Mix Examples - Polynomials Questions in English

(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 $(x-3)(3x-1)(4x+3)$.

(A) To expand the expression $(x+3)(x+8)$,we use the algebraic identity $(x+a)(x+b) = x^2 + (a+b)x + ab$.
Here,$a = 3$ and $b = 8$.
Substituting these values into the identity:
$(x+3)(x+8) = x^2 + (3+8)x + (3)(8)$
$= x^2 + 11x + 24$.

(A) To expand the expression $(2 x-3)(2 x+5)$,we use the algebraic identity $(x+a)(x+b) = x^2 + (a+b)x + ab$.
Here,let $X = 2x$,$a = -3$,and $b = 5$.
Substituting these values into the identity:
$(2x - 3)(2x + 5) = (2x)^2 + (-3 + 5)(2x) + (-3)(5)$
$= 4x^2 + (2)(2x) - 15$
$= 4x^2 + 4x - 15$.

(D) To expand the expression $(3x - 2)(3x - 6)$,we use the distributive property or the algebraic identity $(x + a)(x + b) = x^2 + (a + b)x + ab$.
Given expression: $(3x - 2)(3x - 6)$
Let $u = 3x$. Then the expression becomes $(u - 2)(u - 6)$.
Expanding this: $u^2 + (-2 - 6)u + (-2)(-6)$
$= u^2 - 8u + 12$
Now,substitute $u = 3x$ back into the expression:
$= (3x)^2 - 8(3x) + 12$
$= 9x^2 - 24x + 12$

(A) To expand the expression $(x+2t)(x-5t)$,we use the algebraic identity $(x+a)(x+b) = x^2 + (a+b)x + ab$.
Here,$a = 2t$ and $b = -5t$.
Substituting these values into the identity:
$(x+2t)(x-5t) = x^2 + (2t - 5t)x + (2t)(-5t)$
$= x^2 + (-3t)x - 10t^2$
$= x^2 - 3xt - 10t^2$.

(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 evaluate $93 \times 95$ without direct multiplication,we can use the algebraic identity $(x + a)(x + b) = x^2 + (a + b)x + ab$.
We can express the numbers as:
$93 = 90 + 3$
$95 = 90 + 5$
Now,substitute these into the identity where $x = 90$,$a = 3$,and $b = 5$:
$(90 + 3)(90 + 5) = (90)^2 + (3 + 5)(90) + (3 \times 5)$
Calculate each term:
$(90)^2 = 8100$
$(3 + 5)(90) = 8 \times 90 = 720$
$(3 \times 5) = 15$
Adding these values together:
$8100 + 720 + 15 = 8835$
Therefore,$93 \times 95 = 8835$.

(B) To evaluate $78 \times 84$ using algebraic identities,we can express the numbers as:
$78 = (80 - 2)$
$84 = (80 + 4)$
Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,where $x = 80$,$a = -2$,and $b = 4$:
$(80 - 2)(80 + 4) = (80)^2 + (-2 + 4)(80) + (-2)(4)$
$= 6400 + (2)(80) - 8$
$= 6400 + 160 - 8$
$= 6560 - 8$
$= 6552$

(A) To expand the expression $(2 x+5 y)^{2}$,we use the algebraic identity $(a+b)^{2} = a^{2}+2 a b+b^{2}$.
Here,$a = 2 x$ and $b = 5 y$.
Substituting these values into the identity:
$(2 x+5 y)^{2} = (2 x)^{2} + 2(2 x)(5 y) + (5 y)^{2}$
$= 4 x^{2} + 20 x y + 25 y^{2}$.

(A) To expand the expression $(3a - 4)^2$,we use the algebraic identity $(x - y)^2 = x^2 - 2xy + y^2$.
Here,$x = 3a$ and $y = 4$.
Substituting these values into the identity:
$(3a - 4)^2 = (3a)^2 - 2(3a)(4) + (4)^2$
$= 9a^2 - 24a + 16$.

(A) The given expression is $16 x^{2}+40 x y+25 y^{2}$.
We can rewrite this expression in the form of the algebraic identity $(a+b)^{2} = a^{2}+2ab+b^{2}$.
Here,$a^{2} = 16 x^{2} = (4 x)^{2}$,so $a = 4 x$.
And $b^{2} = 25 y^{2} = (5 y)^{2}$,so $b = 5 y$.
Now,check the middle term: $2ab = 2(4 x)(5 y) = 40 x y$.
Since the expression matches the identity,we can write it as $(4 x+5 y)^{2}$.
Thus,the factors are $(4 x+5 y)(4 x+5 y)$.

(B) To factorise the expression $49 x^{2}-42 x+9$,we observe that it follows the algebraic identity $a^{2}-2ab+b^{2} = (a-b)^{2}$.
Here,$a^{2} = 49 x^{2} = (7 x)^{2}$,so $a = 7 x$.
And $b^{2} = 9 = (3)^{2}$,so $b = 3$.
Now,check the middle term: $-2ab = -2(7 x)(3) = -42 x$.
Since the expression matches the identity,we can write:
$49 x^{2}-42 x+9 = (7 x)^{2}-2(7 x)(3)+(3)^{2} = (7 x-3)^{2}$.
Thus,the factors are $(7 x-3)(7 x-3)$.

(C) To evaluate $(107)^{2}$,we can express $107$ as $(100 + 7)$.
Using the algebraic identity $(a + b)^{2} = a^{2} + 2ab + b^{2}$,where $a = 100$ and $b = 7$:
$(100 + 7)^{2} = (100)^{2} + 2(100)(7) + (7)^{2}$
$= 10000 + 1400 + 49$
$= 11449$

(A) The given expression is in the form of the algebraic identity $(a + b)(a - b) = a^2 - b^2$.
Here,$a = 3x$ and $b = 7y$.
Applying the identity:
$(3x + 7y)(3x - 7y) = (3x)^2 - (7y)^2$
$= 9x^2 - 49y^2$

(A) The given expression is $121 x^{2}-289 y^{2}$.
We know that $121 = 11^{2}$ and $289 = 17^{2}$.
So,the expression can be written as $(11 x)^{2}-(17 y)^{2}$.
Using the algebraic identity $a^{2}-b^{2}=(a-b)(a+b)$,where $a=11x$ and $b=17y$,we get:
$(11 x)^{2}-(17 y)^{2}=(11 x-17 y)(11 x+17 y)$.

(A) To evaluate $66 \times 74$ without direct multiplication,we can express the numbers as $(70 - 4)$ and $(70 + 4)$.
Using the algebraic identity $(a - b)(a + b) = a^2 - b^2$,where $a = 70$ and $b = 4$:
$66 \times 74 = (70 - 4)(70 + 4)$
$= (70)^2 - (4)^2$
$= 4900 - 16$
$= 4884$

(A) To expand $(x+3y-5z)^{2}$,we use the algebraic identity: $(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2ab+2bc+2ca$.
Here,$a = x$,$b = 3y$,and $c = -5z$.
Substituting these values into the identity:
$(x+3y-5z)^{2} = (x)^{2} + (3y)^{2} + (-5z)^{2} + 2(x)(3y) + 2(3y)(-5z) + 2(-5z)(x)$
$= x^{2} + 9y^{2} + 25z^{2} + 6xy - 30yz - 10zx$.

(A) To expand $(2x - y - 5)^2$,we use the algebraic identity: $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.
Here,$a = 2x$,$b = -y$,and $c = -5$.
Substituting these values into the identity:
$(2x - y - 5)^2 = (2x)^2 + (-y)^2 + (-5)^2 + 2(2x)(-y) + 2(-y)(-5) + 2(-5)(2x)$.
Calculating each term:
$= 4x^2 + y^2 + 25 - 4xy + 10y - 20x$.

(A) We use the algebraic identity: $(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2ab+2bc+2ca$.
Given expression: $x^{2}+9 y^{2}+4+6 x y+12 y+4 x$.
We can rewrite this as:
$= (x)^{2} + (3y)^{2} + (2)^{2} + 2(x)(3y) + 2(3y)(2) + 2(2)(x)$.
Comparing this with the identity,where $a=x$,$b=3y$,and $c=2$,we get:
$= (x+3y+2)^{2}$.
Thus,the factors are $(x+3y+2)(x+3y+2)$.

(A) The given expression is $x^{2}+4 y^{2}+9 z^{2}-4 x y-12 y z+6 z x$.
We use the algebraic identity: $(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2ab+2bc+2ca$.
Comparing the given expression with the identity,we can rewrite it as:
$x^{2}+(-2 y)^{2}+(3 z)^{2}+2(x)(-2 y)+2(-2 y)(3 z)+2(3 z)(x)$.
Here,$a = x$,$b = -2y$,and $c = 3z$.
Thus,the expression simplifies to $(x-2y+3z)^{2}$.
Therefore,the factors are $(x-2y+3z)(x-2y+3z)$.

(C) To evaluate $(132)^{2}$ using algebraic identities,we can express $132$ as $(130 + 2)$.
Using the identity $(a + b)^{2} = a^{2} + 2ab + b^{2}$,where $a = 130$ and $b = 2$:
$(130 + 2)^{2} = (130)^{2} + 2(130)(2) + (2)^{2}$
$= 16900 + 520 + 4$
$= 17424$.

To expand $(3x + 2y)^3$,we use the algebraic identity: $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$.
Here,$a = 3x$ and $b = 2y$.
Substituting these values into the identity:
$(3x + 2y)^3 = (3x)^3 + (2y)^3 + 3(3x)(2y)(3x + 2y)$
$= 27x^3 + 8y^3 + 18xy(3x + 2y)$
$= 27x^3 + 8y^3 + 54x^2y + 36xy^2$

(N/A) To expand the expression $(2a - 5b)^3$,we use the algebraic identity:
$(x - y)^3 = x^3 - y^3 - 3xy(x - y)$
Here,$x = 2a$ and $y = 5b$.
Substituting these values into the identity:
$= (2a)^3 - (5b)^3 - 3(2a)(5b)(2a - 5b)$
$= 8a^3 - 125b^3 - 30ab(2a - 5b)$
$= 8a^3 - 125b^3 - 60a^2b + 150ab^2$

(N/A) The given expression is $8 x^{3}+27 y^{3}+36 x^{2} y+54 x y^{2}$.
We can rewrite this expression using the algebraic identity $(a+b)^{3} = a^{3} + b^{3} + 3a^{2}b + 3ab^{2}$.
Here,let $a = 2x$ and $b = 3y$.
Then,$a^{3} = (2x)^{3} = 8x^{3}$ and $b^{3} = (3y)^{3} = 27y^{3}$.
The middle terms are $3a^{2}b = 3(2x)^{2}(3y) = 3(4x^{2})(3y) = 36x^{2}y$ and $3ab^{2} = 3(2x)(3y)^{2} = 3(2x)(9y^{2}) = 54xy^{2}$.
Thus,the expression is of the form $(2x)^{3} + (3y)^{3} + 3(2x)^{2}(3y) + 3(2x)(3y)^{2}$.
This simplifies to $(2x + 3y)^{3}$.
Therefore,the factorised form is $(2x + 3y)(2x + 3y)(2x + 3y)$.

(N/A) The given expression is $27 x^{3}-64-108 x^{2}+144 x$.
We can rewrite this expression in the form $a^{3} + b^{3} + 3a^{2}b + 3ab^{2} = (a + b)^{3}$.
Here,$a = 3x$ and $b = -4$.
Substituting these values into the identity:
$27 x^{3}-64-108 x^{2}+144 x = (3x)^{3} + (-4)^{3} + 3(3x)^{2}(-4) + 3(3x)(-4)^{2}$.
This simplifies to $(3x - 4)^{3}$.
Thus,the factorised form is $(3x - 4)(3x - 4)(3x - 4)$.

(D) To evaluate $(998)^{3}$,we can express $998$ as $(1000 - 2)$.
Using the algebraic identity $(a - b)^{3} = a^{3} - b^{3} - 3ab(a - b)$,where $a = 1000$ and $b = 2$:
$(1000 - 2)^{3} = (1000)^{3} - (2)^{3} - 3(1000)(2)(1000 - 2)$
$= 1,000,000,000 - 8 - 6000(998)$
$= 1,000,000,000 - 8 - 5,988,000$
$= 994,011,992$

(D) The given expression is $8 x^{3}+y^{3}-27 z^{3}+18 x y z$.
We use the algebraic identity: $a^{3}+b^{3}+c^{3}-3 a b c = (a+b+c)(a^{2}+b^{2}+c^{2}-a b-b c-c a)$.
Rewrite the expression as: $(2 x)^{3}+(y)^{3}+(-3 z)^{3}-3(2 x)(y)(-3 z)$.
Here,$a = 2 x$,$b = y$,and $c = -3 z$.
Applying the identity:
$= (2 x+y-3 z)((2 x)^{2}+(y)^{2}+(-3 z)^{2}-(2 x)(y)-(y)(-3 z)-(-3 z)(2 x))$.
Simplifying the terms:
$= (2 x+y-3 z)(4 x^{2}+y^{2}+9 z^{2}-2 x y+3 y z+6 z x)$.

(B) Let $a = 21$,$b = 15$,and $c = -36$.
First,calculate the sum of $a, b,$ and $c$:
$a + b + c = 21 + 15 + (-36) = 36 - 36 = 0$.
We know the algebraic identity: If $a + b + c = 0$,then $a^{3} + b^{3} + c^{3} = 3abc$.
Substituting the values into the identity:
$(21)^{3} + (15)^{3} + (-36)^{3} = 3 \times (21) \times (15) \times (-36)$.
Calculating the product:
$3 \times 21 = 63$
$15 \times (-36) = -540$
$63 \times (-540) = -34020$.
Therefore,the value is $-34020$.

(B) To expand the expression $(x+4)(x+9)$,we use the distributive property ($FOIL$ method):
$(x+4)(x+9) = x(x+9) + 4(x+9)$
$= x^{2} + 9x + 4x + 36$
$= x^{2} + 13x + 36$

(A) To expand the expression $(3x - 1)(3x + 4)$,we use the distributive property ($FOIL$ method):
$(3x - 1)(3x + 4) = (3x)(3x) + (3x)(4) + (-1)(3x) + (-1)(4)$
$= 9x^2 + 12x - 3x - 4$
$= 9x^2 + 9x - 4$

(A) To expand the expression $(2x - 7)(2x - 5)$,we use the distributive property ($FOIL$ method):
$1$. Multiply the first terms: $(2x) \times (2x) = 4x^2$.
$2$. Multiply the outer terms: $(2x) \times (-5) = -10x$.
$3$. Multiply the inner terms: $(-7) \times (2x) = -14x$.
$4$. Multiply the last terms: $(-7) \times (-5) = 35$.
Combining these results: $4x^2 - 10x - 14x + 35 = 4x^2 - 24x + 35$.

(N/A) To expand the expression $(2a + 3b)(2a - 5b)$,we use the distributive property ($FOIL$ method):
$1$. Multiply the first terms: $(2a) \times (2a) = 4a^2$.
$2$. Multiply the outer terms: $(2a) \times (-5b) = -10ab$.
$3$. Multiply the inner terms: $(3b) \times (2a) = 6ab$.
$4$. Multiply the last terms: $(3b) \times (-5b) = -15b^2$.
Combining these results: $4a^2 - 10ab + 6ab - 15b^2$.
Finally,simplify the middle terms: $-10ab + 6ab = -4ab$.
Thus,the expanded form is $4a^2 - 4ab - 15b^2$.

(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 $(5 x + 3)$:
$(5 x + 3)(5 x + 2)$.