Factorisation of Polynomials Questions in English

(A) To determine if $(x+1)$ is a factor of $P(x) = 6x^3 + 11x^2 - 5x - 12$,we use the Factor Theorem.
According to the Factor Theorem,$(x+1)$ is a factor of $P(x)$ if and only if $P(-1) = 0$.
Substitute $x = -1$ into the polynomial:
$P(-1) = 6(-1)^3 + 11(-1)^2 - 5(-1) - 12$
$P(-1) = 6(-1) + 11(1) + 5 - 12$
$P(-1) = -6 + 11 + 5 - 12$
$P(-1) = 5 + 5 - 12$
$P(-1) = 10 - 12$
$P(-1) = -2$
Since $P(-1) \neq 0$,$(x+1)$ is not a factor of the given polynomial.

(A) To factorise the quadratic polynomial $x^{2}+10x+16$ by splitting the middle term,we need to find two numbers whose sum is $10$ (the coefficient of $x$) and whose product is $16$ (the constant term).
$1$. Identify the factors of $16$: $(1, 16), (2, 8), (4, 4)$.
$2$. Among these,the pair $(2, 8)$ adds up to $10$.
$3$. Rewrite the middle term $10x$ as $2x + 8x$:
$x^{2} + 2x + 8x + 16$
$4$. Group the terms and factor out the common elements:
$x(x + 2) + 8(x + 2)$
$5$. Factor out the common binomial $(x + 2)$:
$(x + 2)(x + 8)$

(A) To factorise the quadratic polynomial $x^{2}-12x+20$,we need to find two numbers whose product is $20$ and whose sum is $-12$.
These two numbers are $-2$ and $-10$,since $(-2) \times (-10) = 20$ and $(-2) + (-10) = -12$.
Now,split the middle term $-12x$ as $-2x - 10x$:
$x^{2} - 2x - 10x + 20$
Group the terms:
$(x^{2} - 2x) - (10x - 20)$
Factor out the common terms from each group:
$x(x - 2) - 10(x - 2)$
Finally,factor out the common binomial $(x - 2)$:
$(x - 2)(x - 10)$

(A) To factorise $6x^2 + 19x + 10$ by splitting the middle term,we need to find two numbers whose product is $6 \times 10 = 60$ and whose sum is $19$.
These two numbers are $15$ and $4$,since $15 \times 4 = 60$ and $15 + 4 = 19$.
Now,rewrite the middle term $19x$ as $15x + 4x$:
$6x^2 + 15x + 4x + 10$
Group the terms to factor out common factors:
$(6x^2 + 15x) + (4x + 10)$
$3x(2x + 5) + 2(2x + 5)$
Finally,factor out the common binomial $(2x + 5)$:
$(2x + 5)(3x + 2)$

(N/A) To factorise the quadratic polynomial $x^{2}+14x+33$ by splitting the middle term,we need to find two numbers whose sum is $14$ and whose product is $33$.
$1$. Identify the coefficients: $a=1, b=14, c=33$.
$2$. Find two numbers $p$ and $q$ such that $p+q=14$ and $p \times q=33$.
$3$. The factors of $33$ are $(1, 33)$ and $(3, 11)$.
$4$. Since $3+11=14$,the numbers are $3$ and $11$.
$5$. Rewrite the middle term: $x^{2}+3x+11x+33$.
$6$. Factor by grouping: $x(x+3)+11(x+3)$.
$7$. Final factored form: $(x+3)(x+11)$.

(A) To factorise the quadratic polynomial $x^{2}-3x-40$,we need to find two numbers whose product is $-40$ and whose sum is $-3$.
$1$. The factors of $-40$ are $(1, -40), (-1, 40), (2, -20), (-2, 20), (4, -10), (-4, 10), (5, -8), (-5, 8)$.
$2$. Among these pairs,the pair $(5, -8)$ satisfies the condition: $5 + (-8) = -3$.
$3$. Now,split the middle term $-3x$ as $5x - 8x$:
$x^{2} + 5x - 8x - 40$
$4$. Group the terms and factor out the common factors:
$x(x + 5) - 8(x + 5)$
$5$. Factor out the common binomial $(x + 5)$:
$(x + 5)(x - 8)$

(A) To factorise $6 x^{2}+7 x-20$ by splitting the middle term,we need to find two numbers whose product is $6 \times (-20) = -120$ and whose sum is $7$.
These two numbers are $15$ and $-8$,since $15 \times (-8) = -120$ and $15 + (-8) = 7$.
Now,rewrite the middle term $7x$ as $15x - 8x$:
$6 x^{2} + 15 x - 8 x - 20$
Group the terms to factor out common factors:
$(6 x^{2} + 15 x) - (8 x + 20)$
$3 x(2 x + 5) - 4(2 x + 5)$
Taking $(2 x + 5)$ as a common factor:
$(2 x + 5)(3 x - 4)$

(A) To factorise the quadratic polynomial $15x^2 + 7x - 2$ by splitting the middle term,we look for two numbers whose product is $15 \times (-2) = -30$ and whose sum is $7$.
The two numbers that satisfy these conditions are $10$ and $-3$.
Now,rewrite the middle term $7x$ as $10x - 3x$:
$15x^2 + 10x - 3x - 2$
Group the terms:
$(15x^2 + 10x) - (3x + 2)$
Factor out the common terms from each group:
$5x(3x + 2) - 1(3x + 2)$
Finally,factor out the common binomial $(3x + 2)$:
$(3x + 2)(5x - 1)$

(A) To factorise the quadratic polynomial $x^{2}-4x-77$,we need to find two numbers whose product is $-77$ and whose sum is $-4$.
$1$. The factors of $-77$ are $(1, -77), (-1, 77), (7, -11), (-7, 11)$.
$2$. Among these,the pair $(7, -11)$ satisfies the condition $7 + (-11) = -4$.
$3$. Now,split the middle term $-4x$ as $7x - 11x$:
$x^{2} + 7x - 11x - 77$
$4$. Group the terms and factor out the common elements:
$x(x + 7) - 11(x + 7)$
$5$. Factor out the common binomial $(x + 7)$:
$(x + 7)(x - 11)$

(A) To factorise the quadratic polynomial $x^{2}+2x-143$,we need to find two numbers whose product is $-143$ and whose sum is $2$.
Let the two numbers be $a$ and $b$.
We have $a \times b = -143$ and $a + b = 2$.
Factors of $143$ are $1, 11, 13, 143$.
Considering the signs,we look for $13$ and $-11$ because $13 \times (-11) = -143$ and $13 + (-11) = 2$.
Now,split the middle term $2x$ as $13x - 11x$:
$x^{2} + 13x - 11x - 143$
Group the terms:
$(x^{2} + 13x) - (11x + 143)$
Factor out the common terms:
$x(x + 13) - 11(x + 13)$
Finally,factor out $(x + 13)$:
$(x + 13)(x - 11)$

(N/A) To factorise $12x^2 + 23x + 5$ by splitting the middle term,we need to find two numbers whose product is $12 \times 5 = 60$ and whose sum is $23$.
These two numbers are $20$ and $3$.
Now,rewrite the middle term $23x$ as $20x + 3x$:
$12x^2 + 20x + 3x + 5$
Group the terms:
$(12x^2 + 20x) + (3x + 5)$
Factor out the common terms from each group:
$4x(3x + 5) + 1(3x + 5)$
Finally,factor out the common binomial $(3x + 5)$:
$(4x + 1)(3x + 5)$

(A) Let $p(x) = x^{3}+8x^{2}+9x-18$.
By the factor theorem,we test values to find a root. If $x=1$,then $p(1) = 1^{3} + 8(1)^{2} + 9(1) - 18 = 1 + 8 + 9 - 18 = 0$.
Since $p(1) = 0$,$(x-1)$ is a factor of $p(x)$.
Now,divide $x^{3}+8x^{2}+9x-18$ by $(x-1)$ using synthetic division or long division:
$x^{3}+8x^{2}+9x-18 = (x-1)(x^{2}+9x+18)$.
Next,factorize the quadratic expression $x^{2}+9x+18$:
$x^{2}+6x+3x+18 = x(x+6) + 3(x+6) = (x+3)(x+6)$.
Thus,the complete factorization is $(x-1)(x+3)(x+6)$.

(A) Let $p(x) = x^{3}+12 x^{2}+39 x+28$.
By the Factor Theorem,if $(x+a)$ is a factor,then $p(-a) = 0$.
Testing $x = -1$: $p(-1) = (-1)^{3} + 12(-1)^{2} + 39(-1) + 28 = -1 + 12 - 39 + 28 = 0$.
Since $p(-1) = 0$,$(x+1)$ is a factor of $p(x)$.
Now,divide $p(x)$ by $(x+1)$ using synthetic division or long division:
$x^{3}+12 x^{2}+39 x+28 = (x+1)(x^{2}+11 x+28)$.
Factor the quadratic expression $x^{2}+11 x+28$ by splitting the middle term:
$x^{2} + 7x + 4x + 28 = x(x+7) + 4(x+7) = (x+4)(x+7)$.
Thus,the complete factorization is $(x+1)(x+4)(x+7)$.

(N/A) Let $p(x) = x^{3}+2x^{2}-13x+10$.
By the Factor Theorem,we test the factors of the constant term $10$ (i.e.,$\pm 1, \pm 2, \pm 5, \pm 10$).
For $x = 1$,$p(1) = (1)^{3} + 2(1)^{2} - 13(1) + 10 = 1 + 2 - 13 + 10 = 0$.
Since $p(1) = 0$,$(x-1)$ is a factor of $p(x)$.
Dividing $p(x)$ by $(x-1)$,we get $x^{2}(x-1) + 3x(x-1) - 10(x-1) = (x-1)(x^{2}+3x-10)$.
Now,factorise the quadratic expression $x^{2}+3x-10$ by splitting the middle term:
$x^{2} + 5x - 2x - 10 = x(x+5) - 2(x+5) = (x-2)(x+5)$.
Thus,the factors are $(x-1)(x-2)(x+5)$.

(A) Let $p(x) = x^{3}-11 x^{2}+20 x+32$.
By the Factor Theorem,we test values to find a root. Let $x = -1$:
$p(-1) = (-1)^{3} - 11(-1)^{2} + 20(-1) + 32 = -1 - 11 - 20 + 32 = 0$.
Since $p(-1) = 0$,$(x+1)$ is a factor of $p(x)$.
Now,divide $x^{3}-11 x^{2}+20 x+32$ by $(x+1)$ using synthetic division or long division:
$x^{3}+x^{2} - 12x^{2}-12x + 32x+32 = x^{2}(x+1) - 12x(x+1) + 32(x+1) = (x+1)(x^{2}-12x+32)$.
Factor the quadratic $x^{2}-12x+32$:
$x^{2}-8x-4x+32 = x(x-8)-4(x-8) = (x-4)(x-8)$.
Thus,the factors are $(x+1)(x-4)(x-8)$.

(A) Let $p(x) = 6x^{3}-23x^{2}+29x-12$.
By the Factor Theorem,we test values to find a root.
For $x = 1$,$p(1) = 6(1)^{3}-23(1)^{2}+29(1)-12 = 6-23+29-12 = 0$.
Since $p(1) = 0$,$(x-1)$ is a factor.
Dividing $6x^{3}-23x^{2}+29x-12$ by $(x-1)$,we get the quotient $6x^{2}-17x+12$.
Now,factorise the quadratic $6x^{2}-17x+12$:
$6x^{2}-9x-8x+12 = 3x(2x-3)-4(2x-3) = (3x-4)(2x-3)$.
Thus,the factors are $(x-1)(2x-3)(3x-4)$.

(N/A) Let $p(x) = 6x^{3} + 7x^{2} - 14x - 15$.
By the Factor Theorem,we test values for $x$. Let $x = -1$:
$p(-1) = 6(-1)^{3} + 7(-1)^{2} - 14(-1) - 15 = -6 + 7 + 14 - 15 = 0$.
Since $p(-1) = 0$,$(x + 1)$ is a factor.
Now,divide $6x^{3} + 7x^{2} - 14x - 15$ by $(x + 1)$ using long division or synthetic division:
$6x^{3} + 7x^{2} - 14x - 15 = (x + 1)(6x^{2} + x - 15)$.
Next,factorize the quadratic $6x^{2} + x - 15$ by splitting the middle term:
$6x^{2} + 10x - 9x - 15 = 2x(3x + 5) - 3(3x + 5) = (2x - 3)(3x + 5)$.
Thus,the complete factorization is $(x + 1)(3x + 5)(2x - 3)$.

(N/A) Let $p(x) = 8x^{3} - 26x^{2} + 13x + 5$.
By the factor theorem,we test values to find a root. Let $x = 1$:
$p(1) = 8(1)^{3} - 26(1)^{2} + 13(1) + 5 = 8 - 26 + 13 + 5 = 0$.
Since $p(1) = 0$,$(x - 1)$ is a factor.
Now,divide $8x^{3} - 26x^{2} + 13x + 5$ by $(x - 1)$ using long division or synthetic division:
$8x^{3} - 26x^{2} + 13x + 5 = (x - 1)(8x^{2} - 18x - 5)$.
Next,factorize the quadratic $8x^{2} - 18x - 5$ by splitting the middle term:
$8x^{2} - 20x + 2x - 5 = 4x(2x - 5) + 1(2x - 5) = (4x + 1)(2x - 5)$.
Thus,the factors are $(x - 1)(4x + 1)(2x - 5)$.

(A) Let $p(x) = 12x^3 + 17x^2 + 3x - 2$.
By the Factor Theorem,we test values to find a root. Let $x = -1$:
$p(-1) = 12(-1)^3 + 17(-1)^2 + 3(-1) - 2 = -12 + 17 - 3 - 2 = 0$.
Since $p(-1) = 0$,$(x+1)$ is a factor.
Now,divide $12x^3 + 17x^2 + 3x - 2$ by $(x+1)$ using long division or synthetic division:
$12x^3 + 17x^2 + 3x - 2 = (x+1)(12x^2 + 5x - 2)$.
Next,factorize the quadratic $12x^2 + 5x - 2$ by splitting the middle term:
$12x^2 + 8x - 3x - 2 = 4x(3x + 2) - 1(3x + 2) = (4x - 1)(3x + 2)$.
Thus,the complete factorization is $(x+1)(4x-1)(3x+2)$.

(A) According to the Factor Theorem,if $(x-c)$ is a factor of a polynomial $p(x)$,then $p(c) = 0$.
Here,$p(x) = x^{3} + 7x^{2} + ax - 3$ and the factor is $(x-1)$,so $c = 1$.
Setting $p(1) = 0$:
$(1)^{3} + 7(1)^{2} + a(1) - 3 = 0$
$1 + 7 + a - 3 = 0$
$8 + a - 3 = 0$
$5 + a = 0$
$a = -5$

(A) Let $p(x) = 6x^3 + 19x^2 + 16x + 4$. According to the factor theorem,$(x+2)$ is a factor if $p(-2) = 0$.
$p(-2) = 6(-2)^3 + 19(-2)^2 + 16(-2) + 4 = 6(-8) + 19(4) - 32 + 4 = -48 + 76 - 32 + 4 = 0$.
Since $p(-2) = 0$,$(x+2)$ is a factor.
Now,divide $6x^3 + 19x^2 + 16x + 4$ by $(x+2)$ to get the quadratic $6x^2 + 7x + 2$.
Factorising $6x^2 + 7x + 2 = 6x^2 + 4x + 3x + 2 = 2x(3x+2) + 1(3x+2) = (2x+1)(3x+2)$.
Thus,the complete factorisation is $(x+2)(2x+1)(3x+2)$.

(A) Step $1$: To show $(x-3)$ is a factor,we evaluate $p(3)$.
$p(3) = 12(3)^3 - 31(3)^2 - 18(3) + 9$
$p(3) = 12(27) - 31(9) - 54 + 9$
$p(3) = 324 - 279 - 54 + 9 = 0$.
Since $p(3) = 0$,by the factor theorem,$(x-3)$ is a factor of $p(x)$.
Step $2$: Divide $p(x)$ by $(x-3)$ to get the quadratic factor.
$12x^3 - 31x^2 - 18x + 9 = (x-3)(12x^2 + 5x - 3)$.
Step $3$: Factorise the quadratic $12x^2 + 5x - 3$.
$12x^2 + 9x - 4x - 3 = 3x(4x + 3) - 1(4x + 3) = (3x - 1)(4x + 3)$.
Thus,the complete factorisation is $(x-3)(3x-1)(4x+3)$.

(B) To factorise the quadratic expression $4 x^{2}+4 x y-3 y^{2}$,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 $4$.
These two numbers are $6$ and $-2$.
Now,rewrite the middle term $4xy$ as $6xy - 2xy$:
$4 x^{2}+6 x y-2 x y-3 y^{2}$
Group the terms:
$= (4 x^{2}+6 x y) - (2 x y+3 y^{2})$
Factor out the common terms from each group:
$= 2 x(2 x+3 y)-y(2 x+3 y)$
Finally,factor out the common binomial $(2 x+3 y)$:
$= (2 x+3 y)(2 x-y)$

(A) To factorise the quadratic expression $16 x^{2}-16 x-21$,we look for two numbers whose product is $16 \times (-21) = -336$ and whose sum is $-16$.
These two numbers are $-28$ and $12$,because $(-28) \times 12 = -336$ and $(-28) + 12 = -16$.
Now,split the middle term $-16 x$ as $-28 x + 12 x$:
$16 x^{2} - 28 x + 12 x - 21$
Group the terms:
$(16 x^{2} - 28 x) + (12 x - 21)$
Factor out the common terms from each group:
$4 x(4 x - 7) + 3(4 x - 7)$
Finally,factor out the common binomial $(4 x - 7)$:
$(4 x - 7)(4 x + 3)$.

(A) To factorise the quadratic expression $25 x^{2}+25 x+6$,we look for two numbers whose product is $25 \times 6 = 150$ and whose sum is $25$.
These two numbers are $15$ and $10$,since $15 \times 10 = 150$ and $15 + 10 = 25$.
Now,split the middle term $25 x$ as $15 x + 10 x$:
$25 x^{2} + 15 x + 10 x + 6$
Group the terms:
$(25 x^{2} + 15 x) + (10 x + 6)$
Factor out the common terms from each group:
$5 x(5 x + 3) + 2(5 x + 3)$
Finally,factor out the common binomial $(5 x + 3)$:
$(5 x + 3)(5 x + 2)$.

(A) To factorise the quadratic expression $9 x^{2}-21 x y+10 y^{2}$,we use the splitting the middle term method.
We need to find two numbers whose product is $9 \times 10 = 90$ and whose sum is $-21$.
The two numbers are $-15$ and $-6$,since $(-15) \times (-6) = 90$ and $(-15) + (-6) = -21$.
Now,rewrite the middle term $-21 x y$ as $-15 x y - 6 x y$:
$9 x^{2} - 15 x y - 6 x y + 10 y^{2}$
Group the terms:
$(9 x^{2} - 15 x y) - (6 x y - 10 y^{2})$
Factor out the common terms from each group:
$3 x(3 x - 5 y) - 2 y(3 x - 5 y)$
Finally,factor out the common binomial $(3 x - 5 y)$:
$(3 x - 5 y)(3 x - 2 y)$

(A) To factorise the quadratic expression $49 x^{2}-35 x+6$,we use the splitting the middle term method.
We need to find two numbers such that their product is $49 \times 6 = 294$ and their sum is $-35$.
The two numbers are $-21$ and $-14$,since $(-21) \times (-14) = 294$ and $(-21) + (-14) = -35$.
Now,rewrite the middle term:
$49 x^{2}-21 x-14 x+6$
Group the terms:
$(49 x^{2}-21 x) - (14 x-6)$
Factor out the common terms from each group:
$7 x(7 x-3) - 2(7 x-3)$
Finally,factor out the common binomial $(7 x-3)$:
$(7 x-3)(7 x-2)$

(A) To factorise the expression $144 x^{2}-289 y^{2}$,we use the algebraic identity $a^{2}-b^{2}=(a-b)(a+b)$.
First,express the given terms as squares:
$144 x^{2} = (12 x)^{2}$
$289 y^{2} = (17 y)^{2}$
Now,substitute these into the identity:
$144 x^{2}-289 y^{2} = (12 x)^{2}-(17 y)^{2}$
Using the identity $a^{2}-b^{2}=(a-b)(a+b)$,where $a=12 x$ and $b=17 y$,we get:
$(12 x-17 y)(12 x+17 y)$.

(A) To factorise $16 x^{4}-y^{4}$,we use the difference of squares identity: $a^{2}-b^{2}=(a-b)(a+b)$.
First,rewrite the expression as $(4 x^{2})^{2}-(y^{2})^{2}$.
Applying the identity,we get $(4 x^{2}-y^{2})(4 x^{2}+y^{2})$.
Now,factorise the term $(4 x^{2}-y^{2})$ further by writing it as $(2 x)^{2}-(y)^{2}$.
Using the identity again,$(4 x^{2}-y^{2}) = (2 x-y)(2 x+y)$.
Combining these,the final factorised form is $(2 x-y)(2 x+y)(4 x^{2}+y^{2})$.

(A) The area of a rectangle is given by the product of its length and breadth: $\text{Area} = \text{Length} \times \text{Breadth}$.
Given area = $20 x^{2} + 22 x + 6$.
To find the length and breadth,we factorize the quadratic expression:
$20 x^{2} + 22 x + 6 = 20 x^{2} + 10 x + 12 x + 6$.
Grouping the terms: $(20 x^{2} + 10 x) + (12 x + 6)$.
Factor out common terms: $10 x(2 x + 1) + 6(2 x + 1)$.
Thus,the factors are $(10 x + 6)(2 x + 1)$ or $(5 x + 3)(4 x + 2)$.
Therefore,the possible expressions for length and breadth are $(5 x + 3)$ units and $(4 x + 2)$ units.

(A) The volume of a cuboid is given by the product of its dimensions: $V = \text{length} \times \text{breadth} \times \text{height}$.
Given $V = 2x^3 + 15x^2 + 33x + 20$.
To find the dimensions, we factorize the polynomial.
Let $p(x) = 2x^3 + 15x^2 + 33x + 20$.
By testing values, if $x = -1$, $p(-1) = 2(-1)^3 + 15(-1)^2 + 33(-1) + 20 = -2 + 15 - 33 + 20 = 0$.
Thus, $(x + 1)$ is a factor.
Dividing $2x^3 + 15x^2 + 33x + 20$ by $(x + 1)$ using synthetic division or long division gives $2x^2 + 13x + 20$.
Now, factorize the quadratic $2x^2 + 13x + 20$:
$2x^2 + 8x + 5x + 20 = 2x(x + 4) + 5(x + 4) = (2x + 5)(x + 4)$.
Therefore, the dimensions are $(2x + 5), (x + 4), (x + 1)$.

(D) Let the polynomial be $p(x) = x^{2} - 8x + 10$. Let the value to be added be $k$. The new polynomial is $g(x) = x^{2} - 8x + 10 + k$. For $g(x)$ to be divisible by $x - 3$,the remainder when $g(x)$ is divided by $x - 3$ must be $0$. According to the Remainder Theorem,the remainder is $g(3)$. Thus,$g(3) = 0$. Substituting $x = 3$ into $g(x)$: $(3)^{2} - 8(3) + 10 + k = 0$. $9 - 24 + 10 + k = 0$. $-15 + 10 + k = 0$. $-5 + k = 0$. Therefore,$k = 5$.

(A) To verify the statement,we expand the right-hand side:
$(x-6)(x-2) = x(x-2) - 6(x-2)$
$= x^{2} - 2x - 6x + 12$
$= x^{2} - 8x + 12$
Since the expanded form matches the left-hand side,the statement is True.

(C) According to the Factor Theorem,if $p(a) = 0$ for a polynomial $p(x),$ then $(x - a)$ is a factor of $p(x).$
Given that $p(7) = 0,$ we substitute $a = 7$ into the theorem.
Therefore,$(x - 7)$ is a factor of $p(x).$

(D) Given the quadratic expression $x^{2}-10x+21 = (x+m)(x+n)$.
Expanding the right side,we get $(x+m)(x+n) = x^{2} + nx + mx + mn = x^{2} + (m+n)x + mn$.
Comparing the coefficients of $x$ and the constant term with the original expression $x^{2}-10x+21$,we get:
$m+n = -10$
$mn = 21$
Therefore,the value of $m+n$ is $-10$.

(A) Let $p(x) = x^{3}-3x^{2}+ax+24$.
Since $(x-2)$ is a factor of $p(x)$,by the Factor Theorem,$p(2) = 0$.
Substituting $x = 2$ into the polynomial:
$p(2) = (2)^{3} - 3(2)^{2} + a(2) + 24 = 0$.
$8 - 3(4) + 2a + 24 = 0$.
$8 - 12 + 2a + 24 = 0$.
$20 + 2a = 0$.
$2a = -20$.
$a = -10$.

(C) To factorize the quadratic polynomial $x^{2}-23x+120$,we need to find two numbers whose sum is $-23$ and whose product is $120$.
Let the two numbers be $a$ and $b$.
We have $a+b = -23$ and $ab = 120$.
By testing factors of $120$,we find that $-15$ and $-8$ satisfy these conditions because $(-15) + (-8) = -23$ and $(-15) \times (-8) = 120$.
Now,rewrite the middle term $-23x$ as $-15x - 8x$:
$x^{2}-15x-8x+120$
Group the terms:
$(x^{2}-15x) - (8x-120)$
Factor out the common terms:
$x(x-15) - 8(x-15)$
$(x-15)(x-8)$
Thus,the factors are $(x-15)(x-8)$.

(D) To find the factor of the polynomial $p(x) = x^3 - 3x^2 + 7x - 5$,we use the Factor Theorem.
According to the Factor Theorem,$(x-a)$ is a factor of $p(x)$ if $p(a) = 0$.
Let us test the given options:
For option $D$,$x-1 = 0$ implies $x = 1$.
$p(1) = (1)^3 - 3(1)^2 + 7(1) - 5$
$p(1) = 1 - 3 + 7 - 5$
$p(1) = 8 - 8 = 0$.
Since $p(1) = 0$,$(x-1)$ is a factor of the polynomial $p(x)$.

(A) According to the Remainder Theorem,if a polynomial $p(x)$ is divided by $(x - a)$,the remainder is $p(a)$.
Here,we are dividing $p(x) = x^{2} + 12x + 36$ by $(x + 5)$,which is equivalent to $(x - (-5))$.
Therefore,the remainder is $p(-5)$.
Substitute $x = -5$ into the polynomial:
$p(-5) = (-5)^{2} + 12(-5) + 36$
$p(-5) = 25 - 60 + 36$
$p(-5) = 61 - 60$
$p(-5) = 1$
Thus,the remainder is $1$.

(B) Given the quadratic expression $x^{2}-8x-20 = (x+a)(x+b)$.
Expanding the right side,we get $(x+a)(x+b) = x^{2} + (a+b)x + ab$.
Comparing this with the given expression $x^{2}-8x-20$,we equate the constant terms.
The constant term on the left side is $-20$.
The constant term on the right side is $ab$.
Therefore,$ab = -20$.

(D) Let $p(x) = x^{3} + 12x^{2} + ax + 60$.
Since $(x+3)$ is a factor of $p(x)$,by the Factor Theorem,$p(-3) = 0$.
Substituting $x = -3$ into the polynomial:
$(-3)^{3} + 12(-3)^{2} + a(-3) + 60 = 0$
$-27 + 12(9) - 3a + 60 = 0$
$-27 + 108 - 3a + 60 = 0$
$141 - 3a = 0$
$3a = 141$
$a = 47$.

(B) According to the Factor Theorem,if $p(a) = 0$,then $(x - a)$ is a factor of the polynomial $p(x)$.
Given that $p(-3) = 0$,we substitute $a = -3$ into the expression $(x - a)$.
This gives us $(x - (-3)) = x + 3$.
Therefore,$(x + 3)$ is a factor of $p(x)$.

(B) According to the Factor Theorem,if $p(a) = 0$ for a polynomial $p(x)$,then $(x - a)$ is a factor of $p(x)$.
Given that $p(7) = 0$,we substitute $a = 7$ into the theorem.
Therefore,$(x - 7)$ is a factor of $p(x)$.

(D) To find the remainder when $p(x) = 3x^3 - 6x^2 + 5x - 10$ is divided by $(x - 2)$,we 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,$a = 2$.
Substitute $x = 2$ into the polynomial $p(x)$:
$p(2) = 3(2)^3 - 6(2)^2 + 5(2) - 10$
$p(2) = 3(8) - 6(4) + 10 - 10$
$p(2) = 24 - 24 + 10 - 10$
$p(2) = 0$
Therefore,the remainder is $0$.

(A) To determine if $(x-1)$ is a factor of the polynomial $p(x) = 3x^2 + 7x - 10$,we use the Factor Theorem.
According to the Factor Theorem,$(x-a)$ is a factor of $p(x)$ if $p(a) = 0$.
Here,$a = 1$.
Substitute $x = 1$ into the polynomial:
$p(1) = 3(1)^2 + 7(1) - 10$
$p(1) = 3(1) + 7 - 10$
$p(1) = 3 + 7 - 10$
$p(1) = 10 - 10 = 0$.
Since $p(1) = 0$,$(x-1)$ is indeed a factor of $3x^2 + 7x - 10$.

(B) To find the remainder when $16x^2 - 24x + 9$ is divided by $4x - 3$,we can use the Remainder Theorem or perform polynomial long division.
Method $1$: Factorization
Notice that $16x^2 - 24x + 9$ is a perfect square trinomial.
$16x^2 - 24x + 9 = (4x)^2 - 2(4x)(3) + (3)^2 = (4x - 3)^2$.
Since $(4x - 3)^2$ is exactly divisible by $(4x - 3)$,the remainder is $0$.
Method $2$: Remainder Theorem
Set the divisor equal to zero: $4x - 3 = 0 \implies x = 3/4$.
Substitute $x = 3/4$ into the polynomial $P(x) = 16x^2 - 24x + 9$:
$P(3/4) = 16(3/4)^2 - 24(3/4) + 9$
$P(3/4) = 16(9/16) - 18 + 9$
$P(3/4) = 9 - 18 + 9 = 0$.
Thus,the remainder is $0$.

(A) To determine if $(x+1)$ is a factor of the polynomial $p(x) = 4x^3 + 7x^2 - 2x - 5$,we use the Factor Theorem.
According to the Factor Theorem,$(x-a)$ is a factor of $p(x)$ if $p(a) = 0$.
Here,we set $x+1 = 0$,which gives $x = -1$.
Now,we calculate $p(-1)$:
$p(-1) = 4(-1)^3 + 7(-1)^2 - 2(-1) - 5$
$p(-1) = 4(-1) + 7(1) + 2 - 5$
$p(-1) = -4 + 7 + 2 - 5$
$p(-1) = 0$
Since $p(-1) = 0$,$(x+1)$ is indeed a factor of the given polynomial.

(A) To factorise the expression $x^{3}-125$,we use the algebraic identity for the difference of two cubes: $a^{3}-b^{3} = (a-b)(a^{2}+ab+b^{2})$.
Here,$x^{3}-125$ can be written as $x^{3}-5^{3}$.
Comparing this with $a^{3}-b^{3}$,we have $a = x$ and $b = 5$.
Substituting these values into the identity: $x^{3}-5^{3} = (x-5)(x^{2} + x(5) + 5^{2})$.
Simplifying the expression,we get: $(x-5)(x^{2}+5x+25)$.