(a)

\begin{align} \text{Let } f(x) & = x^4 - 4x^3 + 3x^2 + x + 3 \\ \\ \text{By } & \text{Remainder theorem,} \\ f(-1) & = (-1)^4 - 4(-1)^3 + 3(-1)^2 + (-1) + 3 \\ & = 10 \\ \\ \therefore \text{Remainder} & = 10 \end{align}

(b)

\begin{align} \text{Let } f(x) & = x(x - 1)(1 - 2x)^2 + x^2 - 3 \\ \\ \text{By } & \text{Remainder theorem,} \\ f(2) & = (2)(2 - 1)[1 - 2(2)]^2 + (2)^2 - 3 \\ & = 19 \\ \\ \therefore \text{Remainder} & = 19 \end{align}

(c)

\begin{align} \text{Let } f(x) & = 3(x + 4)^2 - (1 - x)^3 \\ \\ \text{By } & \text{Remainder theorem,} \\ f(0) & = 3(0 + 4)^2 - (1 - 0)^3 \\ & = 47 \\ \\ \therefore \text{Remainder} & = 47 \end{align}

\begin{align} \text{Let } f(x) & = x^3 + 3x^2 - kx + 4 \\ \\ \text{By } & \text{Remainder theorem,} \\ f(2) & = k \\ (2)^3 + 3(2)^2 - k(2) + 4 & = k \\ 8 + 12 - 2k + 4 & = k \\ 24 - 2k & = k \\ 24 & = k + 2k \\ 24 & = 3k \\ {24 \over 3} & = k \\ 8 & = k \end{align}

\begin{align} \text{Let } f(x) & = x^4 + x^3 + 2ax^2 - 14a^4 \\ \\ \text{By } & \text{Remainder theorem,} \\ f(-2a) & = 32 \\ (-2a)^4 + (-2a)^3 + 2a(-2a)^2 - 14a^4 & = 32 \\ 16a^4 - 8a^3 + 8a^3 - 14a^4 & = 32 \\ 2a^4 & = 32 \\ a^4 & = {32 \over 2} \\ a^4 & = 16 \\ a & = \pm \sqrt[4]{16} \\ a & = \pm 2 \end{align}

(i)

\begin{align} f(x) & = ax^2 + bx - 6 \\ \\ \text{By } & \text{Remainder theorem,} \\ f(-3) & = 9 \\ a(-3)^2 + b(-3) - 6 & = 9 \\ 9a - 3b - 6 & = 9 \\ -3b & = 9 + 6 - 9a \\ -3b & = 15 - 9a \\ 3b & = 9a - 15 \\ b & = 3a - 5 \end{align}

(ii)

\begin{align} \text{Let } g(x) & = 2x^3 - bx^2 + 2ax - 4 \\ \\ \text{By } & \text{Remainder theorem,} \\ g(2) & = 2(2)^3 - b(2)^2 + 2a(2) - 4 \\ & = 16 - 4b + 4a - 4 \\ \\ \text{Since } b & = 3a - 5, \\ g(2) & = 16 - 4(3a - 5) + 4a - 4 \\ & = 16 - 12a + 20 + 4a - 4 \\ & = 32 - 8a \\ \\ \therefore \text{Remainder} & = 32 - 8a \end{align}

\begin{align} \text{Let } f(x) & = 8x^3 + ax^2 + bx - 9 \\ \\ \text{By } & \text{Remainder theorem,} \\ f(-2) & = -95 \\ 8(-2)^3 + a(-2)^2 + b(-2) - 9 & = -95 \\ -64 + 4a - 2b - 9 & = -95 \\ 4a - 2b & = -95 + 64 + 9 \\ 4a - 2b & = -22 \\ -2b & = -22 - 4a \\ 2b & = 22 + 4a \\ b & = 11 + 2a \phantom{000} \text{ --- (1)} \\ \\ \text{By } & \text{Remainder theorem,} \\ f\left(3 \over 2\right) & = 3 \\ 8\left(3 \over 2\right)^3 + a\left(3 \over 2\right)^2 + b\left(3 \over 2\right) - 9 & = 3 \\ 27 + {9 \over 4}a + {3 \over 2}b - 9 & = 3 \\ 108 + 9a + 6b - 36 & = 12 \\ 9a + 6b & = 12 - 108 + 36 \\ 9a + 6b & = -60 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 9a + 6(11 + 2a) & = -60 \\ 9a + 66 + 12a & = -60 \\ 21a & = -60 - 66 \\ 21a & = -126 \\ a & = {-126 \over 21} \\ & = -6 \\ \\ \text{Substitute } a & = -6 \text{ into (1),} \\ b & = 11 + 2(-6) \\ & = -1 \\ \\ \therefore a & = -6, b = -1 \end{align}

(i)

\begin{align} \text{Let } f(x) & = x^3 + ax^2 + bx - 3 \\ \\ \text{By } & \text{Remainder theorem,} \\ f(2) & = 27 \\ (2)^3 + a(2)^2 + b(2) - 3 & = 27 \\ 8 + 4a + 2b - 3 & = 27 \\ 4a + 2b & = 27 - 8 + 3 \\ 4a + 2b & = 22 \\ 2b & = 22 - 4a \\ b & = 11 - 2a \phantom{000} \text{ --- (1)} \\ \\ \text{By } & \text{Remainder theorem,} \\ f(-1) & = 3 \\ (-1)^3 + a(-1)^2 + b(-1) - 3 & = 3 \\ -1 + a - b - 3 & = 3 \\ a - b & = 3 + 1 + 3 \\ a - b & = 7 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ a - (11 - 2a) & = 7 \\ a - 11 + 2a & = 7 \\ 3a & = 7 + 11 \\ & = 18 \\ a & = {18 \over 3} \\ & = 6 \\ \\ \text{Substitute } a & = 6 \text{ into (1),} \\ b & = 11 - 2(6) \\ & = -1 \\ \\ \therefore a & = 6, b = - 1 \end{align}

(ii)

\begin{align} f(x) & = x^3 + ax^2 + bx - 3 \\ & = x^3 + 6x^2 - x - 3 \\ \\ \text{By } & \text{Remainder theorem,} \\ f(1) & = (1)^3 + 6(1)^2 - (1) - 3 \\ & = 3 \\ \\ \therefore \text{Remainder} & = 3 \end{align}

(i)

\begin{align} \text{Let } f(x) & = x^3 + ax^2 + 7 \\ \\ \text{By } & \text{Remainder theorem,} \\ f(-1) & = 2p \\ (-1)^3 + a(-1)^2 + 7 & = 2p \\ -1 + a + 7 & = 2p \\ a + 6 & = 2p \\ a & = 2p - 6 \phantom{000} \text{ --- (1)} \\ \\ \text{By } & \text{Remainder theorem,} \\ f(2) & = p + 5 \\ (2)^3 + a(2)^2 + 7 & = p + 5 \\ 8 + 4a + 7 & = p + 5 \\ 4a & = p + 5 - 8 - 7 \\ 4a & = p - 10 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 4(2p - 6) & = p - 10 \\ 8p - 24 & = p - 10 \\ 8p - p & = -10 + 24 \\ 7p & = 14 \\ p & = {14 \over 7} \\ & = 2 \\ \\ \text{Substitute } p & = 2 \text{ into (1),} \\ a & = 2(2) - 6 \\ & = -2 \\ \\ \therefore a & = -2, p = 2 \end{align}

(ii)

\begin{align} f(x) & = x^3 + ax^2 + 7 \\ & = x^3 - 2x^2 + 7 \\ \\ \text{By } & \text{Remainder theorem,} \\ f(-2) & = (-2)^3 - 2(-2)^2 + 7 \\ & = -9 \\ \\ \therefore \text{Remainder} & = -9 \end{align}

\begin{align} \text{Let } f(x) & = ax^3 + bx^2 + 2x + c \\ \\ \text{By } & \text{Remainder theorem,} \\ f(1) & = a(1)^3 + b(1)^2 + 2(1) + c \\ & = a + b + 2 + c \\ \\ \text{By } & \text{Remainder theorem,} \\ f(-1) & = a(-1)^3 + b(-1)^2 + 2(-1) + c \\ & = -a + b - 2 + c \\ \\ f(1) & = 2f(-1) \\ a + b + 2 + c & = 2(-a + b - 2 + c) \\ a + b + 2 + c & = -2a + 2b - 4 + 2c \\ a + 2a + b - 2b + 2 + 4 & = 2c - c \\ \\ 3a - b + 6 & = c \text{ (Shown)} \end{align}

\begin{align} \text{Let } f(x) & = 2x^2 + 6x + 3 \\ \\ \text{By } & \text{Remainder theorem,} \\ f(-p) & = 2(-p)^2 + 6(-p) + 3 \\ & = 2p^2 - 6p + 3 \\ \\ \text{By } & \text{Remainder theorem,} \\ f(2q) & = 2(2q)^2 + 6(2q) + 3 \\ & = 8q^2 + 12q + 3 \\ \\ f(-p) & = f(2q) \\ 2p^2 - 6p + 3 & = 8q^2 + 12q + 3 \\ 2p^2 - 6p & = 8q^2 + 12q \\ p^2 - 3p & = 4q^2 + 6q \\ p^2 - 4q^2 - 3p - 6q & = 0 \\ (p + 2q)(p - 2q) - 3(p + 2q) & = 0 \\ (p + 2q) [ (p - 2q) - 3 ] & = 0 \end{align} \begin{align} p + 2q & = 0 \phantom{00}&\text{or}\phantom{000} (p - 2q) - 3 & = 0 \\ p & = -2q \text{ (Reject)} & p - 2q & = 3 \end{align}

\begin{align} [f(x)][g(x)] & = (x^2 - 2x + p)(px^2 + x + 3) \\ \\ \text{Let } h(x) & = (x^2 - 2x + p)(px^2 + x + 3) \\ \\ \text{By } & \text{Remainder theorem,} \\ h(-1) & = 20 \\ [(-1)^2 - 2(-1) + p][p(-1)^2 + (-1) + 3] & = 20 \\ (1 + 2 + p)(p - 1 + 3) & = 20 \\ (3 + p)(p + 2) & = 20 \\ 3p + 6 + p^2 + 2p & = 20 \\ p^2 + 5p + 6 & = 20 \\ p^2 + 5p + 6 - 20 & = 0 \\ p^2 + 5p - 14 & = 0 \\ (p + 7)(p - 2) & = 0 \\ \\ p + 7 = 0 \phantom{000}&\text{or}\phantom{000} p - 2 = 0 \\ p = - 7 \phantom{0.}&\phantom{or000-2} p = 2 \end{align}

(i)

\begin{align} P(x) = 9x^3 - 3x^2 - 5x - 2 & = (x - 1)(3x + 2)Q(x) + ax + b \\ \\ P(1) = 9(1)^3 - 3(1)^2 - 5(1) - 2 & = (1 - 1)[3(1) + 2]Q(1) + a(1) + b \\ \\ P(1) = -1 & = (0)(5)Q(1) + a + b \\ \\ P(1) = -1 & = a + b \text{ (Shown)} \end{align}

(ii)

\begin{align} P(x) = 9x^3 - 3x^2 - 5x - 2 & = (x - 1)(3x + 2)Q(x) + ax + b \\ \\ P\left(-{2 \over 3}\right) = 9\left(-{2 \over 3}\right)^3 - 3\left(-{2 \over 3}\right)^2 - 5\left(-{2 \over 3}\right) - 2 & = (-{2 \over 3} - 1) \left[3\left(-{2 \over 3}\right) + 2\right]Q\left(-{2 \over 3}\right) + a\left(-{2 \over 3}\right) + b \\ \\ -{8 \over 3} & = \left(-{5 \over 3}\right)(0)Q\left(-{2 \over 3}\right) - {2 \over 3}a + b \\ \\ -{8 \over 3} & = 0 - {2 \over 3}a + b \\ \\ -{8 \over 3} & = -{2a \over 3} + b \text{ (Shown)} \end{align}

(iii)

\begin{align} 3x^2 - x - 2 & = (x - 1)(3x + 2) \end{align} \begin{align} -1 & = a + b \\ -1 - a & = b \phantom{000} \text{ --- (1)} \\ \\ -{8 \over 3} & = -{2a \over 3} + b \\ -8 & = -2a + 3b \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ -8 & = -2a + 3(-1 - a) \\ -8 & = -2a - 3 - 3a \\ -8 + 3 & = -2a - 3a \\ -5 & = -5a \\ {-5 \over -5} & = a \\ 1 & = a \\ \\ \text{Substitute } a & = 1 \text{ into (1),} \\ -1 - (1) & = b \\ -2 & = b \\ \\ \therefore P(x) & = (x - 1)(3x + 2)Q(x) + x - 2 \\ \\ \therefore \text{Remainder} & = x - 2 \end{align}

\begin{align} \text{When divided by } & x - 1, \text{ the remainder is } 4. \\ \\ \therefore f(x) & = (x - 1)Q_1(x) + 4 \\ \\ f(1) & = (1 - 1)Q_1(1) + 4 \\ & = 0 + 4 \\ & = 4 \\ \\ \\ \text{When divided by } & x + 2, \text{ the remainder is } -2. \\ \\ \therefore f(x) & = (x + 2)Q_2(x) - 2 \\ \\ f(-2) & = (-2 + 2)Q_2(-2) - 2 \\ & = 0 - 2 \\ & = -2 \\ \\ \\ \text{When divided by } & x^2 + x - 2 = (x - 1)(x + 2), \\ \\ \therefore f(x) & = (x - 1)(x + 2)Q_3(x) + ax + b \\ \\ f(1) & = (1 - 1)(1 + 2)Q_3(1) + a(1) + b \\ 4 & = (0)(3)Q_3(1) + a + b \\ 4 & = 0 + a + b \\ 4 & = a + b \phantom{000} \text{ --- (1)} \\ \\ f(-2) & = (-2 -1)(-2 + 2)Q_3(-2) + a(-2) + b \\ -2 & = (-3)(0)Q_3(-2) - 2a + b \\ -2 & = 0 - 2a + b \\ 2a - 2 & = b \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ \\ 4 & = a + (2a - 2) \\ 4 + 2 & = a + 2a \\ 6 & = 3a \\ {6 \over 3} & = a \\ 2 & = a \\ \\ \text{Substitute } a & = 2 \text{ into (2),} \\ \\ 2(2) - 2 & = b \\ 2 & = b \\ \\ \\ \therefore f(x) & = (x - 1)(x + 2)Q_3(x) + ax + b \\ & = (x - 1)(x + 2)Q_3(x) + 2x + 2 \\ \\ \text{Remainder} & = 2x + 2 \end{align}

\begin{align} \text{Disagree, as } f(x) \text{ is not a polynomial due to the term } x \sqrt{x} \end{align}