Factorisation of Polynomials Questions in English

(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) Given the equation: $x^{2}+kx+6=(x+2)(x+3)$.
First,expand the right side of the equation:
$(x+2)(x+3) = x(x+3) + 2(x+3)$
$= x^{2} + 3x + 2x + 6$
$= x^{2} + 5x + 6$.
Now,compare this with the left side of the equation: $x^{2}+kx+6 = x^{2}+5x+6$.
By comparing the coefficients of $x$ on both sides,we get $k = 5$.

(A) Let $p(x) = 2x^2 + kx$.
If $(x+1)$ is a factor of $p(x)$,then by the Factor Theorem,$p(-1) = 0$.
Substituting $x = -1$ into the polynomial:
$p(-1) = 2(-1)^2 + k(-1) = 0$
$2(1) - k = 0$
$2 - k = 0$
$k = 2$.
Therefore,the correct option is $A$.

(B) According to the Factor Theorem,if $(x+1)$ is a factor of a polynomial $p(x)$,then $p(-1)$ must be equal to $0$.
$(a)$ Let $p(x) = x^{3}+x^{2}-x+1$.
$p(-1) = (-1)^{3}+(-1)^{2}-(-1)+1 = -1+1+1+1 = 2 \neq 0$.
So,$(x+1)$ is not a factor.
$(b)$ Let $p(x) = x^{3}+x^{2}+x+1$.
$p(-1) = (-1)^{3}+(-1)^{2}+(-1)+1 = -1+1-1+1 = 0$.
Since $p(-1) = 0$,$(x+1)$ is a factor.
$(c)$ Let $p(x) = x^{4}+x^{3}+x^{2}+1$.
$p(-1) = (-1)^{4}+(-1)^{3}+(-1)^{2}+1 = 1-1+1+1 = 2 \neq 0$.
So,$(x+1)$ is not a factor.
$(d)$ Let $p(x) = x^{4}+3x^{3}+3x^{2}+x+1$.
$p(-1) = (-1)^{4}+3(-1)^{3}+3(-1)^{2}+(-1)+1 = 1-3+3-1+1 = 1 \neq 0$.
So,$(x+1)$ is not a factor.
Therefore,$(x+1)$ is a factor of $x^{3}+x^{2}+x+1$. The correct option is $(b)$.

(A) To factorise the quadratic expression $4 x^{2}+8 x+3$,we use the splitting the middle term method.
We need to find two numbers whose product is $4 \times 3 = 12$ and whose sum is $8$.
These two numbers are $6$ and $2$,since $6 \times 2 = 12$ and $6 + 2 = 8$.
Now,rewrite the middle term $8x$ as $6x + 2x$:
$4 x^{2}+6 x+2 x+3$
Group the terms:
$(4 x^{2}+6 x) + (2 x+3)$
Factor out the common terms from each group:
$2 x(2 x+3) + 1(2 x+3)$
Finally,factor out the common binomial $(2 x+3)$:
$(2 x+3)(2 x+1)$
Thus,the correct option is $A$.

(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 $0$.
To find the remainder when $p(x)$ is divided by $g(x) = 2 - 3x$,we set $g(x) = 0$:
$2 - 3x = 0 \implies 3x = 2 \implies x = \frac{2}{3}$.
Now,we evaluate $p\left(\frac{2}{3}\right)$:
$p\left(\frac{2}{3}\right) = \left(\frac{2}{3}\right)^{3} - \left(\frac{2}{3}\right) + 1$
$= \frac{8}{27} - \frac{2}{3} + 1$
$= \frac{8 - 18 + 27}{27} = \frac{17}{27}$.
Since the remainder is $\frac{17}{27} \neq 0$,$p(x)$ is not a multiple of $g(x)$.

(YES) According to the Factor Theorem,$g(x)$ is a factor of $p(x)$ if $p(a) = 0$,where $a$ is the zero of $g(x)$.
First,find the zero of $g(x)$:
$g(x) = \frac{x}{3} - \frac{1}{4} = 0$
$\frac{x}{3} = \frac{1}{4}$
$x = \frac{3}{4}$
Now,evaluate $p\left(\frac{3}{4}\right)$:
$p\left(\frac{3}{4}\right) = 8\left(\frac{3}{4}\right)^3 - 6\left(\frac{3}{4}\right)^2 - 4\left(\frac{3}{4}\right) + 3$
$= 8 \times \frac{27}{64} - 6 \times \frac{9}{16} - 3 + 3$
$= \frac{27}{8} - \frac{54}{16} - 3 + 3$
$= \frac{27}{8} - \frac{27}{8} - 3 + 3 = 0$
Since $p\left(\frac{3}{4}\right) = 0$,by the Factor Theorem,$g(x)$ is a factor of $p(x)$.

(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 $-\pi^{3}+3\pi^{2}-3\pi+1$.

(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 $x^{3} + x^{2} + x + 1$.

(A) According to the Factor Theorem,$g(x) = x - a$ is a factor of $p(x)$ if $p(a) = 0$.
Here,$g(x) = x + 1$,which can be written as $x - (-1)$. So,$a = -1$.
We evaluate $p(-1)$:
$p(-1) = 2(-1)^3 + (-1)^2 - 2(-1) - 1$
$p(-1) = 2(-1) + 1 + 2 - 1$
$p(-1) = -2 + 1 + 2 - 1$
$p(-1) = 0$
Since $p(-1) = 0$,by the Factor Theorem,$g(x)$ is a factor of $p(x)$.

(D) Given the polynomial $p(x) = x^{2} + x + k$.
According to the Factor Theorem,if $(x - 1)$ is a factor of $p(x)$,then $p(1)$ must be equal to $0$.
Substituting $x = 1$ into the polynomial:
$p(1) = (1)^{2} + 1 + k$
$p(1) = 1 + 1 + k$
$p(1) = 2 + k$
Since $p(1) = 0$,we have:
$2 + k = 0$
$k = -2$

(A) Given the polynomial $p(x) = 2x^2 + kx + \sqrt{2}$.
According to the Factor Theorem, if $(x - 1)$ is a factor of $p(x)$, then $p(1) = 0$.
Substituting $x = 1$ into the polynomial:
$p(1) = 2(1)^2 + k(1) + \sqrt{2} = 0$
$2 + k + \sqrt{2} = 0$
Solving for $k$:
$k = -2 - \sqrt{2}$
Therefore, the value of $k$ is $-(2 + \sqrt{2})$.

(B) Given that $p(x) = kx^2 - \sqrt{2}x + 1$ and $(x - 1)$ is a factor of $p(x)$.
According to the Factor Theorem, if $(x - 1)$ is a factor of $p(x)$, then $p(1) = 0$.
Substituting $x = 1$ in the polynomial:
$p(1) = k(1)^2 - \sqrt{2}(1) + 1 = 0$
$k - \sqrt{2} + 1 = 0$
Solving for $k$:
$k = \sqrt{2} - 1$.

(C) Given that $x - 1$ is a factor of $p(x) = kx^2 - 3x + k$.
According to the Factor Theorem,if $(x - a)$ is a factor of $p(x)$,then $p(a) = 0$.
Here,$a = 1$,so $p(1) = 0$.
Substituting $x = 1$ into the polynomial:
$p(1) = k(1)^2 - 3(1) + k = 0$
$k - 3 + k = 0$
$2k - 3 = 0$
$2k = 3$
$k = 3/2$.

(A) To factorise the quadratic polynomial $12 x^{2}-7 x+1$,we use the splitting the middle term method.
Here,the coefficient of $x^{2}$ is $a = 12$,the coefficient of $x$ is $b = -7$,and the constant term is $c = 1$.
We need to find two numbers $l$ and $m$ such that their sum $l + m = b = -7$ and their product $lm = a \times c = 12 \times 1 = 12$.
The two numbers that satisfy these conditions are $-4$ and $-3$,since $(-4) + (-3) = -7$ and $(-4) \times (-3) = 12$.
Now,rewrite the middle term $-7x$ as $-4x - 3x$:
$12 x^{2} - 7 x + 1 = 12 x^{2} - 4 x - 3 x + 1$
Group the terms and factor out the common factors:
$= 4 x(3 x - 1) - 1(3 x - 1)$
Finally,factor out the common binomial $(3 x - 1)$:
$= (3 x - 1)(4 x - 1)$
Thus,the factors are $(3 x - 1)(4 x - 1)$.

(A) To factorise the quadratic expression $2 x^{2}+7 x+3$,we use the splitting the middle term method.
Here,the expression is in the form $ax^{2} + bx + c$,where $a = 2$,$b = 7$,and $c = 3$.
We need to find two numbers $l$ and $m$ such that $l + m = b = 7$ and $l \times m = a \times c = 2 \times 3 = 6$.
The two numbers that satisfy these conditions are $1$ and $6$,since $1 + 6 = 7$ and $1 \times 6 = 6$.
Now,rewrite the middle term $7x$ as $x + 6x$:
$2 x^{2} + 7 x + 3 = 2 x^{2} + x + 6 x + 3$
Group the terms and factor out common factors:
$= x(2 x + 1) + 3(2 x + 1)$
Finally,factor out the common binomial $(2 x + 1)$:
$= (2 x + 1)(x + 3)$
Thus,the factors are $(2 x + 1)(x + 3)$.

(A) To factorise the quadratic polynomial $6x^2 + 5x - 6$,we use the splitting the middle term method.
We need to find two numbers $l$ and $m$ such that $l + m = 5$ (the coefficient of $x$) and $lm = 6 \times (-6) = -36$ (the product of the coefficient of $x^2$ and the constant term).
We observe that $9 + (-4) = 5$ and $9 \times (-4) = -36$.
Now,rewrite the middle term $5x$ as $9x - 4x$:
$6x^2 + 9x - 4x - 6$
Group the terms:
$(6x^2 + 9x) - (4x + 6)$
Factor out the common terms from each group:
$3x(2x + 3) - 2(2x + 3)$
Finally,factor out the common binomial $(2x + 3)$:
$(2x + 3)(3x - 2)$

(A) Given polynomial: $3x^{2}-x-4$
Comparing with the standard quadratic form $ax^{2}+bx+c$,we have $a=3$,$b=-1$,and $c=-4$.
We need to find two numbers $l$ and $m$ such that $l+m = b = -1$ and $lm = a \times c = 3 \times (-4) = -12$.
The two numbers that satisfy these conditions are $-4$ and $3$,since $-4+3 = -1$ and $(-4) \times 3 = -12$.
Now,split the middle term $-x$ as $-4x+3x$:
$3x^{2}-4x+3x-4$
Group the terms:
$= x(3x-4) + 1(3x-4)$
Factor out the common binomial $(3x-4)$:
$= (3x-4)(x+1)$
Thus,the factors are $(3x-4)(x+1)$.

(A) Given expression: $x^{3}-2x^{2}-x+2$
Step $1$: Group the terms to facilitate factoring by grouping.
$x^{3}-2x^{2}-x+2 = (x^{3}-2x^{2}) - (x-2)$
Step $2$: Factor out the common terms from each group.
$= x^{2}(x-2) - 1(x-2)$
Step $3$: Factor out the common binomial $(x-2)$.
$= (x^{2}-1)(x-2)$
Step $4$: Use the identity $a^{2}-b^{2} = (a-b)(a+b)$ to factor $(x^{2}-1)$.
$= (x-1)(x+1)(x-2)$
Thus,the factors are $(x-1)(x+1)(x-2)$.

(A) Let $p(x) = x^{3}-3x^{2}-9x-5$.
By the factor theorem,we test values to find a root. Let $x = -1$:
$p(-1) = (-1)^{3} - 3(-1)^{2} - 9(-1) - 5$
$p(-1) = -1 - 3(1) + 9 - 5$
$p(-1) = -1 - 3 + 9 - 5 = 0$.
Since $p(-1) = 0$,$(x+1)$ is a factor of $p(x)$.
Now,divide $p(x)$ by $(x+1)$ using long division or synthetic division:
$x^{3}-3x^{2}-9x-5 = (x+1)(x^{2}-4x-5)$.
Factorize the quadratic expression $x^{2}-4x-5$ by splitting the middle term:
$x^{2}-5x+x-5 = x(x-5) + 1(x-5) = (x-5)(x+1)$.
Therefore,the complete factorization is:
$x^{3}-3x^{2}-9x-5 = (x+1)(x+1)(x-5) = (x+1)^{2}(x-5)$.

(D) Let $p(x) = x^{3}+13x^{2}+32x+20$.
By trial,let us find $p(1)$:
$p(1) = (1)^{3}+13(1)^{2}+32(1)+20 = 1+13+32+20 = 66 \neq 0$.
Now,let us find $p(-1)$:
$p(-1) = (-1)^{3}+13(-1)^{2}+32(-1)+20 = -1+13-32+20 = 0$.
$\therefore$ By the factor theorem,$(x+1)$ is a factor of $p(x)$.
Now,divide $p(x)$ by $(x+1)$:
$x^{3}+13x^{2}+32x+20 = (x+1)(x^{2}+12x+20)$.
Factorize the quadratic expression $x^{2}+12x+20$ by splitting the middle term:
$x^{2}+12x+20 = x^{2}+2x+10x+20$
$= x(x+2)+10(x+2)$
$= (x+2)(x+10)$.
Therefore,the factors are $(x+1)(x+2)(x+10)$.

(A) Let $p(y) = 2y^{3} + y^{2} - 2y - 1$.
By trial,we check $p(1)$:
$p(1) = 2(1)^{3} + (1)^{2} - 2(1) - 1 = 2 + 1 - 2 - 1 = 0$.
Since $p(1) = 0$,by the Factor Theorem,$(y - 1)$ is a factor of $p(y)$.
Now,divide $p(y)$ by $(y - 1)$:
$2y^{3} + y^{2} - 2y - 1 = 2y^{2}(y - 1) + 3y(y - 1) + 1(y - 1)$
$= (y - 1)(2y^{2} + 3y + 1)$.
Factorize the quadratic expression $2y^{2} + 3y + 1$ by splitting the middle term:
$2y^{2} + 2y + y + 1 = 2y(y + 1) + 1(y + 1) = (y + 1)(2y + 1)$.
Thus,the factors are $(y - 1)(y + 1)(2y + 1)$.

(A) Using the algebraic identity $a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$,we can factorise the expression.
Here,$27 y^{3}+125 z^{3} = (3 y)^{3}+(5 z)^{3}$.
Comparing this with $a^{3}+b^{3}$,we get $a = 3y$ and $b = 5z$.
Substituting these values into the identity:
$(3 y)^{3}+(5 z)^{3} = (3 y+5 z)((3 y)^{2}-(3 y)(5 z)+(5 z)^{2})$
$= (3 y+5 z)(9 y^{2}-15 y z+25 z^{2})$.

(A) The area of a rectangle is given by the formula: $\text{Area} = \text{Length} \times \text{Breadth}$.
Given the area: $25a^2 - 35a + 12$.
To find the length and breadth,we need to factorise the quadratic polynomial $25a^2 - 35a + 12$ by splitting the middle term.
We look for two numbers whose product is $25 \times 12 = 300$ and whose sum is $-35$.
These two numbers are $-20$ and $-15$,since $(-20) \times (-15) = 300$ and $(-20) + (-15) = -35$.
Now,rewrite the expression:
$25a^2 - 20a - 15a + 12$
Factor by grouping:
$= 5a(5a - 4) - 3(5a - 4)$
$= (5a - 4)(5a - 3)$
Thus,the possible expressions for the length and breadth are $(5a - 3)$ and $(5a - 4)$.

(A) The area of a rectangle is given by the formula: $\text{Area} = \text{Length} \times \text{Breadth}$.
Given the area is $35y^2 + 13y - 12$,we need to factorise this quadratic polynomial.
To factorise $35y^2 + 13y - 12$,we split the middle term $13y$ into two parts such that their sum is $13y$ and their product is equal to the product of the coefficient of $y^2$ and the constant term $(35 \times -12 = -420)$.
We find that $28y$ and $-15y$ satisfy these conditions because $28y - 15y = 13y$ and $28y \times (-15y) = -420y^2$.
Now,rewrite the expression:
$35y^2 + 28y - 15y - 12$
Group the terms:
$(35y^2 + 28y) - (15y + 12)$
Factor out the common terms:
$7y(5y + 4) - 3(5y + 4)$
$(7y - 3)(5y + 4)$
Thus,the possible expressions for the length and breadth are $(7y - 3)$ and $(5y + 4)$.

(N/A) The volume of a cuboid is given by the formula: $\text{Volume} = \text{Length} \times \text{Breadth} \times \text{Height}$.
Given volume $= 3x^2 - 12x$.
To find the possible dimensions,we factorize the expression:
$3x^2 - 12x = 3x(x - 4)$.
Thus,the expression can be written as the product of three factors: $3$,$x$,and $(x - 4)$.
Therefore,the possible dimensions of the cuboid are $3$,$x$,and $(x - 4)$ units.

(A) The volume of a cuboid is given by the formula: $\text{Volume} = \text{Length} \times \text{Breadth} \times \text{Height}$.
Given volume $= 12ky^{2} + 8ky - 20k$.
First,factor out the common term $4k$:
$12ky^{2} + 8ky - 20k = 4k(3y^{2} + 2y - 5)$.
Next,factor the quadratic expression $(3y^{2} + 2y - 5)$ by splitting the middle term:
$3y^{2} + 2y - 5 = 3y^{2} + 5y - 3y - 5$
$= y(3y + 5) - 1(3y + 5)$
$= (3y + 5)(y - 1)$.
Combining these,the volume is $4k \times (3y + 5) \times (y - 1)$.
Therefore,the possible dimensions of the cuboid are $4k$,$(3y + 5)$,and $(y - 1)$ units.

(A) Let $p(x) = x^{3} - ax^{2} + 2x + a - 1$.
Since $(x - a)$ is a factor of $p(x)$,by the Factor Theorem,$p(a) = 0$.
Substituting $x = a$ in the polynomial:
$p(a) = (a)^{3} - a(a)^{2} + 2(a) + a - 1 = 0$.
$a^{3} - a^{3} + 2a + a - 1 = 0$.
$3a - 1 = 0$.
$3a = 1$.
Therefore,$a = 1/3$.

(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 $0$.

(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 $\frac{3}{2}$.

(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 $p(2) \neq 0$,$(x-2)$ 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:
$-2a + 3 = 0$
$2a = 3$
$a = \frac{3}{2}$