Textbook - Polynomials Questions in English

(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 $(y - 1)(y + 1)(2y + 1)$.

(N/A) $(i)$ Here,we can use the identity $(x + y)^2 = x^2 + 2xy + y^2$. Substituting $y = 3$ into this identity,we get:
$(x + 3)(x + 3) = (x + 3)^2 = x^2 + 2(x)(3) + (3)^2 = x^2 + 6x + 9$.
$(ii)$ Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,where $a = -3$ and $b = 5$,we have:
$(x - 3)(x + 5) = x^2 + (-3 + 5)x + (-3)(5) = x^2 + 2x - 15$.

(A) To evaluate $105 \times 106$ without direct multiplication,we can express the numbers as $(100 + 5)$ and $(100 + 6)$.
Using the algebraic identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,where $x = 100$,$a = 5$,and $b = 6$:
$105 \times 106 = (100 + 5)(100 + 6)$
$= (100)^2 + (5 + 6)(100) + (5 \times 6)$
$= 10000 + 11(100) + 30$
$= 10000 + 1100 + 30$
$= 11130$

(B) To factorise the expression $49 a^{2}+70 a b+25 b^{2}$,we observe the terms:
$49 a^{2} = (7 a)^{2}$
$25 b^{2} = (5 b)^{2}$
$70 a b = 2(7 a)(5 b)$
Comparing this with the algebraic identity $(x+y)^{2} = x^{2}+2 x y+y^{2}$,we identify $x = 7 a$ and $y = 5 b$.
Substituting these values into the identity,we get:
$49 a^{2}+70 a b+25 b^{2} = (7 a+5 b)^{2} = (7 a+5 b)(7 a+5 b)$.

(A) We have the expression: $\frac{25}{4} x^{2}-\frac{y^{2}}{9}$
This can be written as: $\left(\frac{5}{2} x\right)^{2}-\left(\frac{y}{3}\right)^{2}$
Using the algebraic identity $a^{2}-b^{2} = (a+b)(a-b)$,where $a = \frac{5}{2} x$ and $b = \frac{y}{3}$,we get:
$\frac{25}{4} x^{2}-\frac{y^{2}}{9} = \left(\frac{5}{2} x+\frac{y}{3}\right)\left(\frac{5}{2} x-\frac{y}{3}\right)$

Comparing the given expression with $(x + y + z)^2$,we identify that $x = 3a$,$y = 4b$,and $z = 5c$.
Using the algebraic identity $(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$,we substitute the values:
$(3a + 4b + 5c)^2 = (3a)^2 + (4b)^2 + (5c)^2 + 2(3a)(4b) + 2(4b)(5c) + 2(5c)(3a)$
Calculating each term:
$(3a)^2 = 9a^2$
$(4b)^2 = 16b^2$
$(5c)^2 = 25c^2$
$2(3a)(4b) = 24ab$
$2(4b)(5c) = 40bc$
$2(5c)(3a) = 30ac$
Thus,the expanded form is $9a^2 + 16b^2 + 25c^2 + 24ab + 40bc + 30ac$.

(N/A) Using the algebraic identity $(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$,we have:
$(4a - 2b - 3c)^2 = [4a + (-2b) + (-3c)]^2$
$= (4a)^2 + (-2b)^2 + (-3c)^2 + 2(4a)(-2b) + 2(-2b)(-3c) + 2(-3c)(4a)$
$= 16a^2 + 4b^2 + 9c^2 - 16ab + 12bc - 24ac$

(A) The given expression is $4x^2 + y^2 + z^2 - 4xy - 2yz + 4xz$.
This can be written in the form of the algebraic identity $(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.
We rewrite the expression as:
$(2x)^2 + (-y)^2 + (z)^2 + 2(2x)(-y) + 2(-y)(z) + 2(2x)(z)$.
Comparing this with the identity,we have $a = 2x$,$b = -y$,and $c = z$.
Therefore,the expression becomes $(2x - y + z)^2$.
Thus,the factors are $(2x - y + z)(2x - y + z)$.

(N/A) To expand $(3a + 4b)^3$,we use the algebraic identity: $(x + y)^3 = x^3 + y^3 + 3xy(x + y)$.
Comparing $(3a + 4b)^3$ with $(x + y)^3$,we get $x = 3a$ and $y = 4b$.
Substituting these values into the identity:
$(3a + 4b)^3 = (3a)^3 + (4b)^3 + 3(3a)(4b)(3a + 4b)$
$= 27a^3 + 64b^3 + 36ab(3a + 4b)$
$= 27a^3 + 64b^3 + 108a^2b + 144ab^2$.

(N/A) To expand $(5 p-3 q)^{3}$,we use the algebraic identity: $(x-y)^{3} = x^{3} - y^{3} - 3xy(x-y)$.
Comparing $(5 p-3 q)^{3}$ with $(x-y)^{3}$,we identify $x = 5p$ and $y = 3q$.
Substituting these values into the identity:
$(5 p-3 q)^{3} = (5 p)^{3} - (3 q)^{3} - 3(5 p)(3 q)(5 p-3 q)$
Calculating the cubes and the product:
$= 125 p^{3} - 27 q^{3} - 45 pq(5 p-3 q)$
Distributing $-45 pq$ into the parentheses:
$= 125 p^{3} - 27 q^{3} - 225 p^{2} q + 135 p q^{2}$

(A) We have $(104)^{3} = (100 + 4)^{3}$.
Using the algebraic identity $(x + y)^{3} = x^{3} + y^{3} + 3xy(x + y)$,where $x = 100$ and $y = 4$:
$(100 + 4)^{3} = (100)^{3} + (4)^{3} + 3(100)(4)(100 + 4)$
$= 1000000 + 64 + 1200(104)$
$= 1000000 + 64 + 124800$
$= 1124864$.

(B) We can express $999$ as $(1000 - 1)$.
Using the algebraic identity $(x - y)^{3} = x^{3} - y^{3} - 3xy(x - y)$,where $x = 1000$ and $y = 1$:
$(1000 - 1)^{3} = (1000)^{3} - (1)^{3} - 3(1000)(1)(1000 - 1)$
$= 1,000,000,000 - 1 - 3000(999)$
$= 1,000,000,000 - 1 - 2,997,000$
$= 1,000,000,000 - 2,997,001$
$= 997,002,999$

(A) The given expression is $8 x^{3}+27 y^{3}+36 x^{2} y+54 x y^{2}$.
We can rewrite this expression as:
$(2 x)^{3}+(3 y)^{3}+3(2 x)^{2}(3 y)+3(2 x)(3 y)^{2}$.
This is in the form of the algebraic identity $a^{3}+b^{3}+3 a^{2} b+3 a b^{2} = (a+b)^{3}$,where $a = 2 x$ and $b = 3 y$.
Substituting these values into the identity,we get:
$(2 x+3 y)^{3}$.
Thus,the factorised form is $(2 x+3 y)(2 x+3 y)(2 x+3 y)$.

(N/A) We use the algebraic identity: $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$.
Here,$a = 2x$,$b = y$,and $c = 3z$.
Substituting these values into the identity:
$8x^{3} + y^{3} + 27z^{3} - 18xyz = (2x)^{3} + (y)^{3} + (3z)^{3} - 3(2x)(y)(3z)$.
Applying the formula:
$= (2x + y + 3z)((2x)^{2} + (y)^{2} + (3z)^{2} - (2x)(y) - (y)(3z) - (2x)(3z))$.
Simplifying the terms inside the second bracket:
$= (2x + y + 3z)(4x^{2} + y^{2} + 9z^{2} - 2xy - 3yz - 6xz)$.

(A) To find the product of $(x+4)(x+10)$,we use the algebraic identity:
$(x+a)(x+b) = x^2 + (a+b)x + ab$
Here,$a = 4$ and $b = 10$.
Substituting these values into the identity:
$(x+4)(x+10) = x^2 + (4+10)x + (4 \times 10)$
Calculating the terms:
$= x^2 + 14x + 40$
Thus,the product is $x^2 + 14x + 40$.

(A) To find the product of $(x+8)(x-10)$,we use the algebraic identity: $(x+a)(x+b) = x^{2} + (a+b)x + ab$.
Here,$a = 8$ and $b = -10$.
Substituting these values into the identity:
$(x+8)(x-10) = x^{2} + (8 + (-10))x + (8 \times (-10))$
Simplifying the expression:
$= x^{2} + (-2)x + (-80)$
$= x^{2} - 2x - 80$.

(A) To find the product of $(3x + 4)(3x - 5)$,we use the algebraic identity: $(x + a)(x + b) = x^2 + (a + b)x + ab$.
Here,we let $x = 3x$,$a = 4$,and $b = -5$.
Substituting these values into the identity:
$(3x + 4)(3x - 5) = (3x)^2 + (4 + (-5))(3x) + (4 \times (-5))$
$= 9x^2 + (-1)(3x) - 20$
$= 9x^2 - 3x - 20$.

(A) The given expression is $\left(y^{2}+\frac{3}{2}\right)\left(y^{2}-\frac{3}{2}\right)$.
Using the algebraic identity $(a+b)(a-b) = a^{2}-b^{2}$,where $a = y^{2}$ and $b = \frac{3}{2}$,we get:
$\left(y^{2}+\frac{3}{2}\right)\left(y^{2}-\frac{3}{2}\right) = (y^{2})^{2} - \left(\frac{3}{2}\right)^{2}$
$= y^{4} - \frac{9}{4}$

(A) To find the product of $(3-2 x)(3+2 x)$,we use the algebraic identity: $(a-b)(a+b) = a^{2}-b^{2}$.
Here,$a = 3$ and $b = 2x$.
Substituting these values into the identity:
$(3-2 x)(3+2 x) = (3)^{2} - (2x)^{2}$.
Calculating the squares:
$(3)^{2} = 9$ and $(2x)^{2} = 4x^{2}$.
Therefore,the product is $9 - 4x^{2}$.

(B) We can express the product as: $103 \times 107 = (100 + 3)(100 + 7)$
Using the algebraic identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,where $x = 100$,$a = 3$,and $b = 7$:
$103 \times 107 = (100)^2 + (3 + 7) \times 100 + (3 \times 7)$
$= 10000 + (10 \times 100) + 21$
$= 10000 + 1000 + 21$
$= 11021$

(C) We can express the product as: $95 \times 96 = (100 - 5) \times (100 - 4)$.
Using the algebraic identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,where $x = 100$,$a = -5$,and $b = -4$:
$95 \times 96 = (100)^2 + [(-5) + (-4)] \times 100 + [(-5) \times (-4)]$.
$= 10000 + (-9) \times 100 + 20$.
$= 10000 - 900 + 20$.
$= 9100 + 20 = 9120$.

(D) We have $104 \times 96 = (100 + 4)(100 - 4)$.
Using the algebraic identity $(a + b)(a - b) = a^2 - b^2$,where $a = 100$ and $b = 4$:
$= (100)^2 - (4)^2$
$= 10000 - 16$
$= 9984$.

(A) We have the expression $9x^{2}+6xy+y^{2}$.
This can be rewritten as $(3x)^{2} + 2(3x)(y) + (y)^{2}$.
Using the algebraic identity $a^{2} + 2ab + b^{2} = (a+b)^{2}$,where $a = 3x$ and $b = y$,we get:
$(3x+y)^{2} = (3x+y)(3x+y)$.

(B) We have the expression $4y^{2}-4y+1$.
This can be rewritten as $(2y)^{2}-2(2y)(1)+(1)^{2}$.
Using the algebraic identity $a^{2}-2ab+b^{2}=(a-b)^{2}$,where $a=2y$ and $b=1$,we get:
$(2y-1)^{2}$.
Thus,the factors are $(2y-1)(2y-1)$.

(A) We have the expression $x^{2} - \frac{y^{2}}{100}$.
This can be rewritten as $x^{2} - (\frac{y}{10})^{2}$.
Using the algebraic identity $a^{2} - b^{2} = (a + b)(a - b)$,where $a = x$ and $b = \frac{y}{10}$,we get:
$x^{2} - (\frac{y}{10})^{2} = (x + \frac{y}{10})(x - \frac{y}{10})$.

(A) We use the algebraic identity: $(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$.
Comparing $(x + 2y + 4z)^2$ with the identity,we have $x = x$,$y = 2y$,and $z = 4z$.
Substituting these values into the identity:
$(x + 2y + 4z)^2 = (x)^2 + (2y)^2 + (4z)^2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)$.
Simplifying each term:
$= x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8zx$.

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

(N/A) To expand $(-2x + 3y + 2z)^2$,we use the algebraic identity: $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.
Here,$a = -2x$,$b = 3y$,and $c = 2z$.
Substituting these values into the identity:
$(-2x + 3y + 2z)^2 = (-2x)^2 + (3y)^2 + (2z)^2 + 2(-2x)(3y) + 2(3y)(2z) + 2(2z)(-2x)$
Calculating each term:
$(-2x)^2 = 4x^2$
$(3y)^2 = 9y^2$
$(2z)^2 = 4z^2$
$2(-2x)(3y) = -12xy$
$2(3y)(2z) = 12yz$
$2(2z)(-2x) = -8zx$
Combining these results,we get:
$4x^2 + 9y^2 + 4z^2 - 12xy + 12yz - 8zx$

(N/A) To expand $(3a - 7b - c)^{2}$,we use the algebraic identity: $(x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2zx$.
Here,$x = 3a$,$y = -7b$,and $z = -c$.
Substituting these values into the identity:
$(3a - 7b - c)^{2} = (3a)^{2} + (-7b)^{2} + (-c)^{2} + 2(3a)(-7b) + 2(-7b)(-c) + 2(-c)(3a)$
Calculating each term:
$= 9a^{2} + 49b^{2} + c^{2} + (-42ab) + (14bc) + (-6ca)$
Simplifying the expression:
$= 9a^{2} + 49b^{2} + c^{2} - 42ab + 14bc - 6ca$

(N/A) To expand $(-2x + 5y - 3z)^2$,we use the algebraic identity: $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.
Here,$a = -2x$,$b = 5y$,and $c = -3z$.
Substituting these values into the identity:
$(-2x + 5y - 3z)^2 = (-2x)^2 + (5y)^2 + (-3z)^2 + 2(-2x)(5y) + 2(5y)(-3z) + 2(-3z)(-2x)$
Calculating each term:
$(-2x)^2 = 4x^2$
$(5y)^2 = 25y^2$
$(-3z)^2 = 9z^2$
$2(-2x)(5y) = -20xy$
$2(5y)(-3z) = -30yz$
$2(-3z)(-2x) = 12zx$
Combining these results,we get:
$4x^2 + 25y^2 + 9z^2 - 20xy - 30yz + 12zx$

We use the algebraic identity $(x+y+z)^{2} = x^{2}+y^{2}+z^{2}+2xy+2yz+2zx$.
Here,$x = \frac{1}{4}a$,$y = -\frac{1}{2}b$,and $z = 1$.
Substituting these values into the identity:
$\left[\frac{1}{4} a-\frac{1}{2} b+1\right]^{2} = \left(\frac{1}{4} a\right)^{2} + \left(-\frac{1}{2} b\right)^{2} + (1)^{2} + 2\left(\frac{1}{4} a\right)\left(-\frac{1}{2} b\right) + 2\left(-\frac{1}{2} b\right)(1) + 2(1)\left(\frac{1}{4} a\right)$
$= \frac{1}{16} a^{2} + \frac{1}{4} b^{2} + 1 - \frac{1}{4} ab - b + \frac{1}{2} a$.

(A) The given expression is $4x^{2} + 9y^{2} + 16z^{2} + 12xy - 24yz - 16xz$.
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:
$4x^{2} = (2x)^{2}$
$9y^{2} = (3y)^{2}$
$16z^{2} = (-4z)^{2}$ (Since the terms $-24yz$ and $-16xz$ are negative,the $z$ term must be negative).
Now,rewrite the expression:
$= (2x)^{2} + (3y)^{2} + (-4z)^{2} + 2(2x)(3y) + 2(3y)(-4z) + 2(-4z)(2x)$
Using the identity,this simplifies to:
$= (2x + 3y - 4z)^{2}$
Thus,the factorised form is $(2x + 3y - 4z)(2x + 3y - 4z)$.

(A) The given expression is $2x^2 + y^2 + 8z^2 - 2\sqrt{2}xy + 4\sqrt{2}yz - 8xz$.
We use the algebraic identity: $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.
We can rewrite the expression as:
$(-\sqrt{2}x)^2 + (y)^2 + (2\sqrt{2}z)^2 + 2(-\sqrt{2}x)(y) + 2(y)(2\sqrt{2}z) + 2(2\sqrt{2}z)(-\sqrt{2}x)$
Here,$a = -\sqrt{2}x$,$b = y$,and $c = 2\sqrt{2}z$.
Thus,the expression becomes $(-\sqrt{2}x + y + 2\sqrt{2}z)^2$ or $(y - \sqrt{2}x + 2\sqrt{2}z)^2$.

(A) To expand $(2x + 1)^3$,we use the algebraic identity: $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$.
Here,$a = 2x$ and $b = 1$.
Substituting these values into the identity:
$(2x + 1)^3 = (2x)^3 + (1)^3 + 3(2x)(1)(2x + 1)$
Calculating the terms:
$= 8x^3 + 1 + 6x(2x + 1)$
Distributing $6x$ into the parentheses:
$= 8x^3 + 1 + 12x^2 + 6x$
Rearranging in descending order of powers:
$= 8x^3 + 12x^2 + 6x + 1$

(N/A) To expand $(2a - 3b)^{3}$,we use the algebraic identity: $(x - y)^{3} = x^{3} - y^{3} - 3xy(x - y)$.
Here,$x = 2a$ and $y = 3b$.
Substituting these values into the identity:
$(2a - 3b)^{3} = (2a)^{3} - (3b)^{3} - 3(2a)(3b)(2a - 3b)$
Calculating the cubes and the product:
$= 8a^{3} - 27b^{3} - 18ab(2a - 3b)$
Distributing $-18ab$ inside the parentheses:
$= 8a^{3} - 27b^{3} - (36a^{2}b - 54ab^{2})$
$= 8a^{3} - 27b^{3} - 36a^{2}b + 54ab^{2}$

To expand the expression $\left[\frac{3}{2} x+1\right]^{3}$,we use the algebraic identity $(a+b)^{3} = a^{3} + b^{3} + 3ab(a+b)$.
Here,$a = \frac{3}{2}x$ and $b = 1$.
Substituting these values into the identity:
$\left[\frac{3}{2} x+1\right]^{3} = \left(\frac{3}{2} x\right)^{3} + (1)^{3} + 3\left(\frac{3}{2} x\right)(1)\left(\frac{3}{2} x + 1\right)$
Calculating the powers and the product:
$= \frac{27}{8} x^{3} + 1 + \frac{9}{2} x \left(\frac{3}{2} x + 1\right)$
Distributing $\frac{9}{2}x$ inside the bracket:
$= \frac{27}{8} x^{3} + 1 + \left(\frac{9}{2} x \cdot \frac{3}{2} x\right) + \left(\frac{9}{2} x \cdot 1\right)$
$= \frac{27}{8} x^{3} + 1 + \frac{27}{4} x^{2} + \frac{9}{2} x$
Rearranging the terms in descending order of powers:
$= \frac{27}{8} x^{3} + \frac{27}{4} x^{2} + \frac{9}{2} x + 1$

(N/A) To expand $\left(x-\frac{2}{3} y\right)^{3}$,we use the algebraic identity: $(a-b)^{3} = a^{3} - b^{3} - 3ab(a-b)$.
Here,$a = x$ and $b = \frac{2}{3}y$.
Substituting these values into the identity:
$\left(x-\frac{2}{3} y\right)^{3} = x^{3} - \left(\frac{2}{3} y\right)^{3} - 3(x)\left(\frac{2}{3} y\right)\left(x-\frac{2}{3} y\right)$
$= x^{3} - \frac{8}{27} y^{3} - 2xy\left(x-\frac{2}{3} y\right)$
$= x^{3} - \frac{8}{27} y^{3} - 2x^{2}y + \left(2xy \cdot \frac{2}{3}y\right)$
$= x^{3} - \frac{8}{27} y^{3} - 2x^{2}y + \frac{4}{3}xy^{2}$

(D) We have $99 = 100 - 1$.
Using the algebraic identity $(a - b)^{3} = a^{3} - b^{3} - 3ab(a - b)$,where $a = 100$ and $b = 1$:
$(100 - 1)^{3} = 100^{3} - 1^{3} - 3(100)(1)(100 - 1)$
$= 1000000 - 1 - 300(99)$
$= 1000000 - 1 - 29700$
$= 1000000 - 29701$
$= 970299$