(a)

$$ \require{enclose} \begin{array}{rll} 3x + 4 \phantom{00000}\\ 3x - 2 \enclose{longdiv}{9x^2 + 6x + 2 \phantom{0}}\kern-.2ex \\ -\underline{(9x^2 - 6x){\phantom{000.}}} \\ 12x + 2\phantom{0} \\ -\underline{(12x - 8 ){\phantom{}}} \\ 10\phantom{0} \end{array} $$ $$ \text{Quotient is } 3x + 4, \text{ Remainder is } 10 $$

(b)

$$ \require{enclose} \begin{array}{rll} 2x^2 + 3x + 2 \phantom{000000} \\ 2x + 1 \enclose{longdiv}{ 4x^3 + 8x^2 + 7x - 5 \phantom{0}}\kern-.2ex \\ -\underline{( 4x^3 + 2x^2 ){\phantom{0000000.}}} \\ 6x^2 + 7x - 5 \phantom{0} \\ -\underline{ ( 6x^2 + 3x ){\phantom{.000}}} \\ 4x - 5\phantom{0} \\ -\underline{ ( 4x + 2 ){\phantom{}}} \\ -7 \phantom{.} \end{array} $$ $$ \text{Quotient is } 2x^2 + 3x + 2, \text{ Remainder is } - 7 $$

(c)

$$ \require{enclose} \begin{array}{rll} 5x^2 + 8x + 37 \phantom{0000000000} \\ x^2 - 2x - 3 \enclose{longdiv}{ 5x^4 - 2x^3 + 6x^2 + 4x - 3 \phantom{0}}\kern-.2ex \\ -\underline{( 5x^4 - 10x^3 - 15x^2 ){\phantom{000000}}} \\ 8x^3 + 21x^2 + 4x - 3 \phantom{0} \\ -\underline{ ( 8x^3 - 16x^2 - 24x ){\phantom{.00}}} \\ 37x^2 + 28x - 3 \phantom{00} \\ -\underline{ ( 37x^2 - 74x - 111 ){\phantom{}}} \\ 102x + 108 \phantom{.} \end{array} $$ $$ \text{Quotient is } 5x^2 + 8x + 37, \text{ Remainder is } 102x + 108 $$

(a)

$$ \require{enclose} \begin{array}{rll} 2x^2 + 1 \phantom{0000} \\ x - 2 \enclose{longdiv}{ 2x^3 - 4x^2 + x - 2 \phantom{0}}\kern-.2ex \\ -\underline{( 2x^3 - 4x^2 ){\phantom{000000.}}} \\ x - 2 \phantom{0.} \\ -\underline{ ( x - 2 ){\phantom{.}}} \\ 0 \phantom{.0} \end{array} $$ $$ 2x^3 - 4x^2 + x - 2 = (x - 2)(2x^2 + 1) $$

(b)

$$ \require{enclose} \begin{array}{rll} 3x^3 - 3x^2 + 2x + 6 \phantom{00000.} \\ x + 1 \enclose{longdiv}{ 3x^4 + 0x^3 - x^2 + 8x - 4 \phantom{0}}\kern-.2ex \\ -\underline{( 3x^4 + 3x^3 ){\phantom{000000000000}}} \\ -3x^3 - x^2 + 8x - 4\phantom{0.} \\ -\underline{ ( -3x^3 - 3x^2 ){\phantom{0000000}}} \\ 2x^2 + 8x - 4 \phantom{0.} \\ -\underline{ (2x^2 + 2x ) \phantom{000.} } \\ 6x - 4 \phantom{0.} \\ -\underline{ (6x + 6) \phantom{.} } \\ -10 \phantom{.} \end{array} $$ \begin{align} 3x^4 - x^2 + 8x - 4 & = (x + 1)(3x^3 - 3x^2 + 2x + 6) + (- 10) \\ & = (x + 1)(3x^3 - 3x^2 + 2x + 6) - 10 \end{align}

(c)

$$ \require{enclose} \begin{array}{rll} 4x^2 - 4x + 5.5 \phantom{000.} \\ 2x^2 + 2x - 1 \enclose{longdiv}{ 8x^4 + 0x^3 - x^2 + 0x + 5 \phantom{0}}\kern-.2ex \\ -\underline{( 8x^4 + 8x^3 - 4x^2 ){\phantom{0000000}}} \\ -8x^3 + 3x^2 + 0x + 5 \phantom{.} \\ -\underline{ ( -8x^3 - 8x^2 + 4x ){\phantom{000}}} \\ 11x^2 - 4x + 5 \phantom{00.} \\ -\underline{ ( 11x^2 + 11x - 5.5 ) \phantom{} } \\ -15x + 10.5 \phantom{0.} \end{array} $$ \begin{align} 8x^4 - x^2 + 5 & = (2x^2 + 2x - 1)(4x^2 - 4x + 5.5) + (- 15x + 10.5) \\ & = (2x^2 + 2x - 1)\left(4x^2 - 4x + {11 \over 2}\right) - 15x + {21 \over 2} \end{align}

(a) Degree of quotient is 3 and degree of remainder is 0

$$ \require{enclose} \begin{array}{rll} x^3 - 2x + 2 \phantom{000000.} \\ x + 2 \enclose{longdiv}{ x^4 + 2x^3 - 2x^2 - 2x + 4 \phantom{0}}\kern-.2ex \\ -\underline{( x^4 + 2x^3 ){\phantom{0000000000000}}} \\ -2x^2 - 2x + 4 \phantom{0.} \\ -\underline{ ( -2x^2 - 4x ){\phantom{0000}}} \\ 2x + 4 \phantom{00.} \\ -\underline{ ( 2x + 4 ) \phantom{0.} } \\ 0 \phantom{00} \end{array} $$

(b) Degree of quotient is 2 and degree of remainder is 1

$$ \require{enclose} \begin{array}{rll} x^2 + x - 6 \phantom{000000.} \\ x^2 + x + 3 \enclose{longdiv}{ x^4 + 2x^3 - 2x^2 - 2x + 4 \phantom{0}}\kern-.2ex \\ -\underline{( x^4 + x^3 + 3x^2 ){\phantom{00000000.}}} \\ x^3 - 5x^2 - 2x + 4 \phantom{00} \\ -\underline{ ( x^3 + x^2 + 3x ){\phantom{00000}}} \\ -6x^2 - 5x + 4 \phantom{00} \\ -\underline{ ( -6x^2 - 6x - 18 ) \phantom{0.} } \\ x + 22 \phantom{00} \end{array} $$

(c) Degree of quotient is 2 and degree of remainder is 0

$$ \require{enclose} \begin{array}{rll} x^2 + 2x - 1 \phantom{000000.} \\ x^2 - 1 \enclose{longdiv}{ x^4 + 2x^3 - 2x^2 - 2x + 4 \phantom{0}}\kern-.2ex \\ -\underline{( x^4 \phantom{+2x^30} - x^2 ){\phantom{00000000.}}} \\ 2x^3 - x^2 - 2x + 4 \phantom{00} \\ -\underline{ ( 2x^3 \phantom{ - x^20} - 2x ){\phantom{0000}}} \\ - x^2 \phantom{-2x} + 4 \phantom{00} \\ -\underline{ ( -x^2 \phantom{-2x} + 1 ) \phantom{0} } \\ 3 \phantom{00} \end{array} $$

(d) Degree of quotient is 1 and degree of remainder is 1

$$ \require{enclose} \begin{array}{rll} x + 3 \phantom{000000000000000.} \\ x^3 - x^2 + x - 1 \enclose{longdiv}{ x^4 + 2x^3 - 2x^2 - 2x + 4 \phantom{0}}\kern-.2ex \\ -\underline{( x^4 - \phantom{0}x^3 + \phantom{0} x^2 - \phantom{0} x ){\phantom{000.}}} \\ 3x^3 - 3x^2 - x + 4 \phantom{00} \\ -\underline{ ( 3x^3 - 3x^2 + 3x - 3 ){\phantom{}}} \\ - 4x + 7 \phantom{0.} \end{array} $$

(a)

\begin{align} (2x^2 - 1)(3x + 2) & = 6x^3 + 4x^2 - 3x - 2 \end{align} $$ \require{enclose} \begin{array}{rll} 6x^2 + 16x + 29 \phantom{0000.} \\ x - 2 \enclose{longdiv}{ 6x^3 + 4x^2 - 3x - 2 \phantom{0}}\kern-.2ex \\ -\underline{( 6x^3 - 12x^2 ){\phantom{000000.}}} \\ 16x^2 - 3x - 2 \phantom{0} \\ -\underline{ ( 16x^2 - 32x ){\phantom{00.}}} \\ 29x - 2 \phantom{0} \\ -\underline{ (29x - 58) } \\ 56 \phantom{0} \end{array} $$ $$ (2x^2 - 1)(3x + 2) = (x - 2)(6x^2 + 16x + 29) + 56 $$

(b)

\begin{align} (x + 1)(x - 1) & = x^2 - 1 \end{align} $$ \require{enclose} \begin{array}{rll} 3x - 1 \phantom{0000.} \\ x^2 - 1 \enclose{longdiv}{ 3x^3 - x^2 + 2x + 3 \phantom{0}}\kern-.2ex \\ -\underline{( 3x^3 \phantom{-x^20} - 3x ){\phantom{000.}}} \\ -x^2 + 5x + 3 \phantom{0} \\ -\underline{ ( -x^2 \phantom{ + 5x 0} + 1 ){\phantom{}}} \\ 5x + 2 \phantom{.} \end{array} $$ $$ 3x^3 - x^2 + 2x + 3 = (x + 1)(x - 1)(3x - 1) + 5x + 2 $$

(c)

\begin{align} (x + 1)(2x - 3) & = 2x^2 - 3x + 2x - 3 \\ & = 2x^2 - x - 3 \end{align} $$ \require{enclose} \begin{array}{rll} 5x^2 - x + 5 \phantom{00000000.} \\ 2x^2 - x - 3 \enclose{longdiv}{ 10x^4 - 7x^3 - 4x^2 + 8x + 7 \phantom{0}}\kern-.2ex \\ -\underline{( 10x^4 - 5x^3 - 15x^2 ){\phantom{000000.}}} \\ -2x^3 + 11x^2 + 8x + 7 \phantom{0} \\ -\underline{ ( -2x^3 + \phantom{11} x^2 + 3x ){\phantom{000.}}} \\ 10x^2 + 5x + 7 \phantom{0.} \\ -\underline{ ( 10x^2 - 5x - 15 ) } \\ 10x + 22 \phantom{.} \end{array} $$ $$ 10x^4 - 7x^3 - 4x^2 + 8x + 7 = (x + 1)(2x - 3)(5x^2 - x + 5) + 10x + 22 $$

(a)

$$ \require{enclose} \begin{array}{rll} 4x^2 - 11x + 6 \phantom{0} \\ 2x + 1 \enclose{longdiv}{ 8x^3 - 18x^2 + x + 6 \phantom{0}}\kern-.2ex \\ -\underline{( 8x^3 + 4x^2 ){\phantom{0000000.}}} \\ -22x^2 + x + 6 \phantom{0} \\ -\underline{ ( -22x^2 - 11x ){\phantom{0.}}} \\ 12x + 6 \phantom{0} \\ -\underline{ ( 12x + 6 ) } \\ 0 \phantom{0} \end{array} $$ $$ \therefore {8x^3 - 18x^2 + x + 6 \over 2x + 1} = 4x^2 - 11x + 6 $$ $$ \text{Yes, the result is a polynomial}$$

(b)

\begin{align} (x - 2)(3x - 2) & = 3x^2 - 2x - 6x + 4 \\ & = 3x^2 - 8x + 4 \end{align} $$ \require{enclose} \begin{array}{rll} 2x^2 - 3x - 9 \phantom{00000000000000.} \\ 3x^2 - 8x + 4 \enclose{longdiv}{ 6x^4 - 25x^3 + 5x^2 + 60x - 36 \phantom{0}}\kern-.2ex \\ -\underline{( 6x^4 - 16x^3 + 8x^2 ){\phantom{000000000.}}} \\ -9x^3 - 3x^2 + 60x - 36 \phantom{0} \\ -\underline{ ( -9x^3 + 24x^2 - 12x ){\phantom{000.}}} \\ -27x^2 + 72x - 36 \phantom{0} \\ -\underline{ ( -27x^2 + 72x - 36 ) } \\ 0 \phantom{0} \end{array} $$ $$ \therefore {6x^4 - 25x^3 + 5x^2 + 60x - 36 \over (x - 2)(3x - 2)} = 2x^2 - 3x - 9 $$ $$ \text{Yes, the result is a polynomial}$$

(c)

$$ 3 - 2x = -2x + 3 $$ $$ \require{enclose} \begin{array}{rll} -3x^3 - 3x^2 + 9.5x + 14.75 \phantom{00.} \\ -2x + 3 \enclose{longdiv}{ 6x^4 - 3x^3 - 28x^2 - x - 10 \phantom{0}}\kern-.2ex \\ -\underline{( 6x^4 - 9x^3 ){\phantom{00000000000000}}} \\ 6x^3 - 28x^2 - x - 10 \phantom{0} \\ -\underline{ ( 6x^3 - 9x^2 ){\phantom{00000000}}} \\ -19x^2 - x - 10 \phantom{0} \\ -\underline{ ( -19x^2 + 28.5x ) \phantom{00} } \\ -29.5x - 10\phantom{0} \\ -\underline{ (-29.5x + 44.25)} \\ -54.25 \end{array} $$ $$ \therefore {6x^4 - 3x^3 - 28x^2 - x - 10 \over 3 - 2x} = - 3x^3 - 3x^2 + 9.5x + 14.75 - {54.25 \over 3 - 2x} $$ $$ \text{No, the result is not a polynomial, due to the term } {54.25 \over 3 - 2x}$$

$$ \require{enclose} \begin{array}{rll} 2x^3 + 2x^2 + 5x + 10 \phantom{00.} \\ x^2 - x - 2 \enclose{longdiv}{ 2x^5 + 0x^4 - x^3 + x^2 + 0x - 4 \phantom{0}}\kern-.2ex \\ -\underline{( 2x^5 - 2x^4 - 4x^3) \phantom{00000000000} } \\ 2x^4 + 3x^3 + x^2 + 0x - 4 \phantom{.} \\ -\underline{ ( 2x^4 - 2x^3 - 4x^2 ) \phantom{000000} } \\ 5x^3 + 5x^2 + 0x - 4 \\ -\underline{ ( 5x^3 - 5x^2 - 10x ) \phantom{0.} } \\ 10x^2 + 10x - 4 \\ -\underline{ (10x^2 - 10x - 20 )} \\ 20x + 16 \end{array} $$
\begin{align} \text{Since the remainder } & \text{is } 20x + 16, \\ a = 20, & \phantom{0} b = 16 \end{align}

(i)

\begin{align} x^4 - 2x^3 + ax^2 - 2 & = x(x - 1)(x + 1)Q(x) + 4x^2 + bx + c \\ \\ & = x(x^2 - 1)Q(x) + 4x^2 + bx + c \\ \\ & = (x^3 - x)Q(x) + 4x^2 + bx + c \\ \\ \text{Since the degree of the } & \text{polynomial is } 4, \text{ the degree of } Q(x) \text{ is 1.} \end{align}

(ii)

\begin{align} x^4 - 2x^3 + ax^2 - 2 & = (x^3 - x)Q(x) + 4x^2 + bx + c \\ \\ \text{Let } & x = 0, \\ (0)^4 - 2(0)^3 + a(0)^2 - 2 & = (0^3 - 0)Q(0) + 4(0)^2 + b(0) + c \\ 0 - 0 + 0 - 2 & = 0 + 0 + 0 + c \\ -2 & = c \\ \\ x^4 - 2x^3 + ax^2 - 2 & = (x^3 - x)Q(x) + 4x^2 + bx - 2 \\ \\ \text{Let } & x = 1, \\ (1)^4 - 2(1)^3 + a(1)^2 - 2 & = (1^3 - 1)Q(1) + 4(1)^2 + b(1) - 2 \\ 1 - 2 + a - 2 & = 0 + 4 + b - 2 \\ -3 + a & = 2 + b \\ a & = 5 + b \phantom{000} \text{ --- (1)} \\ \\ \text{Let } & x = -1, \\ (-1)^4 - 2(-1)^3 + a(-1)^2 - 2 & = [(-1)^3 - (-1)]Q(-1) + 4(-1)^2 + b(-1) - 2 \\ 1 + 2 + a - 2 & = 0 + 4 - b - 2 \\ a + 1 & = 2 - b \\ a & = 2 - 1 - b \\ a & = 1 - b \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 5 + b & = 1 - b \\ b + b & = 1 - 5 \\ 2b & = -4 \\ b & = {-4 \over 2} \\ & = -2 \\ \\ \text{Substitute } & b = -2 \text{ into (1),} \\ a & = 5 + (-2) \\ & = 3 \\ \\ \therefore a & = 3, b = -2, c = -2 \\ \\ \\ \therefore \text{Remainder} & = 4x^2 + bx + c \\ & = 4x^2 + (-2)x + (-2) \\ & = 4x^2 - 2x - 2 \end{align}

(i)

\begin{align} 4x^3 - 6x^2 + 1 & = (x - 2)Q(x) + ax - 3 \\ \\ \text{Let } & x = 2, \\ 4(2)^3 - 6(2)^2 + 1 & = (2 - 2)Q(2) + a(2) - 3 \\ 9 & = 0 + 2a - 3 \\ 9 + 3 & = 2a \\ 12 & = 2a \\ {12 \over 2} & = a \\ 6 & = a \end{align}

(ii)

\begin{align} 4x^3 - 6x^2 + 1 & = (x - 2)Q(x) + ax - 3 \\ & = (x - 2)Q(x) + 6x - 3 \\ \\ \text{Since the degree of the } & \text{polynomial is 3, the degree of } Q(x) \text{ is 2}. \\ \\ \text{Let } Q(x) & = bx^2 + cx + d \\ \\ 4x^3 - 6x^2 + 1 & = (x - 2)(bx^2 + cx + d) + 6x - 3 \\ & = bx^3 + cx^2 + dx - 2bx^2 - 2cx - 2d + 6x - 3 \\ & = bx^3 + cx^2 - 2bx^2 + dx - 2cx + 6x - 2d - 3 \\ & = bx^3 + (c - 2b)x^2 + (d - 2c + 6)x + (-2d - 3) \\ \\ \text{Comparing} & \text{ coefficients of } x^3, \\ 4 & = b \\ \\ \text{Comparing} & \text{ coefficients of } x^2, \\ -6 & = c - 2b \\ -6 & = c - 2(4) \\ -6 & = c - 8 \\ -6 + 8 & = c \\ 2 & = c \\ \\ \text{Comparing} & \text{ the constants,} \\ 1 & = - 2d - 3 \\ 1 + 3 & = -2d \\ 4 & = -2d \\ {4 \over -2} & = d \\ -2 & = d \\ \\ \therefore Q(x) & = bx^2 + cx + d \\ & = (4)x^2 + (2)x + (-2) \\ & = 4x^2 + 2x - 2 \end{align}

(i)(a)

$$ \require{enclose} \begin{array}{rll} 3x^2 + x + 6 \phantom{00.} \\ x - 2 \enclose{longdiv}{ 3x^3 - 5x^2 + 4x - 9 \phantom{0}}\kern-.2ex \\ -\underline{ ( 3x^3 - 6x^2 ) \phantom{0000000.} } \\ x^2 + 4x - 9 \phantom{0.} \\ -\underline{ ( x^2 - 2x ) \phantom{0000} } \\ 6x - 9 \phantom{0.} \\ -\underline{ ( 6x - 12) } \\ 3 \end{array} $$ $$ \therefore {P(x) \over x - 2} = 3x^2 + x + 6 + {3 \over x - 2} $$
$$ \require{enclose} \begin{array}{rll} 2x^3 + 8x^2 + 16x + 26 \phantom{00.} \\ x - 2 \enclose{longdiv}{ 2x^4 + 4x^3 + 0x^2 - 6x - 5 \phantom{0}}\kern-.2ex \\ -\underline{ ( 2x^4 - 4x^3 ) \phantom{0000000000000} } \\ 8x^3 + 0x^2 - 6x - 5 \phantom{0.} \\ -\underline{ ( 8x^3 - 16x^2 ) \phantom{0000000} } \\ 16x^2 - 6x - 5 \phantom{0.} \\ -\underline{ ( 16x^2 - 32x ) \phantom{000} } \\ 26x - 5 \phantom{0.} \\ -\underline{ ( 26x - 52) } \\ 47 \phantom{0} \end{array} $$ $$ \therefore {Q(x) \over x - 2} = 2x^3 + 8x^2 + 16x + 26 + {47 \over x - 2}$$

(i)(b)

\begin{align} {P(x) + Q(x) \over x - 2} & = {P(x) \over x - 2} + {Q(x) \over x - 2} \\ \text{From } & \text{part (a),} \\ \text{Remainder} & = 3 + 47 \\ & = 50 \\ \\ \\ {2P(x) - Q(x) \over x - 2} & = {2P(x) \over x - 2} - {Q(x) \over x - 2} \\ & = 2\left({P(x) \over x - 2} \right) - {Q(x) \over x - 2} \\ \\ \text{From } & \text{part (a),} \\ \text{Remainder} & = 2(3) - 47 \\ & = -41 \end{align}

(ii)(a)

$$ \require{enclose} \begin{array}{rll} 3x - 2 \phantom{00.} \\ x^2 - x - 2 \enclose{longdiv}{ 3x^3 - 5x^2 + 4x - 9 \phantom{0}}\kern-.2ex \\ -\underline{ ( 3x^3 - 3x^2 - 6x ) \phantom{000.} } \\ -2x^2 + 10x - 9 \phantom{0.} \\ -\underline{ ( -2x^2 + 2x + 4) \phantom{0.} } \\ 8x - 13 \phantom{0.} \end{array} $$ $$ \therefore {P(x) \over x^2 - x - 2} = 3x - 2 + {8x - 13 \over x^2 - x - 2} $$
$$ \require{enclose} \begin{array}{rll} 2x^2 + 6x + 10 \phantom{00.} \\ x^2 - x - 2 \enclose{longdiv}{ 2x^4 + 4x^3 + 0x^2 - 6x - 5 \phantom{0}}\kern-.2ex \\ -\underline{ ( 2x^4 - 2x^3 - 4x^2 ) \phantom{00000000} } \\ 6x^3 + 4x^2 - 6x - 5 \phantom{0.} \\ -\underline{ ( 6x^3 - 6x^2 - 12x ) \phantom{000} } \\ 10x^2 + 6x - 5 \phantom{0.} \\ -\underline{ ( 10x^2 - 10x - 20 ) \phantom{} } \\ 16x + 15 \phantom{0.} \end{array} $$ $$ \therefore {Q(x) \over x^2 - x - 2} = 2x^2 + 6x + 10 + {16x + 15 \over x^2 - x - 2} $$

(ii)(b)

\begin{align} {Q(x) - P(x) \over x^2 - x - 2} & = {Q(x) \over x^2 - x - 2} - {P(x) \over x^2 - x - 2} \\ \\ \text{From } & \text{part (a),} \\ \text{Remainder} & = (16x + 15) - (8x - 13) \\ & = 16x + 15 - 8x + 13 \\ & = 8x + 28 \\ \\ \\ {3P(x) + Q(x) \over x^2 - x - 2} & = {3P(x) \over x^2 - x - 2} + {Q(x) \over x^2 - x - 2} \\ & = 3 \left({P(x) \over x^2 - x - 2}\right) + {Q(x) \over x^2 - x - 2} \\ \\ \text{From } & \text{part (a),} \\ \text{Remainder} & = 3(8x - 13) + (16x + 15) \\ & = 24x - 39 + 16x + 15 \\ & = 40x - 24 \end{align}

(i)

$$ \require{enclose} \begin{array}{rll} 3x^2 + x - 3 \phantom{0000000000.} \\ x^2 + 1 \enclose{longdiv}{ 3x^4 + x^3 + 0x^2 + 6x + 2 \phantom{0}}\kern-.2ex \\ -\underline{ ( 3x^4 \phantom{+ x^3 0} + 3x^2 ) \phantom{0000000.} } \\ x^3 - 3x^2 + 6x + 2 \phantom{0.} \\ -\underline{ ( x^3 \phantom{ - 3x^2 0} + x ) \phantom{0000.} } \\ -3x^2 + 5x + 2 \phantom{0.} \\ -\underline{ (-3x^2 \phantom{+ 5x} - 3) } \phantom{0} \\ 5x + 5 \end{array} $$ \begin{align} \text{Capacity} & = (3x^2 + x - 3) \text{ cm}^3 \\ \\ \text{Amount left} & = (5x + 5) \text{ cm}^3 \end{align}

(ii)

\begin{align} \text{When } & x = 7, \\ \text{Total amount of drink} & = 3(7)^4 + (7)^3 + 6(7) + 2 \\ & = 7590 \text{ cm}^3 \\ \\ \text{No. of bottles} & = (7)^2 + 1 \\ & = 50 \\ \\ \text{Capacity of each bottle} & = 3(7)^2 + (7) - 3 \\ & = 151 \text{ cm}^3 \\ \\ \text{Amount left} & = 5(7) + 5 \\ & = 40 \text{ cm}^3 \end{align}

$$ \require{enclose} \begin{array}{rll} 3x - 4y \phantom{0000000.} \\ 2x + y \enclose{longdiv}{ 6x^2 - 5xy - 4y^2 \phantom{0}}\kern-.2ex \\ -\underline{ ( 6x^2 + 3xy ) \phantom{00000.} } \\ -8xy - 4y^2 \phantom{0.} \\ -\underline{ ( -8xy - 4y^2 ) \phantom{.} } \\ 0 \phantom{0.} \end{array} $$

$$ \require{enclose} \begin{array}{rll} 3x^2 - 5yx + 7y^2 \phantom{00000.} \\ x + 2y \enclose{longdiv}{ 3x^3 + yx^2 - 3y^2x - y^3 \phantom{0}}\kern-.2ex \\ -\underline{ ( 3x^3 + 6yx^2 ) \phantom{000000000} } \\ -5yx^2 - 3y^2x - y^3 \phantom{0.} \\ -\underline{ ( -5yx^2 - 10y^2x ) \phantom{0000} } \\ 7y^2x - y^3 \phantom{00} \\ -\underline{ (7y^2x + 14y^3) } \\ -15y^3 \phantom{0} \end{array} $$ $$ \text{Quotient} = 3x^2 - 5yx + 7y^2, \text{ Remainder} = -15y^3 $$

\begin{align} 3x^3 - ax^2 - bx + 72 & = (x^2 + 5x + c)(dx + e) \\ & = dx^3 + ex^2 + 5dx^2 + 5ex + cdx + ce \\ & = dx^3 + (e + 5d)x^2 + (5e + cd)x + ce \\ \\ \text{Comparing } & \text{coefficients of } x^3, \\ 3 & = d \\ \\ 3x^3 - ax^2 - bx + 72 & = 3x^3 + [e + 5(3)]x^2 + [5e + c(3)]x + ce \\ 3x^3 - ax^2 - bx + 72 & = 3x^3 + (e + 15)x^2 + (5e + 3c)x + ce \\ \\ \text{Comparing } & \text{coefficients of } x^2, \\ -a & = e + 15 \\ a & = -e - 15 \\ \\ \text{Comparing } & \text{coefficients of } x, \\ -b & = 5e + 3c \\ b & = -5e - 3c \\ \\ \text{Comparing } & \text{constants,} \\ 72 & = ce \\ \\ \text{Let } & e = 1, \\ 72 & = c(1) \\ 72 & = c \\ \\ a & = -(1) - 15 \\ a & = -16 \\ \\ b & = -5(1) - 3(72) \\ & = -221 \end{align}