(a)

Yes, the exponent of $x$ is a non-negative integer and $-1$ is a constant term.

(b)

Yes, the exponents of $x$ are non-negative integers.

(c)

No, $\sqrt{x} = x^{1 \over 2}$ has a fractional exponent.

(d)

No, ${3 \over x^2} = 3x^{-2}$ has a negative exponent.

(a)

\begin{align} Q(x) - P(x) & = (2x^2 - 3x + 2) - (x^2 + x + 1) \\ \\ & = 2x^2 - 3x + 2 - x^2 - x - 1 \\ \\ & = 2x^2 - x^2 - 3x - x + 2 - 1 \\ \\ & = x^2 - 4x + 1 \end{align}

(b)

\begin{align} P(x) + 2Q(x) & = (x^2 + x + 1) + 2(2x^2 - 3x + 2) \\ \\ & = x^2 + x + 1 + 4x^2 - 6x + 4 \\ \\ & = x^2 + 4x^2 + x - 6x + 1 + 4 \\ \\ & = 5x^2 - 5x + 5 \end{align}

(a)

\begin{align} (7x - 3)(2x^2 + 4x - 1) & = 14x^3 + 28x^2 - 7x - 6x^2 - 12x + 3 \\ \\ & = 14x^3 + 28x^2 - 6x^2 - 7x - 12x + 3 \\ \\ & = 14x^3 + 22x^2 - 19x + 3 \end{align}

(b)

\begin{align} (5x^2 + 2x - 4)(3x^2 + x - 2) & = 15x^4 + 5x^3 - 10x^2 + 6x^3 + 2x^2 - 4x - 12x^2 - 4x + 8 \\ \\ & = 15x^4 + 5x^3 + 6x^3 - 10x^2 + 2x^2 - 12x^2 - 4x - 4x + 8 \\ \\ & = 15x^4 + 11x^3 - 20x^2 - 8x + 8 \end{align}

(a)

\begin{align} (2x^2 - x + 1)(3x - 2) & = ... - 4x^2 - 3x^2 + ... \\ \\ & = - 7x^2 + ... \\ \\ \therefore \text{Coefficient of } x^2 & = -7 \end{align}

(b)

\begin{align} (x^2 + 3x + 2)(8x^2 - 5x - 4) & = ... - 4x^2 + ... - 15x^2 + ... + 16x^2 + ... \\ \\ & = -3x^2 + ... \\ \\ \therefore \text{Coefficient of } x^2 & = -3 \end{align}

(c)

\begin{align} (2x^2 - 2x + 5)(-x^2 - 3x + 1) & = ... + 2x^2 + ... + 6x^2 + ... - 5x^2 + ... \\ \\ & = 3x^2 + ... \\ \\ \therefore \text{Coefficient of } x^2 & = 3 \end{align}

(d)

\begin{align} (4x^3 - x^2 + 7x - 2)(2x^3 + 3x^2 + 6) & = ... - 6x^2 + ... -6x^2 + ... \\ \\ & = -12x^2 \\ \\ \therefore \text{Coefficient of } x^2 & = -12 \end{align}

(a)

\begin{align} a(x - 2) + b & = 5 - 3x \\ ax - 2a + b & = 5 - 3x \\ (b - 2a) + ax & = 5 - 3x \\ \\ \text{Comparing } & \text{coefficients of }x, \\ a & = -3 \\ \\ \text{Comparing } & \text{the constant term,} \\ b - 2a & = 5 \\ b & = 5 + 2a \\ & = 5 + 2(-3) \\ & = -1 \\ \therefore a & = -3, b = -1 \end{align}

(b)

\begin{align} a(x - 1) + b(x + 3) & = 3x + 1 \\ ax - a + bx + 3b & = 3x + 1 \\ ax + bx + 3b - a & = 3x + 1 \\ (a + b)x + (3b - a) & = 3x + 1 \\ \\ \text{Comparing } & \text{coefficients of }x, \\ a + b & = 3 \\ a & = 3 - b \phantom{000} \text{ --- (1)} \\ \\ \text{Comparing } & \text{the constant term,} \\ 3b - a & = 1 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 3b - (3 - b) & = 1 \\ 3b - 3 + b & = 1 \\ 3b + b & = 1 + 3 \\ 4b & = 4 \\ b & = {4 \over 4} \\ & = 1 \\ \\ \text{Substitute } & b = 1 \text{ into (1),} \\ a & = 3 - 1 \\ & = 2 \\ \\ \therefore a & = 2, b = 1 \end{align}

(c)

\begin{align} a(x - 2) + b(x - 4) & = x + 2 \\ ax - 2a + bx - 4b & = x + 2 \\ ax + bx - 2a - 4b & = x + 2 \\ (a + b)x + (-2a - 4b) & = x + 2 \\ \\ \text{Comparing } & \text{coefficients of }x, \\ a + b & = 1 \\ a & = 1 - b \phantom{000} \text{ --- (1)} \\ \\ \text{Comparing } & \text{the constant term,} \\ \\ -2a - 4b & = 2 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ -2(1 - b) - 4b & = 2 \\ -2 + 2b - 4b & = 2 \\ 2b - 4b & = 2 + 2 \\ -2b & = 4 \\ b & = {4 \over -2} \\ & = -2 \\ \\ \text{Substitute } & b = -2 \text{ into (1),} \\ a & = 1 - (-2) \\ & = 3 \\ \\ \therefore a & = 3, b = -2 \end{align}

(d)

\begin{align} ax^3 - x + 3 & = 4x^3 + bx^2 + c(x - 1) + d \\ ax^3 - x + 3 & = 4x^3 + bx^2 + cx - c + d \\ ax^3 + 0x^2 - x + 3 & = 4x^3 + bx^2 + cx + (d - c) \\ \\ \text{Comparing } & \text{coefficients of }x^3, \\ a & = 4 \\ \\ \text{Comparing } & \text{coefficients of }x^2, \\ 0 & = b \\ \\ \text{Comparing } & \text{coefficients of }x, \\ -1 & = c \\ \\ \text{Comparing } & \text{the constant term,} \\ 3 & = d - c \\ 3 + c & = d \\ 3 + (-1) & = d \\ 2 & = d \end{align}

(e)

\begin{align} x^3 - 6x^2 + 14x - 8 & = (x - 2)^3 + ax \\ \\ \text{Let } & x = 2, \\ (2)^3 - 6(2)^2 + 14(2) - 8 & = (2 - 2)^3 + a(2) \\ 4 & = 0 + 2a \\ 4 & = 2a \\ {4 \over 2} & = a \\ 2 & = a \end{align}

(f)

\begin{align} -2x^2 - 7x + 3 & = a(x - 2)^2 + b(x + 1)^3 + x^3 \\ \\ \text{Let } & x = 2, \\ -2(2)^2 - 7(2) + 3 & = a(2 - 2)^2 + b(2 + 1)^3 + (2)^3 \\ -19 & = 0 + 27b + 8 \\ -19 - 8 & = 27b \\ -27 & = 27b \\ {-27 \over 27} & = b \\ -1 & = b \\ \\ \\ -2x^2 - 7x + 3 & = a(x - 2)^2 - (x + 1)^3 + x^3 \\ \\ \text{Let } & x = -1, \\ -2(-1)^2 - 7(-1) + 3 & = a(-1 - 2)^2 - (-1 + 1)^3 + (-1)^3 \\ 8 & = 9a - 0 - 1 \\ 8 + 1 & = 9a \\ 9 & = 9a \\ {9 \over 9} & = a \\ 1 & = a \\ \\ \therefore a & = 1, b = -1 \end{align}

(i)

\begin{align} A(x) & = {1 \over 2} \times \text{Base} \times \text{Height} \\ \\ & = {1 \over 2}(5x + 10)(3x^2 + 2) \\ \\ & = {1 \over 2}(15x^3 + 10x + 30x^2 + 20) \\ \\ & = {15 \over 2}x^3 + 5x + 15x^2 + 10 \\ \\ & = {15 \over 2}x^3 + 15x^2 + 5x + 10 \\ \\ \text{Yes, } & \text{it is a polynomial as the exponents of } x \text{ are non-negative integers.} \end{align}

(ii)

\begin{align} \text{Let } H(x) & \text{ denote the hypotenuse of the triangle.} \\ \\ \text{By Py} & \text{thgoras theorem,} \\ \\ [H(x)]^2 & = (5x + 10)^2 + (3x^2 + 2)^2 \\ \\ & = (5x)^2 + 2(5x)(10) + (10)^2 + (3x^2)^2 + 2(3x^2)(2) + (2)^2 \\ \\ & = 25x^2 + 100x + 100 + 9x^4 + 12x^2 + 4 \\ \\ & = 9x^4 + 37x^2 + 100x + 104 \\ \\ H(x) & = \sqrt{9x^4 + 37x^2 + 100x + 104} \\ \\ \\ P(x) & = (5x + 10) + (3x^2 + 2) + \sqrt{9x^4 + 37x^2 + 100x + 104} \\ \\ & = 3x^2 + 5x + 12 + \sqrt{9x^4 + 37x^2 + 100x + 104} \\ \\ & = 3x^2 + 5x + 12 + (9x^4 + 37x^2 + 100x + 104)^{1 \over 2} \\ \\ \\ \text{No, } & \text{ it is not a polynomial as the exponent of } (9x^4 + 37x^2 + 100x + 104)^{1 \over 2} \text{ is fractional}. \end{align}

(a)

\begin{align} M(x) & = P(x)Q(x) \\ \\ & = (x + 1)(x^2 - x + 1) \\ \\ & = x^3 - x^2 + x + x^2 - x + 1 \\ \\ & = x^3 + 1 \\ \\ \therefore \text{Coefficient of } x^2 & = 0 \\ \\ \text{Degree of } M(x) & = 3 \end{align}

(b)

\begin{align} 2P(x) - [Q(x)]^2 & = 2(x + 1) - (x^2 - x + 1)^2 \\ \\ & = 2x + 2 - (x^2 - x + 1)(x^2 - x + 1) \\ \\ & = 2x + 2 - (x^4 - x^3 + x^2 - x^3 + x^2 - x + x^2 - x + 1) \\ \\ & = 2x + 2 - (x^4 - 2x^3 + 3x^2 - 2x + 1) \\ \\ & = 2x + 2 - x^4 + 2x^3 - 3x^2 + 2x - 1 \\ \\ & = -x^4 + 2x^3 - 3x^2 + 4x + 1 \\ \\ \therefore \text{Coefficient of } x^2 & = -3 \\ \\ \text{Degree of } 2P(x) - [Q(x)]^2 & = 4 \end{align}

(c)

\begin{align} Q(x) [3x^2 + P(x)] & = (x^2 - x + 1)(3x^2 + x + 1) \\ \\ & = 3x^4 + x^3 + x^2 - 3x^3 - x^2 - x + 3x^2 + x + 1 \\ \\ & = 3x^4 - 2x^3 + 3x^2 + 1 \\ \\ \therefore \text{Coefficient of } x^2 & = 3 \\ \\ \text{Degree of } Q(x)[3x^2 + P(x)] & = 4 \end{align}

(i)

\begin{align} V(x) & = \text{Length} \times \text{Breadth} \times \text{Height} \\ & = (2x + 3)(3x - 1)(2x + 5) \\ & = (6x^2 - 2x + 9x - 3)(2x + 5) \\ & = (6x^2 + 7x - 3)(2x + 5) \\ & = 12x^3 + 30x^2 + 14x^2 + 35x - 6x - 15 \\ & = 12x^3 + 44x^2 + 29x - 15 \end{align}

(ii)

\begin{align} A(x) & = 2(2x + 3)(3x - 1) + 2(2x + 3)(2x + 5) + 2(3x - 1)(2x + 5) \\ & = 2(6x^2 - 2x + 9x - 3) + 2(4x^2 + 10x + 6x + 15) + 2(6x^2 + 15x - 2x - 5) \\ & = 2(6x^2 + 7x - 3) + 2(4x^2 + 16x + 15) + 2(6x^2 + 13x - 5) \\ & = 12x^2 + 14x - 6 + 8x^2 + 32x + 30 + 12x^2 + 26x - 10 \\ & = 32x^2 + 72x + 14 \end{align}

(a)

\begin{align} x^3 - 6x^2 - x + c & = (x - 3)(ax^2 - 3x + b) \\ \\ \text{Let } & x = 3, \\ (3)^3 - 6(3)^2 - (3) + c & = (3 - 3)[a(3)^2 - 3(3) + b] \\ -30 + c & = 0 \\ c & = 30 \\ \\ x^3 - 6x^2 - x + 30 & = (x - 3)(ax^2 - 3x + b) \\ \\ \text{Let } & x = 0, \\ (0)^3 - 6(0)^2 - (0) + 30 & = (0 - 3)[a(0)^2 - 3(0) + b] \\ 30 & = (-3)(b) \\ 30 & = -3b \\ {30 \over -3} & = b \\ -10 & = b \\ \\ x^3 - 6x^2 - x + 30 & = (x - 3)(ax^2 - 3x - 10) \\ \\ \text{Let } & x = 1, \\ (1)^3 - 6(1)^2 - (1) + 30 & = (1 - 3)[a(1)^2 - 3(1) - 10] \\ 24 & = (-2)(a - 13) \\ 24 & = -2a + 26 \\ 24 - 26 & = -2a \\ -2 & = -2a \\ {-2 \over -2} & = a \\ 1 & = a \\ \\ a & = 1, b = -10, c = 30 \end{align}

(b)

\begin{align} x^3 + cx^2 + x + 6 & = (x + 1)(x - 2)(ax + b) \\ \\ \text{Let } & x = 2, \\ (2)^3 + c(2)^2 + (2) + 6 & = (2 + 1)(2 - 2)[a(2) +b] \\ 8 + 4c + 2 + 6 & = 0 \\ 4c + 16 & = 0 \\ 4c & = -16 \\ c & = {-16 \over 4} \\ & = -4 \\ \\ x^3 - 4x^2 + x + 6 & = (x + 1)(x - 2)(ax + b) \\ \\ \text{Let } & x = 0, \\ (0)^3 - 4(0)^2 + (0) + 6 & = (0 + 1)(0 - 2)[a(0) + b] \\ 6 & = (1)(-2)(b) \\ 6 & = -2b \\ {6 \over -2} & = b \\ -3 & = b \\ \\ x^3 - 4x^2 + x + 6 & = (x + 1)(x - 2)(ax - 3) \\ \\ \text{Let } & x = 1, \\ (1)^3 - 4(1)^2 + (1) + 6 & = (1 + 1)(1 - 2)[a(1) - 3] \\ 4 & = (2)(-1)(a - 3) \\ 4 & = -2(a - 3) \\ 4 & = -2a + 6 \\ 4 - 6 & = -2a \\ -2 & = -2a \\ {-2 \over -2} & = a \\ 1 & = a \\ \\ a & = 1, b = -3, c = -4 \end{align}

(c)

\begin{align} 3x^2 - 5x + 4 & = a(x - 2)^2 + b(x - 2) + c \\ \\ \text{Let } & x = 2, \\ 3(2)^2 - 5(2) + 4 & = a(2 - 2)^2 + b(2 - 2) + c \\ 6 & = 0 + 0 + c \\ 6 & = c \\ \\ 3x^2 - 5x + 4 & = a(x - 2)^2 + b(x - 2) + 6 \\ \\ \text{Let } & x = 1, \\ 3(1)^2 - 5(1) + 4 & = a(1 - 2)^2 + b(1 - 2) + 6 \\ 2 & = a(1) + b(-1) + 6 \\ 2 & = a - b + 6 \\ 2 - 6 & = a - b \\ b - 4 & = a \phantom{000} \text{ --- (1)} \\ \\ \text{Let } & x = 0, \\ 3(0)^2 - 5(0) + 4 & = a(0 - 2)^2 + b(0 - 2) + 6 \\ 4 & = a(4) + b(-2) + 6 \\ 4 & = 4a - 2b + 6 \\ 4 - 6 & = 4a - 2b \\ -2 & = 4a - 2b \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ -2 & = 4(b - 4) - 2b \\ -2 & = 4b - 16 - 2b \\ -2 + 16 & = 2b \\ 14 & = 2b \\ {14 \over 2} & = b \\ 7 & = b \\ \\ \text{Substitute } b & = 7 \text{ into (1),} \\ a & = (7) - 4 \\ & = 3 \\ \\ \therefore a & = 3, b = 7, c = 6 \end{align}

(d)

\begin{align} x^3 + 3x^2 - 2x + 16 - c(x + 2) & = ax^2(x - 1) + b(x - 2)^2(x - 1) \\ \\ \text{Let } & x = 1, \\ (1)^3 + 3(1)^2 - 2(1) + 16 - c(1 + 2) & = a(1)^2(1 - 1) + b(1 - 2)^2(1 - 1) \\ 18 - 3c & = 0 + 0 \\ -3c & = -18 \\ c & = {-18 \over -3} \\ & = 6 \\ \\ x^3 + 3x^2 - 2x + 16 - 6(x + 2) & = ax^2(x - 1) + b(x - 2)^2(x - 1) \\ \\ \text{Let } & x = 0, \\ (0)^3 + 3(0)^2 - 2(0) + 16 - 6(0 + 2) & = a(0)^2(0 - 1) + b(0 - 2)^2(0 - 1) \\ 16 - 12 & = 0 + b(4)(-1) \\ 4 & = -4b \\ {4 \over -4} & = b \\ -1 & = b \\ \\ x^3 + 3x^2 - 2x + 16 - 6(x + 2) & = ax^2 (x - 1) - (x - 2)^2(x - 1) \\ \\ \text{Let } & x = 2, \\ (2)^3 + 3(2)^2 - 2(2) + 16 - 6(2 + 2) & = a(2)^2(2 - 1) - (2 - 2)^2 (2 - 1) \\ 8 & = a(4)(1) - 0 \\ 8 & = 4a \\ {8 \over 4} & = a \\ 2 & = a \\ \\ a & = 2, b = -1, c = 6 \end{align}

\begin{align} x^3yz^2 + (x - z)^2 y + (2x - \sqrt{y})^2 + y^2z^4 & = x^3yz^2 + (x^2 - 2xz + z^2)y + (4x^2 - 4x\sqrt{y} + y) + y^2z^4 \\ \\ & = x^3yz^2 + x^2y - 2xyz + yz^2 + 4x^2 - 4x\sqrt{y} + y + y^2z^4 \\ \\ & = x^3yz^2 + x^2y - 2xyz + yz^2 + 4x^2 - 4xy^{1 \over 2} + y + y^2 z^4 \end{align}

(i) Yes, it is a polynomial in $x$ as the exponents of $x$ are non-negative integers. Since the highest power of $x$ is $3$, degree of the polynomial is $3$
(ii) No, it is not a polynomial in $y$ as the exponent of $y^{1 \over 2}$ is fractional
(iii) Yes, it is a polynomial in $z$ as the exponents of $z$ are non-negative integers. Since the highest power of $z$ is $4$, degree of the polynomial is $4$

(i)

\begin{align} B(t) & = bt(t - 2)^2 + 2t^2 + ct + q \\ \\ & = bt(t^2 - 4t + 4) + 2t^2 + ct + q \\ \\ & = bt^3 - 4bt^2 + 4bt + 2t^2 + ct + q \\ \\ & = bt^3 + 2t^2 - 4bt^2 + 4bt + ct + q \\ \\ & = bt^3 + (2 - 4b)t^2 + (4b + c)t + q \end{align}
Yes $B(t)$ is a polynomial in $t$ as the exponents of $t$ are non-negative integers.

(ii)

\begin{align} \text{When } t = 0 & \text{ and } A(t) = 300, \\ 300 & = - (0)^3 + a(0)^2 + p \\ 300 & = 0 + 0 + p \\ 300 & = p \\ \\ \text{When } t = 0 & \text{ and } B(t) = 0, \\ 0 & = b(0)(0 - 2)^2 + 2(0)^2 + c(0) + q \\ 0 & = 0 + 0 + 0 + q \\ 0 & = q \\ \\ p & = 300, q = 0 \end{align}

(iii)

Since the liquid is poured from container $A$ to $B$, the volume of liquid in $B$ at any moment is equals to $300 \text{ cm}^3$ (starting volume of $A$) minus the volume of liquid remaining in $A$ at the same moment.

(iv)

\begin{align} 300 - A(t) & = B(t) \\ \\ 300 -(-t^3 + at^2 + p) & = bt(t - 2)^2 + 2t^2 + ct + q \\ 300 + t^3 - at^2 - p & = bt(t - 2)^2 + 2t^2 + ct + q \\ 300 + t^3 - at^2 - 300 & = bt(t - 2)^2 + 2t^2 + ct \phantom{00000} [p = 300, q = 0] \\ t^3 - at^2 & = bt(t - 2)^2 + 2t^2 + ct \\ \\ t^3 - at^2 & = bt(t^2 - 4t + 4) + 2t^2 + ct \\ t^3 - at^2 & = bt^3 - 4bt^2 + 4bt + 2t^2 + ct \\ t^3 - at^2 & = bt^3 + 2t^2 - 4bt^2 + 4bt + ct \\ t^3 - at^2 & = bt^3 + (2 - 4b)t^2 + (4b + c)t \\ \\ \text{Comparing } & \text{coefficients of } t^3, \\ 1 & = b \\ \\ \text{Comparing } & \text{coefficients of } t^2, \\ -a & = 2 - 4b \\ -a & = 2 - 4(1) \\ -a & = -2 \\ a & = 2 \\ \\ \text{Comparing } & \text{coefficients of } t, \\ 0 & = 4b + c \\ 0 & = 4(1) + c \\ 0 & = 4 + c \\ -4 & = c \\ \\ a & = 2, b = 1, c = -4 \end{align}

(i)(a)

\begin{align} (x - a)(x - b) & = x^2 - bx - ax + ab \\ & = x^2 - a x - bx + ab \\ & = x^2 - (a + b)x + ab \end{align}

(i)(b)

\begin{align} (x - a)(x - b)(x - c) & = (x^2 - bx - ax + ab)(x - c) \\ & = x^3 - c x^2 - b x^2 + bcx - ax^2 + acx + abx - abc \\ & = x^3 - a x^2 - b x^2 - c x^2 + abx + ac x + bcx - abc \\ & = x^3 - (a + b + c)x^2 + (ab + ac + bc)x - abc \end{align}

(ii)

\begin{align} (x - a)(x - b)...(x - z) & = (x - a)(x - b)... (x - x)...(x - z) \\ & = (x - a)(x - b)... (0) ... (x - z) \\ & = 0 \end{align}

You can try to obtain the graphs from Desmos

\begin{align} 3x^3 - 2x^2 + x - 4 & = A(x - 1) + B(x - 1)(x + 1) + Cx(x^2 - 1) + D \\ 3x^3 - 2x^2 + x - 4 & = 3(x - 1) - 2(x - 1)(x + 1) + 5x(x^2 - 1) - 2 \\ \\ \\ \text{Graph 1: } y & = 3x^3 - 2x^2 + x - 4 \\ \\ \text{Graph 2: } y & = 3(x - 1) - 2(x - 1)(x + 1) + 5x(x^2 - 1) - 2 \end{align}

\begin{align} \text{Since graphs 1 and 2 } & \text{have different shape, James' answer is wrong} \\ \\ \\ 3x^3 - 2x^2 + x - 4 & = A(x - 1) + B(x - 1)(x + 1) + Cx(x^2 - 1) + D \\ & = Ax - A + B(x^2 - 1) + Cx^3 - Cx + D \\ & = Ax - A + Bx^2 - B + Cx^3 - Cx + D \\ & = Cx^3 + Bx^2 + Ax - Cx - A - B + D \\ \\ \text{Comparing } & \text{coefficients of } x^3, \\ 3 & = C \\ \\ \text{Comparing } & \text{coefficients of } x^2, \\ -2 & = B \\ \\ \text{Comparing } & \text{coefficients of } x, \\ 1 & = A - C \\ 1 & = A - 3 \\ 4 & = A \\ \\ \text{Comparing } & \text{constants,} \\ -4 & = - A - B + D \\ -4 & = - 4 - (-2) + D \\ -4 & = - 4 + 2 + D \\ -4 & = -2 + D \\ -2 & = D \\ \\ \therefore A & = 4, B = -2, C = 3, D = -2 \end{align}