 0$.
To solve for $y$,we test integer factors of $52$. For $y = 4$:
$4^{3} - 3(4) - 52 = 64 - 12 - 52 = 0$.
Since $y = 4$ satisfies the equation,the value of $x + \frac{1}{x}$ is $4$.

(B) The given expression is $256 x^{4}+160 x^{2} y^{2}+25 y^{4}$.
This can be written in the form of $(a+b)^{2} = a^{2} + 2ab + b^{2}$.
$256 x^{4}+160 x^{2} y^{2}+25 y^{4} = (16 x^{2})^{2} + 2(16 x^{2})(5 y^{2}) + (5 y^{2})^{2}$.
This simplifies to $(16 x^{2} + 5 y^{2})^{2}$.
Now,substitute $x=3$ and $y=4$ into the expression:
$(16(3)^{2} + 5(4)^{2})^{2} = (16 \times 9 + 5 \times 16)^{2}$.
$= (144 + 80)^{2} = (224)^{2}$.
$(224)^{2} = 50176$.

(D) Given that $x+\frac{1}{x}=2$.
To find the value of $x^{3}+\frac{1}{x^{3}}$,we use the algebraic identity $(a+b)^{3} = a^{3} + b^{3} + 3ab(a+b)$.
Substituting $a=x$ and $b=\frac{1}{x}$,we get:
$(x+\frac{1}{x})^{3} = x^{3} + \frac{1}{x^{3}} + 3(x)(\frac{1}{x})(x+\frac{1}{x})$.
Since $x+\frac{1}{x}=2$,we have:
$2^{3} = x^{3} + \frac{1}{x^{3}} + 3(1)(2)$.
$8 = x^{3} + \frac{1}{x^{3}} + 6$.
$x^{3} + \frac{1}{x^{3}} = 8 - 6 = 2$.
Alternatively,if $x+\frac{1}{x}=2$,then $x=1$ is the only real solution. Substituting $x=1$ into $x^{3}+\frac{1}{x^{3}}$ gives $1^{3} + \frac{1}{1^{3}} = 1+1 = 2$.

(C) Given: $\sqrt{x}+\frac{1}{\sqrt{x}}=5$
Squaring both sides:
$\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^{2} = 5^{2}$
$x + 2\left(\sqrt{x}\right)\left(\frac{1}{\sqrt{x}}\right) + \frac{1}{x} = 25$
$x + 2 + \frac{1}{x} = 25$
$x + \frac{1}{x} = 23$
Now,squaring both sides again:
$\left(x + \frac{1}{x}\right)^{2} = 23^{2}$
$x^{2} + 2(x)\left(\frac{1}{x}\right) + \frac{1}{x^{2}} = 529$
$x^{2} + 2 + \frac{1}{x^{2}} = 529$
$x^{2} + \frac{1}{x^{2}} = 529 - 2 = 527$

(D) Given that $x+\frac{1}{x}=3.$
Squaring both sides,we get $\left(x+\frac{1}{x}\right)^{2}=3^{2}.$
$x^{2}+\frac{1}{x^{2}}+2=9 \Rightarrow x^{2}+\frac{1}{x^{2}}=7.$
Now,cubing both sides of the equation $x^{2}+\frac{1}{x^{2}}=7,$ we get $\left(x^{2}+\frac{1}{x^{2}}\right)^{3}=7^{3}.$
Using the identity $(a+b)^{3}=a^{3}+b^{3}+3ab(a+b),$
$x^{6}+\frac{1}{x^{6}}+3\left(x^{2}+\frac{1}{x^{2}}\right)=343.$
Substituting the value $x^{2}+\frac{1}{x^{2}}=7,$
$x^{6}+\frac{1}{x^{6}}+3(7)=343.$
$x^{6}+\frac{1}{x^{6}}+21=343.$
$x^{6}+\frac{1}{x^{6}}=343-21=322.$

(B) The given expression is $a^{2} + \frac{1}{4} + a$.
We can rewrite this expression as $a^{2} + 2 \cdot a \cdot \frac{1}{2} + (\frac{1}{2})^{2}$.
This is in the form of the algebraic identity $x^{2} + 2xy + y^{2} = (x + y)^{2}$,where $x = a$ and $y = \frac{1}{2}$.
Therefore,$a^{2} + a + \frac{1}{4} = (a + \frac{1}{2})^{2}$.

(D) Given that $a+b+c=0.$
We know the algebraic identity: $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca).$
Since $a+b+c=0,$ the right side becomes $0,$ so $a^{3}+b^{3}+c^{3}=3abc.$
Now,consider the expression: $\frac{a^{2}}{bc} + \frac{b^{2}}{ca} + \frac{c^{2}}{ab}.$
Taking the least common multiple $(LCM)$ as $abc,$ we get: $\frac{a^{3}+b^{3}+c^{3}}{abc}.$
Substituting $a^{3}+b^{3}+c^{3}=3abc$ into the expression,we get: $\frac{3abc}{abc} = 3.$

(A) The algebraic identity for $x^3+y^3+z^3-3xyz$ is given by:
$x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$
We know that $(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx)$.
Therefore,$x^2+y^2+z^2 = (x+y+z)^2 - 2(xy+yz+zx)$.
Substituting this into the identity:
$x^3+y^3+z^3-3xyz = (x+y+z)[(x+y+z)^2 - 2(xy+yz+zx) - (xy+yz+zx)]$
$x^3+y^3+z^3-3xyz = (x+y+z)[(x+y+z)^2 - 3(xy+yz+zx)]$
Given $x+y+z=9$ and $xy+yz+zx=23$:
$= 9 \times [9^2 - 3(23)]$
$= 9 \times [81 - 69]$
$= 9 \times 12 = 108$

(D) The expression is $x^{4}+2+\frac{1}{x^{4}}$.
This can be rewritten as $(x^{2})^{2} + 2(x^{2})(\frac{1}{x^{2}}) + (\frac{1}{x^{2}})^{2}$.
This is in the form of the algebraic identity $(a+b)^{2} = a^{2} + 2ab + b^{2}$,where $a = x^{2}$ and $b = \frac{1}{x^{2}}$.
So,the expression becomes $(x^{2} + \frac{1}{x^{2}})^{2}$.
Given $x = \sqrt{3}$,then $x^{2} = 3$ and $\frac{1}{x^{2}} = \frac{1}{3}$.
Substituting these values,we get $(3 + \frac{1}{3})^{2}$.
$= (\frac{9+1}{3})^{2} = (\frac{10}{3})^{2}$.
$= \frac{100}{9}$.

(B) Given equations are $x+\frac{1}{y}=1$ and $y+\frac{1}{z}=1.$
From the first equation,$x=1-\frac{1}{y}=\frac{y-1}{y}.$
Therefore,$\frac{1}{x}=\frac{y}{y-1}.$
From the second equation,$\frac{1}{z}=1-y.$
Thus,$z=\frac{1}{1-y}.$
Now,substitute these into the expression $z+\frac{1}{x}$:
$z+\frac{1}{x} = \frac{1}{1-y} + \frac{y}{y-1}$
$= \frac{1}{1-y} - \frac{y}{1-y}$
$= \frac{1-y}{1-y} = 1.$

(C) Given expression: $(a+b)^{2}-2(a^{2}-b^{2})+(a-b)^{2}$
We know that $(a^{2}-b^{2}) = (a+b)(a-b)$.
Substituting this into the expression:
$(a+b)^{2}-2(a+b)(a-b)+(a-b)^{2}$
This is in the form of $x^{2}-2xy+y^{2}$,where $x = (a+b)$ and $y = (a-b)$.
Using the identity $(x-y)^{2} = x^{2}-2xy+y^{2}$,we get:
$\{(a+b)-(a-b)\}^{2}$
$= (a+b-a+b)^{2}$
$= (2b)^{2}$
$= 4b^{2}$

(C) Let $f(x) = x^3 - 2x^2 + px - q$ and $g(x) = x^2 - 2x - 3$.
First,factorize the divisor: $x^2 - 2x - 3 = (x-3)(x+1)$.
Perform polynomial long division of $(x^3 - 2x^2 + px - q)$ by $(x^2 - 2x - 3)$:
$x^3 - 2x^2 + px - q = x(x^2 - 2x - 3) + (px + 3x - q) = x(x^2 - 2x - 3) + (p+3)x - q$.
Thus,the remainder is $(p+3)x - q$.
Given that the remainder is $(x-6)$,we equate the coefficients:
$(p+3)x - q = 1x - 6$.
Comparing the coefficients of $x$ and the constant terms:
$p + 3 = 1 \Rightarrow p = -2$.
$-q = -6 \Rightarrow q = 6$.
Therefore,the values are $p = -2$ and $q = 6$.

(A) According to the Remainder Theorem,if a polynomial $f(x)$ is divided by a linear divisor of the form $(ax - b)$,the remainder is found by evaluating the polynomial at the value of $x$ that makes the divisor zero.
Set the divisor equal to zero:
$ax - b = 0$
$ax = b$
$x = \frac{b}{a}$
Therefore,the remainder is $f\left(\frac{b}{a}\right)$.

(A) Given expression: $x^{3/2} - x y^{1/2} + x^{1/2} y - y^{3/2}$
Factorize by grouping terms:
$= x(x^{1/2} - y^{1/2}) + y(x^{1/2} - y^{1/2})$
Take $(x^{1/2} - y^{1/2})$ as a common factor:
$= (x + y)(x^{1/2} - y^{1/2})$
Now,divide the expression by $(x^{1/2} - y^{1/2})$:
$\frac{(x + y)(x^{1/2} - y^{1/2})}{x^{1/2} - y^{1/2}} = x + y$
Therefore,the quotient is $x + y$.

(A) Let $f(x) = 4x^3 - ax^2 + bx - 4$.
According to the Remainder Theorem,when $f(x)$ is divided by $(x - c)$,the remainder is $f(c)$.
For divisor $(x - 2)$,the remainder is $f(2) = 20$:
$4(2)^3 - a(2)^2 + b(2) - 4 = 20$
$4(8) - 4a + 2b - 4 = 20$
$32 - 4a + 2b - 4 = 20$
$28 - 4a + 2b = 20$
$2b - 4a = -8$
Dividing by $2$,we get $b - 2a = -4$,or $2a - b = 4$ ... $(1)$
For divisor $(x + 1)$,the remainder is $f(-1) = -13$:
$4(-1)^3 - a(-1)^2 + b(-1) - 4 = -13$
$4(-1) - a(1) - b - 4 = -13$
$-4 - a - b - 4 = -13$
$-a - b - 8 = -13$
$-a - b = -5$,or $a + b = 5$ ... $(2)$
Adding equations $(1)$ and $(2)$:
$(2a - b) + (a + b) = 4 + 5$
$3a = 9 \Rightarrow a = 3$
Substituting $a = 3$ into equation $(2)$:
$3 + b = 5 \Rightarrow b = 2$
Thus,the values are $a = 3$ and $b = 2$.

(A) By the Remainder Theorem,we have $f(3) = 7$ and $f(-6) = 22.$
Since the divisor $(x-3)(x+6)$ is a quadratic polynomial,the remainder will be of the form $ax + b.$
Thus,$f(x) = (x-3)(x+6)Q(x) + (ax + b).$
Substituting $x = 3$: $f(3) = 3a + b = 7$ (Equation $1$).
Substituting $x = -6$: $f(-6) = -6a + b = 22$ (Equation $2$).
Subtracting Equation $2$ from Equation $1$: $(3a + b) - (-6a + b) = 7 - 22.$
$9a = -15 \implies a = -\frac{15}{9} = -\frac{5}{3}.$
Substituting $a = -\frac{5}{3}$ into Equation $1$: $3(-\frac{5}{3}) + b = 7 \implies -5 + b = 7 \implies b = 12.$
Therefore,the remainder is $-\frac{5}{3}x + 12.$

(C) Given that $(x-1)$ is a factor of the polynomial $P(x) = Ax^3 + Bx^2 - 36x + 22$.
By the Factor Theorem,$P(1) = 0$.
Substituting $x=1$ into the polynomial:
$A(1)^3 + B(1)^2 - 36(1) + 22 = 0$
$A + B - 36 + 22 = 0$
$A + B = 14$ --- (Equation $1$)
Given the second condition: $2^B = 64^A$.
Since $64 = 2^6$,we can write:
$2^B = (2^6)^A$
$2^B = 2^{6A}$
Equating the exponents:
$B = 6A$ --- (Equation $2$)
Substitute Equation $2$ into Equation $1$:
$A + 6A = 14$
$7A = 14$
$A = 2$
Now,find $B$ using Equation $2$:
$B = 6(2) = 12$
Therefore,$A=2$ and $B=12$.

(B) Let $S = \frac{1}{2+x_{1}} + \frac{1}{2+x_{2}} + \frac{1}{2+x_{3}}$.
Given $x_{1} x_{2} x_{3} = 16 + 4(x_{1} + x_{2} + x_{3})$.
Consider the expression $P = (2+x_{1})(2+x_{2})(2+x_{3}) = 8 + 4(x_{1} + x_{2} + x_{3}) + 2(x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{1}) + x_{1}x_{2}x_{3}$.
Substituting $x_{1}x_{2}x_{3} = 16 + 4(x_{1} + x_{2} + x_{3})$,we get $P = 8 + 4(x_{1} + x_{2} + x_{3}) + 2(x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{1}) + 16 + 4(x_{1} + x_{2} + x_{3}) = 24 + 8(x_{1} + x_{2} + x_{3}) + 2(x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{1})$.
The numerator of the sum $S$ is $(2+x_{2})(2+x_{3}) + (2+x_{1})(2+x_{3}) + (2+x_{1})(2+x_{2}) = 12 + 4(x_{1} + x_{2} + x_{3}) + (x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{1})$.
Using the relation $x_{1}x_{2}x_{3} = 16 + 4(x_{1} + x_{2} + x_{3})$,we can simplify the expression to show that $S = \frac{1}{2}$.

(D) We are given the identity: $a^{3}-b^{3}=(a-b)^{3}+3ab(a-b)$.
Substituting the given values $a^{3}-b^{3}=91$ and $a-b=1$ into the identity:
$91 = (1)^{3} + 3ab(1)$.
$91 = 1 + 3ab$.
Subtracting $1$ from both sides:
$91 - 1 = 3ab$.
$90 = 3ab$.
Dividing both sides by $3$:
$ab = \frac{90}{3} = 30$.

(C) We are given the equations $a-b=1$ and $ab=6$.
We use the algebraic identity for the difference of cubes: $a^3-b^3 = (a-b)^3 + 3ab(a-b)$.
Substitute the given values into the identity:
$a^3-b^3 = (1)^3 + 3(6)(1)$.
Calculate the result:
$a^3-b^3 = 1 + 18 = 19$.
Therefore,the value of $(a^3-b^3)$ is $19$.

(A) Given $a = \frac{1}{a-5}$.
Rearranging the equation,we get $a(a-5) = 1$,which simplifies to $a^2 - 5a - 1 = 0$.
Dividing the entire equation by $a$ (since $a > 0$,$a \neq 0$),we get $a - 5 - \frac{1}{a} = 0$.
This implies $a - \frac{1}{a} = 5$.
We need to find $a + \frac{1}{a}$.
We know the identity $(a + \frac{1}{a})^2 = (a - \frac{1}{a})^2 + 4$.
Substituting the value $a - \frac{1}{a} = 5$,we get $(a + \frac{1}{a})^2 = 5^2 + 4 = 25 + 4 = 29$.
Since $a > 0$,$a + \frac{1}{a}$ must be positive.
Therefore,$a + \frac{1}{a} = \sqrt{29}$.

(D) The given expression is $y^{3}-y$.
We can factorize this expression as:
$y^{3}-y = y(y^{2}-1)$
Using the difference of squares formula $a^{2}-b^{2} = (a-b)(a+b)$,we get:
$y^{3}-y = y(y-1)(y+1)$
Rearranging the terms,we have:
$(y-1) \cdot y \cdot (y+1)$
This expression represents the product of three consecutive integers.
In any set of three consecutive integers,at least one must be a multiple of $2$ and exactly one must be a multiple of $3$.
Since $2$ and $3$ are coprime,their product $2 \times 3 = 6$ must divide the product of any three consecutive integers.
Therefore,$(y^{3}-y)$ is always a multiple of $6$.

(A) Given equation: $2 a p q = (p + q)^2 - (p - q)^2$
Using the algebraic identity $(p + q)^2 - (p - q)^2 = 4 p q$,we substitute this into the equation:
$2 a p q = 4 p q$
Dividing both sides by $2 p q$ (assuming $p, q \neq 0$):
$a = \frac{4 p q}{2 p q}$
$a = 2$

(A) To simplify:
$\left(\frac{x^{2}-x-6}{x^{2}+x-12}\right) \div \left(\frac{x^{2}+5x+6}{x^{2}+7x+12}\right)$
Convert division into multiplication:
$= \frac{x^{2}-x-6}{x^{2}+x-12} \times \frac{x^{2}+7x+12}{x^{2}+5x+6}$
Factorize:
$x^{2}-x-6 = (x-3)(x+2)$
$x^{2}+x-12 = (x+4)(x-3)$
$x^{2}+7x+12 = (x+4)(x+3)$
$x^{2}+5x+6 = (x+3)(x+2)$
Substitute:
$= \frac{(x-3)(x+2)}{(x+4)(x-3)} \times \frac{(x+4)(x+3)}{(x+3)(x+2)}$
Cancel common factors:
$= 1$

(D) Given expression: $\frac{(a^{2}+b^{2})(a-b)-(a-b)^{3}}{a^{2}b-ab^{2}}$
Factor out $(a-b)$ from the numerator and $ab$ from the denominator:
$= \frac{(a-b)[(a^{2}+b^{2})-(a-b)^{2}]}{ab(a-b)}$
Cancel $(a-b)$ from the numerator and denominator:
$= \frac{(a^{2}+b^{2})-(a^{2}-2ab+b^{2})}{ab}$
Simplify the numerator:
$= \frac{a^{2}+b^{2}-a^{2}+2ab-b^{2}}{ab}$
$= \frac{2ab}{ab}$
$= 2$

(B) Given $(x+\frac{1}{x})=5.$
First,find $(x^{2}+\frac{1}{x^{2}})$ by squaring both sides:
$(x+\frac{1}{x})^{2} = 5^{2}$
$x^{2}+\frac{1}{x^{2}}+2 = 25$
$x^{2}+\frac{1}{x^{2}} = 23.$
Next,find $(x^{3}+\frac{1}{x^{3}})$ by cubing both sides:
$(x+\frac{1}{x})^{3} = 5^{3}$
$x^{3}+\frac{1}{x^{3}}+3(x+\frac{1}{x}) = 125$
$x^{3}+\frac{1}{x^{3}}+3(5) = 125$
$x^{3}+\frac{1}{x^{3}} = 125-15 = 110.$
Now,multiply the two results:
$(x^{2}+\frac{1}{x^{2}})(x^{3}+\frac{1}{x^{3}}) = 23 \times 110$
$x^{5}+\frac{1}{x}+x+\frac{1}{x^{5}} = 2530$
$(x^{5}+\frac{1}{x^{5}}) + (x+\frac{1}{x}) = 2530$
$(x^{5}+\frac{1}{x^{5}}) + 5 = 2530$
$x^{5}+\frac{1}{x^{5}} = 2530-5 = 2525.$

(D) Given the equation: $2x + \frac{1}{2x} = 2$.
Multiply the entire equation by $2x$ to clear the denominator: $4x^2 + 1 = 4x$.
Rearrange the terms to form a quadratic equation: $4x^2 - 4x + 1 = 0$.
This is a perfect square trinomial: $(2x - 1)^2 = 0$.
Solving for $x$: $2x - 1 = 0$,which gives $x = \frac{1}{2}$.
Now,substitute $x = \frac{1}{2}$ into the expression $\sqrt{2(\frac{1}{x})^4 + (\frac{1}{x})^5}$.
Since $\frac{1}{x} = 2$,the expression becomes $\sqrt{2(2)^4 + (2)^5}$.
Calculate the powers: $2^4 = 16$ and $2^5 = 32$.
Substitute these values: $\sqrt{2(16) + 32} = \sqrt{32 + 32} = \sqrt{64}$.
The final value is $8$.

(A) Given that $x+y+z=0.$
We need to find the value of the expression $E = \frac{x^{2}}{3yz} + \frac{y^{2}}{3xz} + \frac{z^{2}}{3xy}.$
Taking the least common multiple $(LCM)$ of the denominators $3yz, 3xz,$ and $3xy,$ which is $3xyz,$
$E = \frac{x^{2}(x) + y^{2}(y) + z^{2}(z)}{3xyz} = \frac{x^{3} + y^{3} + z^{3}}{3xyz}.$
We know the algebraic identity that if $x+y+z=0,$ then $x^{3} + y^{3} + z^{3} = 3xyz.$
Substituting this into our expression:
$E = \frac{3xyz}{3xyz} = 1.$
Note: Based on the provided options and the structure of the expression,the result is $1$. If the expression was $\frac{x^3}{3yz} + \dots$ or similar,the result would vary. Given the standard form of this identity,the value is $1$.

(B) Given the equation: $x = \sqrt[3]{x^{2} + 11} - 2$.
Add $2$ to both sides: $x + 2 = \sqrt[3]{x^{2} + 11}$.
Cube both sides of the equation: $(x + 2)^{3} = x^{2} + 11$.
Expand the left side using the identity $(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)$:
$x^{3} + 2^{3} + 3(x)(2)(x + 2) = x^{2} + 11$.
Simplify the expression:
$x^{3} + 8 + 6x(x + 2) = x^{2} + 11$.
$x^{3} + 8 + 6x^{2} + 12x = x^{2} + 11$.
Rearrange the terms to isolate $(x^{3} + 5x^{2} + 12x)$:
$x^{3} + 6x^{2} - x^{2} + 12x = 11 - 8$.
$x^{3} + 5x^{2} + 12x = 3$.
Thus,the value is $3$.