(a)

\begin{align} \text{Let } f(x) & = x^3 - 2x^2 - 4x + 8 \\ \\ f(2) & = (2)^3 - 2(2)^2 - 4(2) + 8 \\ & = 0 \\ \\ \text{By Factor theorem, } & x - 2 \text{ is a factor of } f(x). \end{align} $$ \require{enclose} \begin{array}{rll} x^2 - 4 \phantom{000000}\\ x - 2 \enclose{longdiv}{ x^3 - 2x^2 - 4x + 8 \phantom{0}}\kern-.2ex \\ -\underline{(x^3 - 2x^2){\phantom{0000000.}}} \\ -4x + 8 \phantom{0} \\ -\underline{(-4x + 8){\phantom{}}} \\ 0 \phantom{0} \end{array} $$ \begin{align} f(x) & = x^3 - 2x^2 - 4x + 8 \\ \\ \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ f(x) & = (x - 2)(x^2 - 4) + 0 \\ & = (x - 2)(x^2 - 4) \\ & = (x - 2)(x + 2)(x - 2) \\ & = (x - 2)^2(x + 2) \end{align}

(b)

\begin{align} \text{Let } f(x) & = 3x^3 - 10x^2 + 9x - 2 \\ \\ f(1) & = 3(1)^3 - 10(1) + 9(1) - 2 \\ & = 0 \\ \\ \text{By Factor theorem, } & x - 1 \text{ is a factor of } f(x). \end{align} $$ \require{enclose} \begin{array}{rll} 3x^2 - 7x + 2 \phantom{000000}\\ x - 1 \enclose{longdiv}{ 3x^3 - 10x^2 + 9x - 2 \phantom{0}}\kern-.2ex \\ -\underline{( 3x^3 - 3x^2 ){\phantom{00000000.}}} \\ -7x^2 + 9x - 2 \phantom{0.} \\ -\underline{( -7x^2 + 7x ){\phantom{0000}}} \\ 2x - 2 \phantom{0.} \\ -\underline{( 2x - 2) \phantom{.} } \\ 0 \phantom{0} \end{array} $$ \begin{align} f(x) & = 3x^3 - 10x^2 + 9x - 2 \\ \\ \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ f(x) & = (x - 1)(3x^2 - 7x + 2) + 0 \\ & = (x - 1)(3x^2 - 7x + 2) \\ & = (x - 1)(3x - 1)(x - 2) \end{align}

(i)

\begin{align} f(x) & = x^3 - 3x^2 - 4x \\ 0 & = x^3 - 3x^2 - 4x \\ 0 & = x(x^2 - 3x - 4) \\ 0 & = x(x + 1)(x - 4) \\ \\ x = 0 \phantom{000}\text{or}\phantom{00} x + 1 & = 0 \phantom{000}\text{or}\phantom{00} x - 4 = 0 \\ x & = -1 \phantom{000or0004} x = 4 \end{align}

(ii)

\begin{align} f(x) & = x^3 - 3x^2 - 4x \\ -12 & = x^3 - 3x^2 - 4x \\ 0 & = x^3 - 3x^2 - 4x + 12 \\ \\ \text{Let } g(x) & = x^3 - 3x^2 - 4x + 12 \\ \\ g(2) & = (2)^3 - 3(2)^2 - 4(2) + 12 \\ & = 0 \\ \\ \text{By Factor theorem, } & x - 2 \text{ is a factor of } g(x). \end{align} $$ \require{enclose} \begin{array}{rll} x^2 - x - 6 \phantom{000000}\\ x - 2 \enclose{longdiv}{ x^3 - 3x^2 - 4x + 12 \phantom{0}}\kern-.2ex \\ -\underline{( x^3 - 2x^2 ){\phantom{00000000.}}} \\ -x^2 - 4x + 12 \phantom{0.} \\ -\underline{( -x^2 + 2x ){\phantom{00000}}} \\ -6x + 12 \phantom{0.} \\ -\underline{( -6x + 12 ) \phantom{.} } \\ 0 \phantom{0} \end{array} $$ \begin{align} g(x) & = x^3 - 3x^2 - 4x + 12 \\ \\ \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ g(x) & = (x - 2)(x^2 - x - 6) + 0 \\ & = (x - 2)(x^2 - x - 6) \\ & = (x - 2)(x + 2)(x - 3) \\ \\ 0 & = (x - 2)(x + 2)(x - 3) \\ \\ x - 2 = 0 \phantom{00}\text{or}\phantom{00} x + 2 & = 0 \phantom{000}\text{or}\phantom{00} x - 3 = 0 \\ x = 2 \phantom{000000000} x & = - 2 \phantom{00000000.} x = 3 \end{align}

(a)

\begin{align} \text{Let } f(x) & = x^3 - 4x^2 + x + 6 \\ \\ f(-1) & = (-1)^3 - 4(-1)^2 + (-1) + 6 \\ & = 0 \\ \\ \text{By Factor theorem, } & x + 1 \text{ is a factor of } f(x). \end{align} $$ \require{enclose} \begin{array}{rll} x^2 - 5x + 6 \phantom{0000}\\ x + 1 \enclose{longdiv}{ x^3 - 4x^2 + x + 6 \phantom{0}}\kern-.2ex \\ -\underline{( x^3 + x^2 ){\phantom{0000000.}}} \\ -5x^2 + x + 6 \phantom{0.} \\ -\underline{( -5x^2 - 5x ){\phantom{000}}} \\ 6x + 6 \phantom{0.} \\ -\underline{( 6x + 6 ) \phantom{.} } \\ 0 \phantom{0} \end{array} $$ \begin{align} f(x) & = x^3 - 4x^2 + x + 6 \\ \\ \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ f(x) & = (x + 1)(x^2 - 5x + 6) + 0 \\ & = (x + 1)(x^2 - 5x + 6) \\ & = (x + 1)(x - 2)(x - 3) \\ \\ \therefore 0 & = (x + 1)(x - 2)(x - 3) \\ \\ x + 1 = 0 \phantom{00}\text{or}\phantom{00} x - 2 & = 0 \phantom{00}\text{or}\phantom{00} x - 3 = 0 \\ x = - 1 \phantom{0000000.} x & = 2 \phantom{000000000} x = 3 \end{align}

(b)

\begin{align} 4x^3 + 18 & = 7x^2 + 21x \\ 0 & = -4x^3 + 7x^2 + 21x - 18 \\ 0 & = 4x^3 - 7x^2 - 21x + 18 \\ \\ \text{Let } f(x) & = 4x^3 - 7x^2 - 21x + 18 \\ \\ f(-2) & = 4(-2)^3 - 7(-2)^2 - 21(-2) + 18 \\ & = 0 \\ \\ \text{By Factor theorem, } & x + 2 \text{ is a factor of } f(x). \end{align} $$ \require{enclose} \begin{array}{rll} 4x^2 - 15x + 9 \phantom{0000}\\ x + 2 \enclose{longdiv}{ 4x^3 - 7x^2 - 21x + 18 \phantom{0}}\kern-.2ex \\ -\underline{( 4x^3 + 8x^2 ){\phantom{000000000.}}} \\ -15x^2 - 21x + 18 \phantom{0} \\ -\underline{( -15x^2 - 30x ){\phantom{0000.}}} \\ 9x + 18 \phantom{0.} \\ -\underline{( 9x + 18 ) \phantom{.} } \\ 0 \phantom{0} \end{array} $$ \begin{align} f(x) & = 4x^3 - 7x^2 - 21x + 18 \\ \\ \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ f(x) & = (x + 2)(4x^2 - 15x + 9) + 0 \\ & = (x + 2)(4x^2 - 15x + 9) \\ & = (x + 2)(x - 3)(4x - 3) \\ \\ \therefore 0 & = (x + 2)(x - 3)(4x - 3) \\ \\ x + 2 = 0 \phantom{00}\text{or}\phantom{00} x - 3 & = 0 \phantom{00}\text{or}\phantom{00} 4x - 3 = 0 \\ x = -2 \phantom{0000000.} x & = 3 \phantom{000000000} 4x = 3 \\ & \phantom{00000000000000} x = {3 \over 4} \end{align}

(c)

\begin{align} x^3 + 4 & = x(x + 4) \\ x^3 + 4 & = x^2 + 4x \\ 0 & = -x^3 + x^2 + 4x - 4 \\ 0 & = x^3 - x^2 - 4x + 4 \\ \\ \text{Let } f(x) & = x^3 - x^2 - 4x + 4 \\ \\ f(1) & = (1)^3 - (1)^2 - 4(1) + 4 \\ & = 0 \\ \\ \text{By Factor theorem, } & x - 1 \text{ is a factor of } f(x). \end{align} $$ \require{enclose} \begin{array}{rll} x^2 - 4 \phantom{00000}\\ x - 1 \enclose{longdiv}{ x^3 - x^2 - 4x + 4 \phantom{0}}\kern-.2ex \\ -\underline{( x^3 - x^2 ){\phantom{00000000}}} \\ -4x + 4 \phantom{0} \\ -\underline{( -4x + 4 ){\phantom{}}} \\ 0 \phantom{0.} \end{array} $$ \begin{align} f(x) & = x^3 - x^2 - 4x + 4 \\ \\ \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ f(x) & = (x - 1)(x^2 - 4) + 0 \\ & = (x - 1)(x^2 - 4) \\ & = (x - 1)(x - 2)(x + 2) \\ \\ \therefore 0 & = (x - 1)(x - 2)(x + 2) \\ \\ x - 1 = 0 \phantom{00}\text{or}\phantom{00} x - 2 & = 0 \phantom{00}\text{or}\phantom{00} x + 2 = 0 \\ x = 1 \phantom{000000000} x & = 2 \phantom{000000000} x = - 2 \end{align}

(d)

\begin{align} x(x + 3)(x - 1) & = x + 8 \\ x(x^2 - x + 3x - 3) & = x + 8 \\ x(x^2 + 2x - 3) & = x + 8 \\ x^3 + 2x^2 - 3x & = x + 8 \\ 0 & = -x^3 - 2x^2 + 3x + x + 8 \\ 0 & = -x^3 - 2x^2 + 4x + 8 \\ 0 & = x^3 + 2x^2 - 4x - 8 \\ \\ \text{Let } f(x) & = x^3 + 2x^2 - 4x - 8 \\ \\ f(2) & = (2)^3 + 2(2)^2 - 4(2) - 8 \\ & = 0 \\ \\ \text{By Factor theorem, } & x - 2 \text{ is a factor of } f(x). \end{align} $$ \require{enclose} \begin{array}{rll} x^2 + 4x + 4 \phantom{00000}\\ x - 2 \enclose{longdiv}{ x^3 + 2x^2 - 4x - 8 \phantom{0}}\kern-.2ex \\ -\underline{( x^3 - 2x^2 ){\phantom{0000000.}}} \\ 4x^2 - 4x - 8 \phantom{0} \\ -\underline{( 4x^2 - 8x ){\phantom{000.}}} \\ 4x - 8 \phantom{0} \\ -\underline{( 4x - 8 ){\phantom{.}}} \\ 0 \phantom{0} \\ \end{array} $$ \begin{align} f(x) & = x^3 + 2x^2 - 4x - 8 \\ \\ \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ f(x) & = (x - 2)(x^2 + 4x + 4) + 0 \\ & = (x - 2)(x^2 + 4x + 4) \\ & = (x - 2)(x + 2)^2 \\ \\ \therefore 0 & = (x - 2)(x + 2)^2 \\ \\ x - 2 = 0 \phantom{00}&\text{or}\phantom{00} x + 2 = 0 \\ x = 2 \phantom{00} & \phantom{or00+2} x = - 2 \end{align}

\begin{align} \text{Since the roots are } -4, -1 & \text{ and } 3, \\ (x + 4)(x + 1)(x - 3) & = 0 \\ (x^2 + x + 4x + 4)(x - 3) & = 0 \\ (x^2 + 5x + 4)(x - 3) & = 0 \\ x^3 - 3x^2 + 5x^2 - 15x + 4x - 12 & = 0 \\ x^3 + 2x^2 - 11x - 12 & = 0 \\ \\ \text{Since the coefficient of } & x^5 \text{ is 5}, \\ 5(x^3 + 2x^2 - 11x - 12) & = 5(0) \\ 5x^3 + 10x^2 - 55x - 60 & = 0 \\ \\ \therefore P(x) = 5x^3 + 10x^2 & - 55x - 60 \end{align}

(a)

\begin{align} \text{Let } f(x) & = x^3 - 9x^2 + 25x - 21 \\ \\ f(3) & = (3)^3 - 9(3)^2 + 25(3) - 21 \\ & = 0 \\ \\ \text{By Factor theorem, } & x - 3 \text{ is a factor of } f(x). \end{align} $$ \require{enclose} \begin{array}{rll} x^2 - 6x + 7 \phantom{00000000}\\ x - 3 \enclose{longdiv}{ x^3 - 9x^2 + 25x - 21 \phantom{0}}\kern-.2ex \\ -\underline{(x^3 - 3x^2){\phantom{000000000.}}} \\ -6x^2 + 25x - 21\phantom{0} \\ -\underline{(-6x^2 + 18x){\phantom{0000.}}} \\ 7x - 21\phantom{0} \\ -\underline{(7x - 21)} \\ 0\phantom{0} \end{array} $$ \begin{align} f(x) & = x^3 - 9x^2 + 25x - 21 \\ \\ \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ f(x) & = (x - 3)(x^2 - 6x + 7) + 0 \\ & = (x - 3)(x^2 - 6x + 7) \\ \\ \therefore 0 & = (x - 3)(x^2 - 6x + 7) \end{align} \begin{align} x - 3 & = 0 \phantom{000}& \text{or}\phantom{00000} x^2 - 6x + 7 & = 0 \\ \\ x & = 3 & x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & & & = {-(-6) \pm \sqrt{(-6)^2 - 4(1)(7)} \over 2(1)} \\ & & & = {6 \pm \sqrt{8} \over 2} \\ & & & = {6 \pm \sqrt{4}\sqrt{2} \over 2} \\ & & & = {6 \pm 2\sqrt{2} \over 2} \\ & & & = {6 \over 2} \pm {2\sqrt{2} \over 2} \\ & & & = 3 \pm \sqrt{2} \end{align}

(b)

\begin{align} 3x^3 + 5x^2 & = 3x + 2 \\ 3x^3 + 5x^2 - 3x - 2 & = 0 \\ \\ \text{Let } f(x) & = 3x^3 + 5x^2 - 3x - 2 \\ \\ f(-2) & = 3(-2)^3 + 5(-2)^2 - 3(-2) - 2 \\ & = 0 \\ \\ \text{By Factor theorem, } & x + 2 \text{ is a factor of } f(x). \end{align} $$ \require{enclose} \begin{array}{rll} 3x^2 - x - 1 \phantom{000000}\\ x + 2 \enclose{longdiv}{ 3x^3 + 5x^2 - 3x - 2 \phantom{0}}\kern-.2ex \\ -\underline{(3x^3 + 6x^2){\phantom{0000000.}}} \\ -x^2 - 3x - 2\phantom{0} \\ -\underline{(-x^2 - 2x){\phantom{000.}}} \\ -x - 2\phantom{0} \\ -\underline{( -x - 2 ) \phantom{.}} \\ 0\phantom{0} \end{array} $$ \begin{align} f(x) & = 3x^3 + 5x^2 - 3x - 2 \\ \\ \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ f(x) & = (x + 2)(3x^2 - x - 1) + 0 \\ & = (x + 2)(3x^2 - x - 1) \\ \\ \therefore 0 & = (x + 2)(3x^2 - x - 1) \end{align} \begin{align} x + 2 & = 0 \phantom{000}&\text{or}\phantom{000} 3x^2 - x - 1 & = 0 \\ \\ x & = -2 & x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & & & = {-(-1) \pm \sqrt{(-1)^2 - 4(3)(-1)} \over 2(3)} \\ & & & = {1 \pm \sqrt{13} \over 6} \end{align}

(c)

\begin{align} 2x^3 + 6x - 6 & = (13x - 6)(x - 1) \\ 2x^3 + 6x - 6 & = 13x^2 - 13x - 6x + 6 \\ 2x^3 - 13x^2 + 6x + 13x + 6x - 6 - 6 & = 0 \\ 2x^3 - 13x^2 + 25x - 12 & = 0 \\ \\ \text{Let } f(x) & = 2x^3 - 13x^2 + 25x - 12 \\ \\ f(3) & = 2(3)^3 - 13(3)^2 + 25(3) - 12 \\ & = 0 \\ \\ \text{By Factor theorem, } & x - 3 \text{ is a factor of } f(x). \end{align} $$ \require{enclose} \begin{array}{rll} 2x^2 - 7x + 4 \phantom{000000}\\ x - 3 \enclose{longdiv}{ 2x^3 - 13x^2 + 25x - 12 \phantom{0}}\kern-.2ex \\ -\underline{( 2x^3 - 6x^2 ){\phantom{0000000000.}}} \\ -7x^2 + 25x - 12 \phantom{0} \\ -\underline{(- 7x^2 + 21x){\phantom{0000.}}} \\ 4x - 12\phantom{0} \\ -\underline{( 4x - 12 ) \phantom{}} \\ 0\phantom{0} \end{array} $$ \begin{align} f(x) & = 2x^3 - 13x^2 + 25x - 12 \\ \\ \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ f(x) & = (x - 3)(2x^2 - 7x + 4) + 0 \\ & = (x - 3)(2x^2 - 7x + 4) \\ \\ \therefore 0 & = (x - 3)(2x^2 - 7x + 4) \end{align} \begin{align} x - 3 & = 0 \phantom{000}&\text{or}\phantom{000} 2x^2 - 7x + 4 & = 0 \\ \\ x & = 3 & x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & & & = {-(-7) \pm \sqrt{(-7)^2 - 4(2)(4)} \over 2(2)} \\ & & & = {7 \pm \sqrt{17} \over 4} \end{align}

(i)

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

(ii)

\begin{align} 2x^3 + 4 & = (x - 1)(x - 2) \\ 2x^3 + 4 & = x^2 - 2x - x + 2 \\ 2x^3 + 4 & = x^2 - 3x + 2 \\ 0 & = - 2x^3 + x^2 - 3x + 2 - 4 \\ 0 & = - 2x^3 + x^2 - 3x - 2 \\ 0 & = 2x^3 - x^2 + 3x + 2 \\ \\ \text{From part (i), } 2x + 1 & \text{ is a factor of } 2x^3 - x^2 + 3x + 2, \end{align} $$ \require{enclose} \begin{array}{rll} x^2 - x + 2 \phantom{000000}\\ 2x + 1 \enclose{longdiv}{ 2x^3 - x^2 + 3x + 2 \phantom{0}}\kern-.2ex \\ -\underline{( 2x^3 + x^2 ){\phantom{0000000.}}} \\ -2x^2 + 3x + 2 \phantom{0} \\ -\underline{(- 2x^2 - x){\phantom{0000.}}} \\ 4x + 2\phantom{0} \\ -\underline{( 4x + 2 ) \phantom{}} \\ 0\phantom{0} \end{array} $$ \begin{align} f(x) & = 2x^3 - 13x^2 + 25x - 12 \\ \\ \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ 0 & = 2x^3 - x^2 + 3x + 2 \\ 0 & = (2x + 1)(x^2 - x + 2) + 0 \\ 0 & = (2x + 1)(x^2 - x + 2) \end{align} \begin{align} 2x + 1 & = 0 \phantom{000}&\text{or}\phantom{000} x^2 - x + 2 & = 0 \\ \\ 2x & = -1 & b^2 - 4ac & = (-1)^2 - 4(1)(2) \\ x & = -{1 \over 2} & & = -7 < 0 \text{ (No real roots)} \\ \\ \\ \therefore \text{Equation } &\text{only has 1 real root.} \end{align}

\begin{align} \text{When } & y = 0 \text{ and } x = -2 \text{ or } 1 \text{ or } 2, \phantom{00000} [x\text{-intercepts -2, 1, 2}] \\ \therefore 0 & = (x + 2)(x - 1)(x - 2) \\ \\ y & = k(x + 2)(x - 1)(x - 2) \phantom{0000000.} [k \text{ is a constant}] \\ \\ \text{When } & x = 0 \text{ and } y = 12, \phantom{000000000000.} [\text{Graph passes through (0, 12)}] \\ 12 & = k(0 + 2)(0 - 1)(0 - 2) \\ 12 & = k(2)(-1)(-2) \\ 12 & = 4k \\ {12 \over 4} & = k \\ 3 & = k \\ \\ \therefore y & = k(x + 2)(x - 1)(x - 2) \\ & = 3(x + 2)(x - 1)(x - 2) \\ \\ \therefore P(x) & = 3(x + 2)(x - 1)(x - 2) \end{align}

\begin{align} \text{When } & N(x) = 35, \\ 35 & = {1 \over 6}x^3 - {1 \over 2}x^2 + {1 \over 3}x \\ 0 & = {1 \over 6}x^3 - {1 \over 2}x^2 + {1 \over 3}x - 35 \\ 0 & = x^3 - 3x^2 + 2x - 210 \\ \\ \text{Let } f(x) & = x^3 - 3x^2 + 2x - 210 \\ \\ f(7) & = (7)^3 - 3(7)^2 + 2(7) - 210 \\ & = 0 \\ \\ \text{By Factor theorem, } & x - 7 \text{ is a factor of } f(x). \end{align} $$ \require{enclose} \begin{array}{rll} x^2 + 4x + 30 \phantom{000000}\\ x - 7 \enclose{longdiv}{ x^3 - 3x^2 + 2x - 210 \phantom{0}}\kern-.2ex \\ -\underline{(x^3 - 7x^2){\phantom{000000000.}}} \\ 4x^2 + 2x - 210\phantom{0} \\ -\underline{(4x^2 - 28x){\phantom{0000.}}} \\ 30x - 210 \phantom{0} \\ -\underline{(30x - 210)} \\ 0\phantom{0} \end{array} $$ \begin{align} f(x) & = x^3 - 3x^2 + 2x - 210 \\ \\ \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ f(x) & = (x - 7)(x^2 + 4x + 30) + 0 \\ & = (x - 7)(x^2 + 4x + 30) \\ \\ \therefore 0 & = (x - 7)(x^2 + 4x + 30) \end{align} \begin{align} x - 7 & = 0 \phantom{000}&\text{or}\phantom{000} x^2 + 4x + 30 & = 0 \\ \\ x & = 7 & b^2 - 4ac & = (4)^2 - 4(1)(30) \\ & & & = -104 < 0 \text{ (No real roots)} \end{align}
$$ \therefore x = 7 $$

(i)

\begin{align} f(x) & = 2x^3 + 9x^2 + 7x + 3 \\ \\ \text{By } & \text{Remainder theorem,} \\ \\ f(k) & = 2(k)^3 + 9(k)^2 + 7(k) + 3 \\ 9 & = 2k^3 + 9k^2 + 7k + 3 \\ \\ 0 & = 2k^3 + 9k^2 + 7k - 6 \text{ (Shown)} \end{align}

(ii)

\begin{align} \text{Let } g(k) & = 2k^3 + 9k^2 + 7k - 6 \\ \\ g(-2) & = 2(-2)^3 + 9(-2)^2 + 7(-2) - 6 \\ & = 0 \\ \\ \text{By Factor} & \text{ theorem, } k + 2 \text{ is a factor of } g(k). \end{align} $$ \require{enclose} \begin{array}{rll} 2k^2 + 5k - 3 \phantom{0000000}\\ k + 2 \enclose{longdiv}{ 2k^3 + 9k^2 + 7k - 6 \phantom{0}}\kern-.2ex \\ -\underline{( 2k^3 + 4k^2 ){\phantom{00000000}}} \\ 5k^2 + 7k - 6 \phantom{0} \\ -\underline{( 5k^2 + 10k ){\phantom{00.}}} \\ -3k - 6 \phantom{0} \\ -\underline{(-3k - 6)} \\ 0\phantom{0} \end{array} $$ \begin{align} f(x) & = 2k^3 + 9k^2 + 7k - 6 \\ \\ \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ g(k) & = (k + 2)(2k^2 + 5k - 3) + 0 \\ & = (k + 2)(2k^2 + 5k - 3) \\ & = (k + 2)(2k - 1)(k + 3) \\ \\ \therefore 0 & = (k + 2)(2k - 1)(k + 3) \\ \\ k + 2 = 0 \phantom{00000} 2k - 1 & = 0 \phantom{00000} k + 3 = 0 \\ k = -2 \phantom{0000000} 2k & = 1 \phantom{00000+3} k = -3 \\ k & = {1 \over 2} \end{align}

\begin{align} y & = 2x^3 \phantom{000} \text{ --- (1)} \\ \\ y & = (2 - x)(5x + 6) \\ & = 10x + 12 - 5x^2 - 6x \\ & = -5x^2 + 10x - 6x + 12 \\ & = -5x^2 + 4x + 12 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2x^3 & = -5x^2 + 4x + 12 \\ 0 & = -2x^3 - 5x^2 + 4x + 12 \\ 0 & = 2x^3 + 5x^2 - 4x - 12 \\ \\ \text{Let } f(x) & = 2x^3 + 5x^2 - 4x - 12 \\ \\ \\ f(-2) & = 2(-2)^3 + 5(-2)^2 - 4(-2) - 12 \\ & = 0 \\ \\ \text{By Factor} & \text{ theorem, } x + 2 \text{ is a factor of } f(x). \end{align} $$ \require{enclose} \begin{array}{rll} 2x^2 + x - 6 \phantom{0000000}\\ x + 2 \enclose{longdiv}{ 2x^3 + 5x^2 - 4x - 12 \phantom{0}}\kern-.2ex \\ -\underline{( 2x^3 + 4x^2 ){\phantom{00000000.}}} \\ x^2 - 4x - 12 \phantom{0.} \\ -\underline{( x^2 + 2x ){\phantom{00000}}} \\ -6x - 12 \phantom{0.} \\ -\underline{(-6x - 12) \phantom{.}} \\ 0\phantom{0} \end{array} $$ \begin{align} f(x) & = 2x^3 + 5x^2 - 4x - 12 \\ \\ \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ f(x) & = (x + 2)(2x^2 + x - 6) + 0 \\ & = (x + 2)(2x^2 + x - 6) \\ & = (x + 2)(x + 2)(2x - 3) \\ & = (x + 2)^2 (2x - 3) \\ \\ \therefore 0 & = (x + 2)^2 (2x - 3) \end{align} \begin{align} x + 2 & = 0 \phantom{000}&\text{or}\phantom{000} 2x - 3 & = 0 \\ x & = -2 & 2x & = 3 \\ & & x & = {3 \over 2} \\ \\ \text{Substitute} & \text{ into (1),} & \text{Substitute} & \text{ into (1),} \\ y & = 2(-2)^3 & y & = 2\left(3 \over 2\right)^3 \\ & = -16 & & = {27 \over 4} \end{align} $$ \therefore \text{Points of intersection are } (-2, -16) \text{ and } \left({3 \over 2}, {27 \over 4}\right) $$

(i)

\begin{align} \text{Since } P(x) \text{ is exactly divisible by } & (x - 2)(x + 1), \phantom{0} x - 2 \text{ and } x + 1 \text{ are factors.} \\ \\ \text{By } & \text{Factor theorem,} \\ P(2) & = (2)^4 + a(2)^3 - (2)^2 + b(2) - 12 \\ 0 & = 16 + 8a - 4 + 2b - 12 \\ 0 & = 8a + 2b \\ -2b & = 8a \\ b & = {8a \over -2} \\ b & = -4a \phantom{000} \text{ --- (1)} \\ \\ \text{By } & \text{Factor theorem,} \\ P(-1) & = (-1)^4 + a(-1)^3 - (-1)^2 + b(-1) - 12 \\ 0 & = 1 - a - 1 - b - 12 \\ 0 & = - a - b - 12 \\ a + b & = - 12 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ a + (-4a) & = -12 \\ -3a & = -12 \\ a & = {-12 \over -3} \\ & = 4 \\ \\ \text{Substitute } a & = 4 \text{ into (1),} \\ b & = -4(4) \\ & = -16 \\ \\ \therefore a & = 4, b = -16 \end{align}

(ii)

\begin{align} P(x) & = x^4 + ax^3 - x^2 + bx - 12 \\ & = x^4 + 4x^3 - x^2 - 16x - 12 \\ \\ (x - 2)(x + 1) & = x^2 + x - 2x - 2 \\ & = x^2 - x - 2 \end{align} $$ \require{enclose} \begin{array}{rll} x^2 + 5x + 6 \phantom{00000000000}\\ x^2 - x - 2 \enclose{longdiv}{ x^4 + 4x^3 - x^2 - 16x - 12 \phantom{0}}\kern-.2ex \\ -\underline{( x^4 - x^3 - 2x^2 ){\phantom{000000000.}}} \\ 5x^3 + x^2 - 16x - 12 \phantom{0.} \\ -\underline{( 5x^3 - 5x^2 - 10x ){\phantom{0000}}} \\ 6x^2 - 6x - 12 \phantom{0.} \\ -\underline{(6x^2 - 6x - 12) \phantom{.}} \\ 0\phantom{0} \end{array} $$ \begin{align} P(x) & = x^4 + 4x^3 - x^2 - 16x -12 \\ \\ \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ P(x) & = (x^2 + 5x + 6)(x^2 - x - 2) + 0 \\ & = (x^2 + 5x + 6)(x^2 - x - 2) \\ & = (x + 2)(x + 3)(x - 2)(x + 1) \\ \\ \text{Since } & P(x) = 0, \\ 0 & = (x + 2)(x + 3)(x - 2)(x + 1) \\ \\ x + 2 = 0 \phantom{000}&\text{or}\phantom{000} x + 3 = 0 \phantom{000}\text{or}\phantom{000} x - 2 = 0 \phantom{000}\text{or} \phantom{000} x + 1 = 0 \\ x = - 2 \phantom{0.}&\phantom{or000-0} x = -3 \phantom{0000o-20.} x = 2 \phantom{0000000or00} x = - 1 \end{align}

(i)

\begin{align} f(-a) & = 2(-a)^3 + a(-a)^2 - 7a^2(-a) - 6a^3 \\ & = -2a^3 + a^3 + 7a^3 - 6a^3 \\ & = 0 \\ \\ \text{By } & \text{Factor theorem, } x + a \text{ is a factor of } f(x). \end{align}

(ii)

$$ \require{enclose} \begin{array}{rll} 2x^2 - ax - 6a^2 \phantom{00000000}\\ x + a \enclose{longdiv}{ 2x^3 + ax^2 - 7a^2 x - 6a^3 \phantom{0}}\kern-.2ex \\ -\underline{( 2x^3 + 2ax^2 ){\phantom{0000000000.}}} \\ -ax^2 - 7a^2x - 6a^3 \phantom{0.} \\ -\underline{( -ax^2 - a^2x ){\phantom{000000.}}} \\ -6a^2 x - 6a^3 \phantom{0.} \\ -\underline{( -6a^2 x - 6a^3 ) \phantom{.}} \\ 0\phantom{00} \end{array} $$ \begin{align} f(x) & = 2x^3 + ax^2 - 7a^2x - 6a^3 \\ \\ \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ f(x) & = (x + a)(2x^2 - ax - 6a^2) + 0 \\ & = (x + a)(2x^2 - ax - 6a^2) \\ & = (x + a)(x - 2a)(2x + 3a) \\ \\ \text{Since } & f(x) = 0, \\ 0 & = (x + a)(x - 2a)(2x + 3a) \\ \\ x + a = 0 \phantom{000}&\text{or}\phantom{000} x - 2a = 0 \phantom{000}\text{or}\phantom{000} 2x + 3a = 0 \\ x = - a \phantom{0.} & \phantom{or000-2a} x = 2a \phantom{0or00000000} 2x = -3a \\ & \phantom{00000000000000000000000000.} x = -{3 \over 2}a \end{align}

(i)

\begin{align} x^2(9 - 2x) & = 2 \\ 9x^2 - 2x^3 & = 2 \\ 0 & = 2x^3 - 9x^2 + 2 \\ \\ \text{Let } & f(x) = 2x^3 - 9x^2 + 2 \\ \\ f\left(1 \over 2\right) & = 2\left(1 \over 2\right)^3 - 9\left(1 \over 2\right)^2 + 2 \\ & = 0 \\ \\ \text{By Factor} & \text{ theorem, } 2x - 1 \text{ is a factor of } f(x). \end{align} $$ \require{enclose} \begin{array}{rll} x^2 - 4x - 2 \phantom{000000}\\ 2x - 1 \enclose{longdiv}{ 2x^3 - 9x^2 + 0x + 2 \phantom{0}}\kern-.2ex \\ -\underline{( 2x^3 - x^2 ){\phantom{00000000.}}} \\ -8x^2 + 0x + 2 \phantom{0.} \\ -\underline{( -8x^2 + 4x ){\phantom{0000}}} \\ -4x + 2 \phantom{0.} \\ -\underline{( -4x + 2 ) \phantom{.}} \\ 0\phantom{0.} \end{array} $$ \begin{align} f(x) & = 2x^3 - 9x^2 + 2 \\ \\ \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ f(x) & = (2x - 1)(x^2 - 4x - 2) + 0 \\ & = (2x - 1)(x^2 - 4x - 2) \\ \\ \therefore 0 & = (2x - 1)(x^2 - 4x - 2) \end{align} \begin{align} 2x - 1 & = 0 \phantom{00}&\text{or}\phantom{000} x^2 - 4x - 2 & = 0 \\ \\ 2x & = 1 & x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ x& = {1 \over 2} & & = {-(-4) \pm \sqrt{(-4)^2 - 4(1)(-2)} \over 2(1)} \\ & & & = {4 \pm \sqrt{24} \over 2} \\ & & & = {4 \pm \sqrt{4} \sqrt{6} \over 2} \\ & & & = {4 \pm 2 \sqrt{6} \over 2} \\ & & & = 2 \pm \sqrt{6} \end{align}

(ii)

\begin{align} x^2(3 + \sqrt{2x})(3 - \sqrt{2x}) & = 2 \\ x^2 \left[ (3)^2 - (\sqrt{2x})^2 \right] & = 2 \\ x^2 (9 - 2x) & = 2 \\ \\ \text{From part } & (i), \\ \therefore x = {1 \over 2}, \phantom{.} 2 + \sqrt{6} , \phantom{.} 2 - \sqrt{6} & \text{ (Reject, since } \sqrt{2x} \text{ is undefined}) \end{align}

\begin{align} \text{Since } y = P(x) \text{ cuts the } x\text{-axis} & \text{ only twice}, \text{ one of the roots is repeated.} \\ \\ \therefore P(x) = a(x + 1)^2(x - 3) \phantom{00} & \text{ OR } \phantom{00} P(x) = a(x + 1)(x - 3)^2 \\ \\ \\ \text{For } P(x) = a(x + 1)^2 & (x - 3), \\ P(1) & = a(1 + 1)^2(1 - 3) \\ 16 & = -8a \\ {16 \over -8} & = a \\ -2 & = a \\ \\ P(x) & = a(x + 1)^2(x - 3) \\ & = -2(x + 1)^2(x - 3) \\ \\ P(-3) & = -2(-3 + 1)^2(-3 - 3) \\ & = 48 \\ \\ \\ \text{For } P(x) = a(x + 1) & (x - 3)^2 , \\ P(1) & = a(1 + 1)(1 - 3)^2 \\ 16 & = 8a \\ {16 \over 8} & = a \\ 2 & = a \\ \\ P(x) & = a(x + 1)(x - 3)^2 \\ & = 2(x + 1)(x - 3)^2 \\ \\ P(-3) & = 2(-3 + 1)(-3 - 3)^2 \\ & = -144 \end{align}

(i)

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

(ii)

\begin{align} P(x) & = 4x^3 + 2ax^2 - 2x^2 + ax - a \\ & = 4x^3 + (2a - 2)x^2 + ax - a \end{align} $$ \require{enclose} \begin{array}{rll} 2x^2 + ax + a \phantom{000000}\\ 2x - 1 \enclose{longdiv}{ 4x^3 + (2a - 2)x^2 + ax - a \phantom{0}}\kern-.2ex \\ -\underline{( 4x^3 - 2x^2 ){\phantom{000000000000}}} \\ 2ax^2 + ax - a \phantom{0.} \\ -\underline{( 2ax^2 - ax ){\phantom{0000}}} \\ 2ax - a \phantom{0.} \\ -\underline{( 2ax - a ) \phantom{.}} \\ 0\phantom{0.} \end{array} $$ \begin{align} P(x) & = 4x^3 + (2a -2)x^2 + ax - a \\ \\ \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ \\ P(x) & = (2x - 1)(2x^2 + ax + a) + 0 \\ & = (2x - 1)(2x^2 + ax + a) \end{align}

(iii)

\begin{align} \text{Since } & P(x) = 0, \\ \\ 0 & = (2x - 1)(2x^2 + ax + a) \\ \\ 2x - 1 & = 0 \phantom{00}\text{or}\phantom{00} 2x^2 + ax + a = 0 \\ 2x & = 1 \\ x & = {1 \over 2} \end{align}
For the equation to have only 1 real root, the equation $2x^2 + ax + a = 0$ must have no real roots

\begin{align} 2x^2 + ax + a & = 0 \\ \\ b^2 - 4ac & < 0 \phantom{000} [\text{No real roots}] \\ (a)^2 - 4(2)(a) & < 0 \\ a^2 - 8a & < 0 \\ a(a - 8) & < 0 \end{align}
$$ 0 < a < 8 $$

\begin{align} 5 - 32x^2 & = 3x(4x^2 + 1) \\ 5 - 32x^2 & = 12x^3 + 3x \\ 0 & = 12x^3 + 32x^2 + 3x - 5 \\ \\ \text{Let } f(x) & = 12x^3 + 32x^2 + 3x - 5 \\ \\ f\left(-{1 \over 2}\right) & = 12\left(-{1 \over 2}\right)^3 + 32\left(-{1 \over 2}\right)^2 + 3\left(-{1 \over 2}\right) - 5 \\ & = 0 \\ \\ \text{By } & \text{Factor theorem, } 2x + 1 \text{ is a factor of } f(x). \end{align} $$ \require{enclose} \begin{array}{rll} 6x^2 + 13x - 5 \phantom{000000}\\ 2x + 1 \enclose{longdiv}{ 12x^3 + 32x^2 + 3x - 5 \phantom{0}}\kern-.2ex \\ -\underline{( 12x^3 + 6x^2 ){\phantom{00000000.}}} \\ 26x^2 + 3x - 5 \phantom{0.} \\ -\underline{( 26x^2 + 13x ){\phantom{000}}} \\ -10x - 5 \phantom{0.} \\ -\underline{( -10x - 5 ) \phantom{.}} \\ 0\phantom{0.} \end{array} $$ \begin{align} f(x) & = 12x^3 + 32x^2 + 3x - 5 \\ \\ \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ f(x) & = (2x + 1)(6x^2 + 13x - 5) + 0 \\ & = (2x + 1)(6x^2 + 13x - 5) \\ & = (2x + 1)(3x - 1)(2x + 5) \\ \\ \therefore 0 & = (2x + 1)(3x - 1)(2x + 5) \\ \\ 2x + 1 & = 0 \phantom{000} \text{or}\phantom{000} 3x - 1 = 0 \phantom{000}\text{or}\phantom{000} 2x + 5 = 0 \\ 2x & = -1 \phantom{000or00001} 3x = 1 \phantom{000or000+5} 2x = -5 \\ x & = -{1 \over 2} \phantom{000or00001} x = {1 \over 3} \phantom{000or000+5} x = -{5 \over 2} \end{align}

(i)

\begin{align} \text{Let } PR & = y \text{ cm} \\ \\ PQ & = 21 - y \phantom{000000} [\text{folded edge} = PR = y]\\ \\ A(x) & = {1 \over 2}x(21 - y) \\ & = {1 \over 2}(21x - xy) \\ \\ \text{By Pyth} & \text{agoras theorem,} \\ PR^2 & = QR^2 + PQ^2 \\ y^2 & = x^2 + (21 - y)^2 \\ y^2 & = x^2 + 21^2 - 2(21)(y) + y^2 \\ y^2 & = x^2 + 441 - 42y + y^2 \\ 0 & = x^2 + 441 - 42y \\ 42y & = x^2 + 441 \\ y & = {1 \over 42}(x^2 + 441) \\ \\ \therefore A(x) & = {1 \over 2} \left[ 21 x - {1 \over 42}x(x^2 + 441) \right] \\ & = {1 \over 84} [882x - x(x^2 + 441)] \\ & = {1 \over 84} (882x - x^3 - 441x) \\ & = {1 \over 84} (441x - x^3) \\ & = {441x - x^3 \over 84} \\ & = {x(441 - x^2) \over 84} \phantom{00} \text{ (Shown)} \end{align}

(ii)

\begin{align} \text{If } QR = 21 \text{ cm, the entire width of the paper will be folded and there will be no triangle } PQR \end{align}

(iii)

\begin{align} 15{3 \over 7} & = {x(441 - x^2) \over 84} \\ 84\left( 15{3 \over 7} \right) & = x(441 - x^2) \\ 1296 & = 441x - x^3 \\ 0 & = - x^3 + 441x - 1296 \\ 0 & = x^3 - 441x + 1296 \\ \\ \text{Let } & f(x) = x^3 - 441x + 1296 \\ \\ f(3) & = (3)^3 - 441(3) + 1296 \\ & = 0 \\ \\ \text{By Factor theorem, } & x - 3 \text{ is a factor of } f(x) \\ \\ x^3 - 441x + 1296 & = (x - 3)(x^2 + ax + b) \\ x^3 - 441x + 1296 & = x^3 + ax^2 + bx - 3x^2 - 3ax - 3b \\ x^3 - 441x + 1296 & = x^3 + ax^2 - 3x^2 + bx - 3 ax - 3b \\ x^3 + 0x^2 - 441x + 1296& = x^3 + (a - 3)x^2 + (b - 3a)x - 3b \\ \\ \text{Compare } & \text{coefficients of } x^2, \\ 0 & = a - 3 \\ 3 & = a \\ \\ \text{Compare } & \text{constants,} \\ 1296 & = -3b \\ {1296 \over -3} & = b \\ -432 & = b \\ \\ 0 & = (x - 3)(x^2 + 3x - 432) \end{align} \begin{align} x - 3 & = 0 && \text{ or } & x^2 + 3x - 432 & = 0 \\ x & = 3 &&& x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ &&&& & = {-3 \pm \sqrt{(3)^2 - 4(1)(-432)} \over 2(1)} \\ &&&& & = 19.338 \text{ or } -22.338 \text{ (Reject, since } x > 0 ) \\ &&&& & \approx 19.3 \end{align}

(iv)

\begin{align} \text{Max. area} & = 42.435 \text{ cm}^2 \end{align}