(i)

\begin{align} \text{By }& \text{Remainder theorem,} \\ g(2) & = (2)^3 + a(2)^2 + (2) + 5 \\ 31 & = 8 + 4a + 2 + 5 \\ 31 & = 4a + 15 \\ 31 - 15 & = 4a \\ 16 & = 4a \\ {16 \over 4} & = a \\ 4 & = a \end{align}

(ii)

\begin{align} g(x) & = x^3 + ax^2 + x + 5 \\ & = x^3 + 4x^2 + x + 5 \\ \\ \text{Since } g(x) & = x(x - 1)(x - b) + cx + 5, \\ x^3 + 4x^2 + x + 5 & = x(x - 1)(x - b) + cx + 5 \\ \\ \text{Let } & x = 1, \\ (1)^3 + 4(1)^2 + (1) + 5 & = (1)(1 - 1)(1 - b) + c(1) + 5 \\ 11 & = (1)(0)(1 - b) + c + 5 \\ 11 & = c + 5 \\ 11 - 5 & = c \\ 6 & = c \\ \\ \text{Let } & x = 2, \\ (2)^3 + 4(2)^2 + (2) + 5 & = (2)(2 - 1)(2 - b) + (6)(2) + 5 \\ 31 & = (2)(1)(2 - b) + 12 + 5 \\ 31 & = 2(2 - b) + 17 \\ 31 - 17 & = 2(2 - b) \\ 14 & = 4 - 2b \\ 2b & = 4 - 14 \\ & = -10 \\ b & = {-10 \over 2} \\ & = -5 \\ \\ b & = -5, c = 6 \end{align}

(i)

\begin{align} f(-1) & = (-1)^3 + (k - 2)(-1)^2 + (k - 7)(-1) - 4 \\ & = -1 + (k - 2)(1) + (-k + 7) - 4 \\ & = -1 + k - 2 - k + 7 - 4 \\ & = k - k - 1 - 2 + 7 - 4 \\ & = 0 \\ \\ \text{By Factor} & \text{ theorem, } x + 1 \text{ is a factor of } f(x). \end{align}

(ii)

\begin{align} \text{By } & \text{Factor theorem,} \\ f(-2) & = (-2)^3 + (k - 2)(-2)^4 + (k - 7)(-2) - 4 \\ 0 & = -8 + (k - 2)(4) + (-2k + 14) - 4 \\ 0 & = -8 + 4k - 8 - 2k + 14 - 4 \\ 8 + 8 - 14 + 4 & = 4k - 2k \\ 6 & = 2k \\ {6 \over 2} & = k \\ 3 & = k \\ \\ \\ f(x) & = x^3 + (3 - 2)x^2 + (3 - 7)x - 4 \\ & = x^3 + x^2 - 4x - 4 \\ \\ (x + 1)(x + 2) & = x^2 + 2x + x + 2 \\ & = x^2 + 3x + 2 \\ \\ \text{Since } (x + 1) \text{ and } (x + 2) \text{ are} & \text{ factors of } f(x), \phantom{.} x^2 + 3x + 2 \text{ is a factor of } f(x). \end{align} $$ \require{enclose} \begin{array}{rll} x - 2 \phantom{0000000000}\\ x^2 + 3x + 2 \enclose{longdiv}{ x^3 + x^2 - 4x - 4 \phantom{0}}\kern-.2ex \\ -\underline{( x^3 + 3x^2 + 2x ){\phantom{000}}} \\ -2x^2 - 6x - 4 \phantom{0} \\ -\underline{( -2x^2 - 6x - 4 ){\phantom{}}} \\ 0\phantom{0} \end{array} $$ \begin{align} \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ x^3 + x^2 - 4x - 4 & = (x^2 + 3x + 2)(x - 2) + 0 \\ & = (x^2 + 3x + 2)(x - 2) \\ \\ \therefore \text{Third factor is } & x - 2 \end{align}

(i)

\begin{align} \text{Since } a^3 - b^3 & = (a - b)(a^2 + ab + b^2), \\ (x + 3)^3 - x^3 & = [(x + 3) - (x)][ (x + 3)^2 + (x + 3)(x) + (x)^2] \\ & = (x + 3 - x)[ x^2 + 2(x)(3) + 3^2 + x^2 + 3x + x^2 ] \\ & = (3)(x^2 + 6x + 9 + x^2 + 3x + x^2) \\ & = 3(3x^2 + 9x + 9) \\ & = 3(3)(x^2 + 3x + 3) \\ & = 9(x^2 + 3x + 3) \end{align}

(ii)

\begin{align} \text{Let } x \text{ denote the length} & \text{ of the cube with the shorter side}. \\ \\ \text{Volume of cube} & = (\text{Length})^3 \\ \\ \text{Difference in volume} & = (x + 3)^3 - (x)^3 \\ 189 & = (x + 3)^3 - x^3 \\ \\ 189 & = 9(x^2 + 3x + 3) \phantom{00000} [\text{Apply result from part (i)}] \\ {189 \over 9} & = x^2 + 3x + 3 \\ 21 & = x^2 + 3x + 3 \\ 0 & = x^2 + 3x + 3 - 21 \\ 0 & = x^2 + 3x - 18 \\ 0 & = (x + 6)(x - 3) \\ \\ x - 3 = 0 \phantom{000}&\text{or}\phantom{000} x + 6 = 0 \\ x = 3 \phantom{000}&\phantom{or000+6} x = -6 \text{ (Reject)} \\ \\ \therefore \text{Length of cube with shorter side} & = 3 \text{ cm} \\ \\ \text{Length of cube with longer side} & = x + 3 \\ & = 3 + 3 \\ & = 6 \text{ cm} \end{align}

(a)

\begin{align} {7x - 3 \over 1 - x^2} & = {7x - 3 \over (1 + x)(1 - x)} \\ & = {A \over 1 + x} + {B \over 1 - x} \\ & = {A(1 - x) \over (1 + x)(1 - x)} + {B(1 + x) \over (1 + x)(1 - x)} \\ & = {A(1 - x) + B(1 + x) \over (1 + x)(1 - x)} \\ \\ \text{Comparing } & \text{the numerator,} \\ 7x - 3 & = A(1 - x) + B(1 + x) \\ \\ \text{Let } & x = 1, \\ 7(1) - 3 & = A(1 - 1) + B(1 + 1) \\ 4 & = A(0) + B(2) \\ 4 & = 2B \\ {4 \over 2} & = B \\ 2 & = B \\ \\ \text{Let } & x = -1, \\ 7(-1) - 3 & = A[1 - (-1)] + (2)[1 + (-1)] \\ -10 & = A(2) + (2)(0) \\ -10 & = 2A \\ {-10 \over 2} & = A \\ -5 & = A \\ \\ \therefore {7x - 3 \over 1 - x^2} & = {A \over 1 + x} + {B \over 1 - x} \\ & = {-5 \over 1 + x} + {2 \over 1 - x} \\ & = -{5 \over 1 + x} + {2 \over 1 - x} \\ & = {2 \over 1 - x} - {5 \over 1 + x} \end{align}

(b)

\begin{align} {10x - 5 \over (x + 1)(x^2 + 4)} & = {A \over x + 1} + {Bx + C \over x^2 + 4} \\ & = {A(x^2 + 4) \over (x + 1)(x^2 + 4)} + {(Bx + C)(x + 1) \over (x + 1)(x^2 + 4)} \\ & = {A(x^2 + 4) + (Bx + C)(x + 1) \over (x + 1)(x^2 + 4)} \\ \\ \text{Comparing } & \text{the numerator,} \\ 10x - 5 & = A(x^2 + 4) + (Bx + C)(x + 1) \\ \\ \text{Let } & x = -1, \\ 10(-1) - 5 & = A[(-1)^2 + 4] + [B(-1) + C](-1 + 1) \\ -15 & = A(5) + (-B + C)(0) \\ -15 & = 5A \\ {-15 \over 5} & = A \\ -3 & = A \\ \\ \text{Let } & x = 0, \\ 10(0) - 5 & = (-3)(0^2 + 4) + [B(0) + C](0 + 1) \\ -5 & = -3(4) + (C)(1) \\ -5 & = -12 + C \\ -5 + 12 & = C \\ 7 & = C \\ \\ \text{Let } & x = 1, \\ 10(1) - 5 & = (-3)(1^2 + 4) + [B(1) + 7](1 + 1) \\ 5 & = -3(5) + (B + 7)(2) \\ 5 & = -15 + 2B + 14 \\ 5 & = -1 + 2B \\ 5 + 1 & = 2B \\ 6 & = 2B \\ {6 \over 2} & = B \\ 3 & = B \\ \\ \therefore {10x - 5 \over (x + 1)(x^2 + 4)} & = {A \over x + 1} + {Bx + C \over x^2 + 4} \\ & = {-3 \over x + 1} + {(3)x + (7) \over x^2 + 4} \\ & = -{3 \over x + 1} + {3x + 7 \over x^2 + 4} \end{align}

(c)

\begin{align} (x - 4)^2 & = x^2 - 2(x)(4) + 4^2 \\ & = x^2 - 8x + 16 \end{align} $$ \require{enclose} \begin{array}{rll} x + 8\phantom{0000000000000}\\ x^2 - 8x + 16 \enclose{longdiv}{ x^3 + 0x^2 + 0x + 0 \phantom{000}}\kern-.2ex \\ -\underline{( x^3 - 8x^2 + 16x ){\phantom{0000.}}} \\ 8x^2 - 16x + 0 \phantom{00.} \\ -\underline{( 8x^2 - 64x + 128 ){\phantom{}}} \\ 48x - 128 \phantom{0} \\ \end{array} $$ \begin{align} \text{Since } {P(x) \over D(x)} & = Q(x) + {R(x) \over D(x)}, \\ \\ {x^3 \over (x - 4)^2} & = x + 8 + {48x - 128 \over (x - 4)^2} \\ \\ {48x - 128 \over (x - 4)^2} & = {A \over x - 4} + {B \over (x - 4)^2} \\ & = {A(x - 4) \over (x - 4)^2} + {B \over (x - 4)^2} \\ & = {A(x - 4) + B \over (x - 4)^2} \\ \\ \text{Comparing } & \text{the numerator,} \\ 48x - 128 & = A(x - 4) + B \\ \\ \text{Let } & x = 4, \\ 48(4) - 128 & = A(4 - 4) + B \\ 64 & = A(0) + B \\ 64 & = B \\ \\ \text{Let } & x = 0, \\ 48(0) - 128 & = A(0 - 4) + (64) \\ -128 & = A(-4) + 64 \\ -128 & = -4A + 64 \\ -128 - 64 & = -4A \\ -192 & = -4A \\ {-192 \over -4} & = A \\ 48 & = A \\ \\ {48x - 128 \over (x - 4)^2} & = {A \over x - 4} + {B \over (x - 4)^2} \\ & = {48 \over x - 4} + {64 \over (x - 4)^2} \\ \\ \therefore {x^3 \over (x - 4)^2} & = x + 8 + {48 \over x - 4} + {64 \over (x - 4)^2} \end{align}

(i)

\begin{align} \text{Let } f(x) & = x^3 + 3x^2 - 6x - 8 \\ \\ f(2) & = (2)^3 + 3(2)^2 - 6(2) - 8 \\ & = 0 \\ \\ \text{By Factor} & \text{ theorem, } x - 2 \text{ is a factor of } f(x). \end{align} $$ \require{enclose} \begin{array}{rll} x^2 + 5x + 4 \phantom{00000}\\ x - 2 \enclose{longdiv}{ x^3 + 3x^2 - 6x - 8 \phantom{} }\kern-.2ex \\ -\underline{( x^3 - 2x^2 ){\phantom{0000000}}} \\ 5x^2 - 6x - 8 \phantom{.} \\ -\underline{( 5x^2 - 10x ){\phantom{00} }} \\ 4x - 8 \phantom{.} \\ -\underline{( 4x - 8 ){\phantom{} }} \\ 0 \phantom{.} \\ \end{array} $$ \begin{align} \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ x^3 + 3x^2 - 6x - 8 & = (x - 2)(x^2 + 5x + 4) + 0 \\ f(x) & = (x - 2)(x^2 + 5x + 4) \\ & = (x - 2)(x + 1)(x + 4) \\ \\ \therefore 0 & = (x - 2)(x + 1)(x + 4) \\ \\ x - 2 = 0 \phantom{000}\text{or}\phantom{000} x + 1 & = 0 \phantom{000}\text{or}\phantom{000} x + 4 = 0 \\ x = 2 \phantom{000or000+1} x & = -1 \phantom{000or--4} x = - 4 \end{align}

(ii)(a)

\begin{align} \text{By } & \text{Remainder theorem,} \\ f(a) & = 22(a)^3 + 15(a)^2 - 4(a) - 6 \\ & = 22a^3 + 15a^2 - 4a - 6 \\ \\ \text{By } & \text{Remainder theorem,} \\ g(a) & = 2 + 14(a) - 12(a)^2 - 5(a)^3 \\ & = 2 + 14a - 12a^2 - 5a^3 \end{align} \begin{align} 22a^3 + 15a^2 - 4a - 6 & = 2 + 14a - 12a^2 - 5a^3 \\ 22a^3 + 5a^3 + 15a^2 + 12a^2 - 4a - 14a - 6 - 2 & = 0 \\ 27a^3 + 27a^2 - 18a - 8 & = 0 \text{ (Shown)} \end{align}

(ii)(b)

\begin{align} 27a^3 + 27a^2 - 18a - 8 & = 0 \\ (3a)^3 + 3(9a^2) - 6(3a) - 8 & = 0 \\ (3a)^3 + 3(3a)^2 - 6(3a) - 8 & = 0 \\ \\ \text{Let } & y = 3a, \\ (y)^3 + 3(y)^2 - 6(y) - 8 & = 0 \\ y^3 + 3y^2 - 6y - 8 & = 0 \\ \\ \text{Using solutions} & \text{ from part (i),} \\ y & = 2, -1, -4 \\ \\ \text{Since } y & = 3a, \\ 3a & = 2, - 1, -4 \\ a & = {2 \over 3}, -{1 \over 3}, -{4 \over 3} \end{align}

(a)(i)

\begin{align} \text{By } & \text{Remainder theorem,} \\ f(4) & = 4(4)^3 - 16(4)^2 - 9(4) + 40 \\ & = 4 \\ \\ \therefore \text{Remainder} & = 4 \end{align}

(a)(ii)

\begin{align} f(x) & = x \\ 4x^3 - 16x^2 - 9x + 40 & = x \\ 4x^3 - 16x^2 - 10x + 40 & = 0 \end{align} \begin{align} \text{Let } g(x) & = 4x^3 - 16x^2 - 10x + 40, \\ \\ g(4) & = 4(4)^3 - 16(4)^2 - 10(4) + 40 \\ & = 0 \\ \\ \text{By Factor} & \text{ theorem, } x - 4 \text{ is a factor of } g(x). \end{align} $$ \require{enclose} \begin{array}{rll} 4x^2 - 10 \phantom{000000}\\ x - 4 \enclose{longdiv}{ 4x^3 - 16x^2 - 10x + 40 \phantom{} }\kern-.2ex \\ -\underline{( 4x^3 - 16x^2 ){\phantom{000000000}}} \\ -10x + 40 \phantom{.} \\ -\underline{( -10x + 40 ){\phantom{} }} \\ 0 \phantom{.} \end{array} $$ \begin{align} \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ g(x) & = (x - 4)(4x^2 - 10) + 0 \\ & = (x - 4)(4x^2 - 10) \\ & = (x - 4)[(2x)^2 - (\sqrt{10})^2] \\ & = (x - 4)(2x + \sqrt{10})(2x - \sqrt{10}) \\ \\ \therefore 0 & = (x - 4)(2x + \sqrt{10})(2x - \sqrt{10}) \\ \\ x - 4 = 0 \phantom{000}\text{or}\phantom{000} 2x + \sqrt{10} & = 0 \phantom{00000}\text{or}\phantom{000} 2x - \sqrt{10} = 0 \\ x = 4 \phantom{000or000+\sqrt{5}} 2x & = -\sqrt{10} \phantom{00000or000-.} 2x = \sqrt{10} \\ x & = -{\sqrt{10} \over 2} \phantom{00000or00002} x = {\sqrt{10} \over 2} \\ x & = -{\sqrt{2} \sqrt{5} \over \sqrt{2}\sqrt{2}} \phantom{00000or0000} x = {\sqrt{2} \sqrt{5} \over \sqrt{2}\sqrt{2}} \\ x & = -{\sqrt{5} \over \sqrt{2}} \phantom{00000or00000.0} x = {\sqrt{5} \over \sqrt{2}} \\ x & = -\sqrt{5 \over 2} \phantom{00000or000000} x = \sqrt{5 \over 2} \end{align}

(b)

\begin{align} P(2y) & = (2y)^3 + 2(2y)^2y - 5(2y)y^2 - 6y^3 \\ & = 8y^3 + 2(4y^2)y - 10y^3 - 6y^3 \\ & = 8y^3 + 8y^3 - 10y^3 - 6y^3 \\ & = 0 \\ \\ \text{By Factor theorem, } & x - 2y \text{ is a factor of } P(x). \end{align} $$ \require{enclose} \begin{array}{rll} x^2 + 4xy + 3y^2 \phantom{000000}\\ x - 2y \enclose{longdiv}{ x^3 + 2x^2y - 5xy^2 - 6y^3 \phantom{} }\kern-.2ex \\ -\underline{( x^3 - 2x^2y ){\phantom{0000000000}}} \\ 4x^2y - 5xy^2 - 6y^3 \phantom{.} \\ -\underline{( 4x^2y - 8xy^2 ){\phantom{0000.} }} \\ 3xy^2 - 6y^3 \phantom{.} \\ -\underline{( 3xy^2 - 6y^3 ){\phantom{} }} \\ 0 \phantom{0} \end{array} $$ \begin{align} \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ \\ P(x) & = (x - 2y)(x^2 + 4xy + 3y^2) + 0 \\ & = (x - 2y)(x^2 + 4xy + 3y^2) \\ & = (x - 2y)(x + y)(x + 3y) \\ \\ \therefore \text{Other} & \text{ factors are } x + y \text{ and } x + 3y. \end{align}

\begin{align} \text{Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder} \\ \\ f(x) & = (x^2 - x - 2)Q(x) + ax + b \\ & = (x + 1)(x - 2)Q(x) + ax + b \\ \\ \text{By } & \text{Remainder theorem,} \\ f(-1) & = (-1 + 1)(-1 - 2)Q(-1) + a(-1) + b \\ -5 & = (0)(-3)Q(-1) - a + b \\ -5 & = -a + b \\ a - 5 & = b \phantom{000} \text{ --- (1)} \\ \\ \text{By } & \text{Remainder theorem,} \\ f(2) & = (2 + 1)(2 - 2)Q(2) + a(2) + b \\ 7 & = (3)(0)Q(2) + 2a + b \\ 7 & = 2a + b \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 7 & = 2a + (a - 5) \\ 7 & = 2a + a - 5 \\ 7 + 5 & = 3a \\ 12 & = 3a \\ {12 \over 3} & = a \\ 4 & = a \\ \\ \text{Substitute } & a = 4 \text{ into (1),} \\ b & = (4) - 5 \\ & = -1 \\ \\ f(x) & = (x + 1)(x - 2)Q(x) + ax + b \\ & = (x + 1)(x - 2)Q(x) + (4)x + (-1) \\ & = (x + 1)(x - 2)Q(x) + 4x - 1 \\ \\ \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ \\ \therefore \text{Remainder} & = 4x - 1 \end{align}

(a)

\begin{align} x^3 + x - 4 & = (x^2 + x - 1)(x - 1) + Ax + B \\ \\ \text{Let } & x = 0, \\ (0)^3 + (0) - 4 & = (0^2 + 0 - 1)(0 - 1) + A(0) + B \\ -4 & = (-1)(-1) + B \\ -4 & = 1 + B \\ -4 - 1 & = B \\ -5 & = B \\ \\ \text{Let } & x = 1, \\ (1)^3 + (1) - 4 & = (1^2 + 1 - 1)(1 - 1) + A(1) + (-5) \\ -2 & = (1)(0) + A - 5 \\ -2 & = A - 5 \\ -2 + 5 & = A\\ 3 & = A \\ \\ \therefore A & = 3, B = -5 \\ \\ \\ \therefore x^3 + x - 4 & = (x^2 + x - 1)(x - 1) + Ax + B \\ & = (x^2 + x - 1)(x - 1) + (3)x + (-5) \\ & = (x^2 + x - 1)(x - 1) + 3x - 5 \\ \\ \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ \\ \text{Remainder} & = 3x - 5 \end{align}

(b)

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

(a)

\begin{align} y & = x(16x - 19) \phantom{000} \text{ --- (1)} \\ \\ y & = 4x^3 - 6 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x(16x - 19) & = 4x^3 - 6 \\ 16x^2 - 19x & = 4x^3 - 6 \\ 0 & = 4x^3 - 16x^2 + 19x - 6 \\ \\ \text{Let } f(x) & = 4x^3 - 16x^2 + 19x - 6 \\ \\ f(2) & = 4(2)^3 - 16(2)^2 + 19(2) - 6 \\ & = 0 \\ \\ \text{By Factor } & \text{theorem, } x - 2 \text{ is a factor of } f(x). \end{align} $$ \require{enclose} \begin{array}{rll} 4x^2 - 8x + 3 \phantom{0000000}\\ x - 2 \enclose{longdiv}{ 4x^3 - 16x^2 + 19x - 6 \phantom{0}}\kern-.2ex \\ -\underline{( 4x^3 - 8x^2){\phantom{000000000.}}} \\ -8x^2 + 19x - 6 \phantom{0.} \\ -\underline{( -8x^2 + 16x ){\phantom{0000}}} \\ 3x - 6 \phantom{0.} \\ -\underline{( 3x - 6) \phantom{.} } \\ 0\phantom{0} \end{array} $$ \begin{align} \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ 4x^3 - 16x^2 + 19x - 6 & = (x - 2)(4x^2 - 8x + 3) + 0 \\ & = (x - 2)(4x^2 - 8x + 3) \\ & = (x - 2)(2x - 1)(2x - 3) \\ \\ \therefore 0 & = (x - 2)(2x - 1)(2x - 3) \\ \\ x - 2 = 0 \phantom{000}\text{or}\phantom{000} 2x - 1 & = 0 \phantom{000}\text{or}\phantom{000} 2x - 3 = 0 \\ x = 2 \phantom{000or000-1} 2x & = 1 \phantom{000or000-3} 2x = 3 \\ x & = {1 \over 2} \phantom{000or00-.32} x = {3 \over 2} \end{align}

(b)

\begin{align} \text{By } & \text{Factor theorem,} \\ g(2b) & = (2b)^4 - (2b)^2b^2 + kb^4 \\ 0 & = 16b^4 - (4b^2)(b^2) + kb^4 \\ 0 & = 16b^4 - 4b^4 + kb^4 \\ 0 & = 12b^4 + kb^4 \\ -12b^4 & = kb^4 \\ -12 & = k \\ \\ g(a) & = a^4 - a^2b^2 + (-12)b^4 \\ & = a^4 - a^2b^2 - 12b^4 \\ & = (a^2 - 4b^2)(a^2 + 3b^2) \\ & = [(a)^2 - (2b)^2] (a^2 + 3b^2) \\ & = (a - 2b)(a + 2b)(a^2 + 3b^2) \end{align}

(i)

\begin{align} \text{Since } a^3 + b^3 & = (a + b)(a^2 - ab + b^2), \\ 81x^3 + 24y^3 & = 3(27x^3 + 8y^3) \\ & = 3[(3x)^3 + (2y)^3] \\ & = 3(3x + 2y)[(3x)^2 - (3x)(2y) + (2y)^2] \\ & = 3(3x + 2y)(9x^2 - 6xy + 4y^2) \end{align}

(ii) Apply the result from part (i) on the numerator

\begin{align} \require{cancel} {81x^3 + 24y^3 \over 9x^2 - 6xy + 4y^2} & = {3(3x + 2y)\cancel{(9x^2 - 6xy + 4y^2)} \over \cancel{9x^2 - 6xy + 4y^2} } \\ \\ & = 3(3x + 2y) \end{align}

(i)(a)

\begin{align} (x + 2)(x - 4) & = x^2 - 4x + 2x - 8 \\ & = x^2 - 2x - 8 \end{align} $$ \require{enclose} \begin{array}{rll} 2 \phantom{0000000000}\\ x^2 - 2x - 8 \enclose{longdiv}{ 2x^2 - 7x + 14 \phantom{0}}\kern-.2ex \\ -\underline{( 2x^2 - 4x - 16 ){\phantom{.}}} \\ -3x + 30 \phantom{0.} \end{array} $$ \begin{align} \text{Since } {P(x) \over D(x)} & = Q(x) + {R(x) \over D(x)}, \\ \\ {2x^2 - 7x + 14 \over (x + 2)(x - 4)} & = 2 + {-3x + 30 \over (x + 2)(x - 4)} \\ \\ {-3x + 30 \over (x + 2)(x - 4)} & = {A \over x + 2} + {B \over x - 4} \\ & = {A(x - 4) \over (x + 2)(x - 4)} + {B(x + 2) \over (x + 2)(x - 4)} \\ & = {A(x - 4) + B(x + 2) \over (x + 2)(x - 4)} \\ \\ \text{Comparing } & \text{the numerator,} \\ -3x + 30 & = A(x - 4) + B(x + 2) \\ \\ \text{Let } & x = 4, \\ -3(4) + 30 & = A(4 - 4) + B(4 + 2) \\ 18 & = A(0) + B(6) \\ 18 & = 6B \\ {18 \over 6} & = B \\ 3 & = B \\ \\ \text{Let } & x = -2, \\ -3(-2) + 30 & = A(-2 - 4) + (3)(-2 + 2) \\ 36 & = A(-6) + (3)(0) \\ 36 & = -6A \\ {36 \over -6} & = A \\ -6 & = A \\ \\ {-3x + 30 \over (x + 2)(x - 4)} & = {A \over x + 2} + {B \over x - 4} \\ & = {-6 \over x + 2} + {3 \over x - 4} \\ & = -{6 \over x + 2} + {3 \over x - 4} \\ \\ \therefore {2x^2 - 7x + 14 \over (x + 2)(x - 4)} & = 2 - {6 \over x + 2} + {3 \over x - 4} \end{align}

(i)(b)

\begin{align} {42 \over (2x - 5)(x + 1)^2} & = {A \over 2x - 5} + {B \over x + 1} + {C \over (x + 1)^2} \\ & = {A(x + 1)^2 \over (2x - 5)(x + 1)^2} + {B(2x - 5)(x + 1) \over (2x - 5)(x + 1)^2} + {C(2x - 5) \over (2x - 5)(x + 1)^2} \\ & = {A(x + 1)^2 + B(2x - 5)(x + 1) + C(2x - 5) \over (2x - 5)(x + 1)^2} \\ \\ \text{Comparing } & \text{the numerator,} \\ \\ 42 & = A(x + 1)^2 + B(2x - 5)(x + 1) + C(2x - 5) \\ \\ \text{Let } & x = -1, \\ 42 & = A(-1 + 1)^2 + B[2(-1) - 5](-1 + 1) + C[2(-1) - 5] \\ 42 & = A(0)^2 + B(-7)(0) + C(-7) \\ 42 & = -7C \\ {42 \over -7} & = C \\ -6 & = C \\ \\ \text{Let } & x = 2.5, \\ 42 & = A(2.5 + 1)^2 + B[2(2.5) - 5](2.5 + 1) + (-6)[2(2.5) - 5] \\ 42 & = A(3.5)^2 + B(0)(3.5) + (-6)(0) \\ 42 & = 12.25A \\ {42 \over 12.25} & = A \\ {24 \over 7} & = A \\ \\ \text{Let } & x = 0, \\ 42 & = {24 \over 7}(0 + 1)^2 + B[2(0) - 5](0 + 1) + (-6)[2(0) - 5] \\ 42 & = {24 \over 7}(1)^2 + B(-5)(1) + (-6)(-5) \\ 42 & = {24 \over 7} - 5B + 30 \\ 5B & = {24 \over 7} + 30 - 42 \\ & = -{60 \over 7} \\ B & = -{60 \over 7} \div 5 \\ & = -{12 \over 7} \\ \\ \therefore {42 \over (2x - 5)(x + 1)^2} & = {A \over 2x - 5} + {B \over x + 1} + {C \over (x + 1)^2} \\ & = {{24 \over 7} \over 2x - 5} + {-{12 \over 7} \over x + 1} + {-6 \over (x + 1)^2} \\ & = {24 \over 7(2x - 5)} - {12 \over 7(x + 1)} - {6 \over (x + 1)^2} \end{align}

(ii)

\begin{align} \text{From } & \text{part (i),} \\ {2x^2 - 7x + 14 \over (x + 2)(x - 4)} & = 2 - {6 \over x + 2} + {3 \over x - 4} \\ \\ \text{Let } & x = x^2, \\ {2(x^2)^2 - 7(x^2) + 14 \over (x^2 + 2)(x^2 - 4)} & = 2 - {6 \over x^2 + 2} + {3 \over x^2 - 4} \\ \\ {2x^4 - 7x^2 + 14 \over (x^2 + 2)(x^2 - 4)} & = 2 - {6 \over x^2 + 2} + {3 \over x^2 - 4} \\ & = 2 - {6 \over x^2 + 2} + {3 \over (x + 2)(x - 2)} \\ \\ {3 \over (x + 2)(x - 2)} & = {A \over x + 2} + {B \over x - 2} \\ & = {A(x - 2) \over (x + 2)(x - 2)} + {B(x + 2) \over (x + 2)(x - 2)} \\ & = {A(x - 2) + B(x + 2) \over (x + 2)(x - 2)} \\ \\ \text{Comparing } & \text{the numerator,} \\ 3 & = A(x - 2) + B(x + 2) \\ \\ \text{Let } & x = 2, \\ 3 & = A(2 - 2) + B(2 + 2) \\ 3 & = A(0) + B(4) \\ 3 & = 4B \\ {3 \over 4} & = B \\ \\ \text{Let } & x = -2, \\ 3 & = A(-2 - 2) + {3 \over 4}(-2 + 2) \\ 3 & = A(-4) + {3 \over 4}(0) \\ 3 & = -4A \\ -{3 \over 4} & = A \\ \\ {3 \over (x + 2)(x - 2)} & = {A \over x + 2} + {B \over x - 2} \\ & = {-{3 \over 4} \over x + 2} + {{3 \over 4} \over x - 2} \\ & = -{3 \over 4(x + 2)} + {3 \over 4(x - 2)} \\ \\ \therefore {2x^4 - 7x^2 + 14 \over (x^2 + 2)(x^2 - 4)} & = 2 - {6 \over x^2 + 2} -{3 \over 4(x + 2)} + {3 \over 4(x - 2)} \end{align}

(a)

\begin{align} \text{Let } f(x) & = x^2 + 5px + p^2 + 5 \\ \\ \text{By } & \text{Factor theorem,} \\ f(-2) & = (-2)^2 + 5p(-2) + p^2 + 5 \\ 0 & = 4 - 10p + p^2 + 5 \\ 0 & = p^2 - 10p + 9 \\ 0 & = (p - 1)(p - 9) \\ \\ p - 1 = 0 \phantom{000}&\text{or}\phantom{000} p - 9 = 0 \\ p = 1 \phantom{000} &\phantom{or000-9} p = 9 \\ \\ \text{For } & p = 1, \\ f(x) & = x^2 + 5(1)x + (1)^2 + 5 \\ & = x^2 + 5x + 1 + 5 \\ & = x^2 + 5x + 6 \\ \\ f(-3) & = (-3)^2 + 5(-3) + 6 \\ & = 0 \\ \\ \text{Since } x + 3 & \text{ is not a factor } \implies p \ne 1 \\ \\ \text{For } & p = 9, \\ f(x) & = x^2 + 5(9)x + (9)^2 + 5 \\ & = x^2 + 45x + 81 + 5 \\ & = x^2 + 45x + 86 \\ \\ f(-3) & = (-3)^2 + 45(-3) + 86 \\ & = -40 \\ \\ \therefore p & = 9 \end{align}

(b)

\begin{align} x^2(x + 3) & = 10x + 24 \\ x^3 + 3x^2 & = 10x + 24 \\ 0 & = -x^3 - 3x^2 + 10x + 24 \\ 0 & = x^3 + 3x^2 - 10x - 24 \\ \\ \text{Let } f(x) & = x^3 + 3x^2 - 10x - 24 \\ \\ f(-2) & = (-2)^3 + 3(-2)^2 - 10(-2) - 24 \\ & = 0 \\ \\ \text{By Factor theorem, } & x + 2 \text{ is a factor of } f(x). \end{align} $$ \require{enclose} \begin{array}{rll} x^2 + x - 12 \phantom{0000000}\\ x + 2 \enclose{longdiv}{ x^3 + 3x^2 - 10x - 24\phantom{0}}\kern-.2ex \\ -\underline{( x^3 + 2x^2){\phantom{0000000000}}} \\ x^2 - 10x - 24 \phantom{0} \\ -\underline{( x^2 + 2x ){\phantom{00000.}}} \\ - 12x - 24 \phantom{0} \\ -\underline{(-12x - 24)} \\ 0\phantom{0} \end{array} $$ \begin{align} \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ f(x) & = (x + 2)(x^2 + x - 12) + 0 \\ & = (x + 2)(x^2 + x - 12) \\ & = (x + 2)(x - 4)(x + 3) \\ \\ \therefore 0 & = (x + 2)(x - 3)(x + 4) \\ \\ x + 2 = 0 \phantom{000}\text{or}\phantom{000} x - 3 & = 0 \phantom{000}\text{or}\phantom{000)} x + 4 = 0 \\ x = - 2 \phantom{000or00004} x & = 3 \phantom{000or000+3.} x = - 4 \end{align}

(b)(i)

\begin{align} 9^x(3^x + 3) & = 10(3^x) + 24 \\ (3^2)^x(3^x + 3) & = 10(3^x) + 24 \\ 3^{2x}(3^x + 3) & = 10(3^x) + 24 \\ \\ \text{Let } u & = 3^x, \\ u^2(u + 3) & = 10u + 24 \\ \\ \text{From } & \text{part (i),} \\ u = - 2 \phantom{000or00004} u & = 3 \phantom{000or000+3.} u = - 4 \\ \\ \text{Since } & u = 3^x, \\ 3^x = - 2 \text{ (Reject)}\phantom{04} 3^x & = 3 \phantom{000o000+3.} 3^x = - 4 \text{ (Reject)} \\ 3^x & = 3^1 \\ \therefore x & = 1 \end{align}

(b)(ii)

\begin{align} 25x^2(5x + 3) & = 50x + 24 \\ (5x)^2 [(5x) + 3] & = 10(5x) + 24 \\ \\ \text{Comparing with } x^2(x + 3) & = 10x + 24 \text{ from part (b)}, \\ \\ \text{let }x & = 5x, \\ \\ 5x = - 2 \phantom{000or00004} 5x & = 3 \phantom{000or000+3.} 5x = - 4 \\ x = -{2 \over 5} \phantom{000o000004} x & = {3 \over 5} \phantom{000or000+35.} x = -{4 \over 5} \end{align}

(i)

\begin{align} \text{When }& f(x) = 0, \\ 0 & = (x + 1)(x - 3)(x - k) \\ \\ \text{Since} & \text{ the coefficient of } x^3 \text{ is 2}, \\ f(x) & = 2(x + 1)(x - 3)(x - k) \\ \\ \text{By } & \text{Remainder theorem,} \\ f(4) & = 2(4 + 1)(4 - 3)(4 - k) \\ 20 & = 2(5)(1)(4 - k) \\ 20 & = 10(4 - k) \\ {20 \over 10} & = 4 - k \\ 2 & = 4 - k \\ k & = 4 - 2 \\ & = 2 \end{align}

(ii)

\begin{align} f(x) & = 2(x + 1)(x - 3)(x - k) \\ & = 2(x + 1)(x - 3)(x - 2) \\ \\ \text{By } & \text{Remainder theorem,} \\ f(-2) & = 2(-2 + 1)(-2 - 3)(-2 -2) \\ & = -40 \\ \\ \therefore \text{Remainder} & = -40 \end{align}

(i)

\begin{align} f(x) - g(x) & = 0 \\ (x^4 - x^3 - 7x^2 + x + 6) - (x^4 - 2x^3 - 10x^2 + 5x + 18) & = 0 \\ x^4 - x^3 - 7x^2 + x + 6 - x^4 + 2x^3 + 10x^2 - 5x - 18 & = 0 \\ x^4 - x^4 - x^3 + 2x^3 - 7x^2 + 10x^2 + x - 5x + 6 - 18 & = 0 \\ x^3 + 3x^2 - 4x - 12 & = 0 \end{align} \begin{align} \text{Let } f(x) & = x^3 + 3x^2 - 4x - 12 \\ \\ f(2) & = (2)^3 + 3(2)^2 - 4(2) - 12 \\ & = 0 \\ \\ \text{By Factor theorem, } & x - 2 \text{ is a factor of } f(x). \end{align} $$ \require{enclose} \begin{array}{rll} x^2 + 5x + 6 \phantom{000000}\\ x - 2 \enclose{longdiv}{ x^3 + 3x^2 - 4x - 12\phantom{0}}\kern-.2ex \\ -\underline{( x^3 - 2x^2){\phantom{000000000}}} \\ 5x^2 - 4x - 12 \phantom{0} \\ -\underline{( 5x^2 - 10x ){\phantom{000.}}} \\ 6x - 12 \phantom{0} \\ -\underline{( 6x - 12)} \\ 0\phantom{0} \end{array} $$ \begin{align} \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ f(x) & = (x - 2)(x^2 + 5x + 6) + 0 \\ & = (x - 2)(x^2 + 5x + 6) \\ & = (x - 2)(x + 2)(x + 3) \\ \\ \therefore 0 & = (x - 2)(x + 2)(x + 3) \\ \\ x - 2 = 0 \phantom{000}\text{or}\phantom{000} x + 2 & = 0 \phantom{000}\text{or}\phantom{000} x + 3 = 0 \\ x = 2 \phantom{000or000+2} x & = - 2 \phantom{000o.00003} x = - 3 \end{align}

(ii)

\begin{align} \text{For the factor } x - \alpha, & \phantom{0} f(\alpha) = 0 \text{ and } g(\alpha) = 0. \\ \\ \therefore f(\alpha) & = g(\alpha) \\ f(\alpha) - g(\alpha) & = 0 \\ \\ \text{From part (i), } & \text{possible values of } \alpha \text{ are } 2, -2, -3 \\ \\ \text{For } & \alpha = 2, \\ f(2) & = (2)^4 - (2)^3 - 7(2)^2 + (2) + 6 \\ & = -12 \\ \\ \therefore x - 2 \text{ is not a factor of } & f(x) \implies \alpha \ne 2 \\ \\ \text{For } & \alpha = - 2, \\ f(-2) & = (-2)^4 - (-2)^3 - 7(-2)^2 + (-2) + 6 \\ & = 0 \\ \\ g(-2) & = (-2)^4 - 2(-2)^3 - 10(-2)^2 + 5(-2) + 18 \\ & = 0 \\ \\ \therefore x + 2 \text{ is a factor of } & f(x) \text{ and } g(x) \implies \alpha = -2 \\ \\ \text{For } & \alpha = -3, \\ f(-3) & = (-3)^4 - (-3)^3 - 7(-3)^2 + (-3) + 6 \\ & = 48 \\ \\ \therefore x + 3 \text{ is not a factor of } & f(x) \implies \alpha \ne -3 \\ \\ \therefore \alpha & = -2 \end{align}