$$ \require{enclose} \begin{array}{rll} 3x - 1 \phantom{000000000.}\\ x^2 + 0x - 1 \enclose{longdiv}{3x^3 - x^2 + 0x + 3\phantom{0}}\kern-.2ex \\ -\underline{(3x^3 + 0x^2 - 3x){\phantom{000}}} \\ -x^2 + 3x + 3\phantom{0} \\ -\underline{(-x^2 - 0x + 1){\phantom{}}} \\ 3x + 2\phantom{0} \end{array} $$
\begin{align} 3x^3 - x^2 + 3 & = (3x - 1)(x^2 - 1) + 3x + 2 \end{align}

\begin{align*} \text{Let } g(x) & = x^3 - 2x^2 + 3x + 6 \\ \\ g(-1) & = (-1)^3 - 2(-1)^2 +3(-1) + 6 \phantom{000000} \left[ x + 1 = 0 \implies x = -1 \right]\\ & = -1 - 2 - 3 + 6 \\ & = 0 \\ \\ \text{By Factor} & \text{ theorem, } (x + 1) \text{ is a factor of } g(x) \end{align*}

\begin{align*} \text{Let } g(x) & = x^3 - 2x^2 - x + 7 \\ \\ g\left(1 \over 2\right) & = \left(1 \over 2\right)^3 - 2\left(1 \over 2\right)^2 - \left(1 \over 2\right) + 7 \phantom{000000} \left[ 2x - 1 = 0 \implies x = {1 \over 2} \right]\\ & = 6.125 \\ \\ \therefore \text{Remainder} & = 6.125 \end{align*}

\begin{align} \text{Let } f(x) & = x^3 - 6x^2 + 11x - 6 \\ \\ f(1) & = (1)^3 - 6(1)^2 + 11(1) - 6 \\ & = 1 - 6 + 11 - 6 \\ & = 0 \\ \\ \text{By Factor} & \text{ theorem, } x - 1 \text{ is a factor of } f(x) \end{align}
\begin{align} \text{Polynomial} & = \text{Quotient} \times \text{Divisor} + \text{Remainder} \\ f(x) = x^3 - 6x^2 + 11x - 6 & = Q(x) \times (x - 1) + 0 \\ & = Q(x) \times (x - 1) \\ \\ [\text{Since degree of } f(x) \text{ is 3 and} & \text{ degree of (x - 1) is 1}, \text{ degree of } Q(x) \text{ is } 2] \\ \\ \text{Let } Q(x) = ax^2 + bx + c, & \text{ where } a, b \text{ and } c \text{ are real constants} \\ \\ x^3 - 6x^2 + 11x - 6 & = (ax^2 + bx + c)(x - 1) \\ \\ \text{By observation, } a & = 1 \\ \\ x^3 - 6x^2 + 11x - 6 & = (x^2 + bx + c)(x - 1) \\ & = x^3 - x^2 + bx^2 - bx + cx - c \\ & = x^3 + (b - 1)x^2 + (c - b)x - c \\ \\ \text{Comparing } & \text{coefficients of } x^2, \\ -6 & = b - 1 \\ -6 + 1 & = b \\ -5 & = b \\ \\ \text{Comparing } & \text{the constants,} \\ -6 & = -c \\ 6 & = c \\ \\ \therefore x^3 - 6x^2 + 11x - 6 & = (x^2 - 5x + 6)(x - 1) \\ & = (x - 2)(x - 3)(x - 1) \end{align}

\begin{align} f(x) & = x^3 + ax^2 + bx + 12 \\ \\ f(-3) & = (-3)^3 + a(-3)^2 + b(-3) + 12 \\ & = -27 + a(9) - 3b + 12 \\ & = 9a - 3b - 15 \\ \\ 0 & = 9a - 3b - 15 \phantom{000000} [x + 3 \text{ is a factor}] \\ 3b & = 9a - 15 \\ b & = 3a - 5 \phantom{00} \text{--- (1)} \\ \\ \\ f(1) & = (1)^3 + a(1)^2 + b(1) + 12 \\ & = 1 + a(1) + b + 12 \\ & = 1 + a + b + 12 \\ & = a + b + 13 \\ \\ 36 & = a + b + 13 \\ 23 & = a + b \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 23 & = a + 3a - 5 \\ 23 & = 4a - 5 \\ 28 & = 4a \\ {28 \over 4} & = a \\ 7 & = a \\ \\ \text{Substitute } & \text{into (1),} \\ b & = 3(7) - 5 \\ b & = 16 \\ \\ \\ \therefore a & = 7, b = 16 \end{align}

\begin{align*} f(1) & = (2p + 1)(1)^2 + p(1) + 2p^2 \\ 0 & = (2p + 1)(1) + p + 2p^2 \\ 0 & = 2p + 1 + p + 2p^2 \\ 0 & = 2p^2 + 3p + 1 \\ 0 & = (2p + 1)(p + 1) \\ \\ 2p + 1 = 0 \phantom{000}&\text{or}\phantom{000} p + 1 = 0 \\ 2p = -1 \phantom{0.} & \phantom{or000+1} p = - 1 \\ p = -{1 \over 2} \phantom{0} & \\ \\ \\ f(-2) & = (2p + 1)(-2)^2 + p(-2) + 2p^2 \\ 0 & = (2p + 1)(4) - 2p + 2p^2 \\ 0 & = 8p + 4 - 2p + 2p^2 \\ 0 & = 2p^2 + 6p + 4 \\ 0 & = p^2 + 3p + 2 \\ 0 & = (p + 2)(p + 1) \\ \\ p + 2 = 0 \phantom{000}&\text{or}\phantom{000} p + 1 = 0 \\ p = - 2 \phantom{0.}&\phantom{or000+1} p = -1 \\ \\ \\ \text{When } p = -1 \text{ or } -{1 \over 2}, & \phantom{0} f(x) \text{ has a factor of } x - 1. \\ \\ \text{When } p = -2 \text{ or } -1, & \phantom{0} f(x) \text{ has a factor of } x + 2. \\ \\ \\ \therefore \text{For } f(x) \text{ to have a factor of } & x + 2 \text{ but not } x - 1, \phantom{0} p = -2 \end{align*}

\begin{align*} 54x^3 - 2y^6 & = 2(27x^3 - y^6) \\ & = 2[(3x)^3 - (y^2)^3] \\ & = 2(3x - y^2)[(3x)^2 + (3x)(y^2) + (y^2)^2] \phantom{000000} [a^3 - b^3 = (a - b)(a^2 + ab + b^2) ] \\ & = 2(3x - y^2)(9x^2 + 3x y^2 + y^4) \end{align*}

\begin{align} x^3 + x + 1 & = 3x^2 \\ x^3 - 3x^2 + x + 1 & =0 \\ \\ \text{Let } f(x) & = x^3 - 3x^2 + x + 1 \\ \\ f(1) & = (1)^3 - 3(1)^2 + (1) + 1 \\ & = 0 \\ \\ \therefore x - 1 & \text{ is a factor of } f(x) \end{align}
$$ \require{enclose} \begin{array}{rll} x^2 - 2x - 1 \phantom{00000}\\ x - 1 \enclose{longdiv}{ x^3 - 3x^2 + x + 1 \phantom{0}}\kern-.2ex \\ -\underline{( x^3 - x^2 ){\phantom{00000000}}} \\ -2x^2 + x + 1 \phantom{0.} \\ -\underline{( -2x^2 + 2x ){\phantom{000}}} \\ -x + 1 \phantom{0.} \\ -\underline{( -x + 1 ) \phantom{.}} \\ 0\phantom{0.} \end{array} $$
\begin{align} x^3 - 3x^2 + x + 1 & = (x - 1)(x^2 - 2x - 1) \\ \\ 0 & = (x - 1)(x^2 - 2x - 1) \end{align} \begin{align} x - 1 & = 0 && \text{ or } & x^2 & - 2x - 1 = 0 \\ x & = 1 &&& x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & &&& & = {-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)} \over 2(1)} \\ & &&& & = {2 \pm \sqrt{8} \over 2} \\ & &&& & = {2 \pm \sqrt{4} \sqrt{2} \over 2} \\ & &&& & = {2 \pm 2 \sqrt{2} \over 2} \\ & &&& & = 1 \pm \sqrt{2} \end{align}

\begin{align} \text{Let } f(x) & = 2x^3 - 5x^2 + 7x + 5 \\ \\ f \left(-{1 \over 2}\right) & = 2\left(-{1 \over 2}\right)^3 - 5\left(-{1 \over 2}\right)^2 + 7\left(-{1 \over 2}\right) + 5 \\ & = 0 \\ \\ \therefore 2x + 1 & \text{ is a factor of } f(x) \end{align}
$$ \require{enclose} \begin{array}{rll} x^2 - 3x + 5 \phantom{000000}\\ 2x + 1 \enclose{longdiv}{ 2x^3 - 5x^2 + 7x + 5 \phantom{0}}\kern-.2ex \\ -\underline{( 2x^3 + x^2 ){\phantom{000000000}}} \\ - 6x^2 + 7x + 5 \phantom{0.} \\ -\underline{( -6x^2 - 3x ){\phantom{0000}}} \\ 10x + 5 \phantom{0.} \\ -\underline{( 10x + 5 ) \phantom{.}} \\ 0\phantom{0.} \end{array} $$
\begin{align} 2x^3 - 5x^2 + 7x + 5 & = (2x + 1)(x^2 - 3x + 5) \\ \\ (2x + 1)(x^2 - 3x + 5) & = 0 \end{align} \begin{align} 2x + 1 & = 0 && \text{ or } & x^2 & - 3x + 5 = 0 \\ 2x & = -1 &&& x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ x & = -{1 \over 2} &&& & = {-(-3) \pm \sqrt{(-3)^2 - 4(1)(5)} \over 2(1)} \\ & &&& & = {3 \pm \sqrt{-11} \over 2} \phantom{0} \text{ (No real roots)} \end{align}
$$ \therefore x = -{1 \over 2} \text{ is the only root of the equation} $$

\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} & \text{ theorem, } x - 2 \text{ is a factor of } f(x) \\ \\ \text{Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder} \\ x^3 + 2x^2 - 4x - 8 & = (x - 2)Q(x) + 0 \\ & = (x - 2)Q(x) \\ \\ \text{Let } Q(x) & = ax^2 + bx + c , \text{ where } a, b \text{ and } c \text{ are real numbers} \\ \\ x^3 + 2x^2 - 4x - 8 & = (x - 2)(ax^2 + bx + c) \\ \\ \text{By obser} & \text{vation, } a = 1 \\ \\ x^3 + 2x^2 - 4x - 8 & = (x - 2)(x^2 + bx + c) \\ & = x^3 + bx^2 + cx - 2x^2 - 2bx - 2c \\ & = x^3 + (b - 2)x^2 + (c - 2b)x - 2c \\ \\ \text{Comparing } & \text{coefficients of } x^2, \\ 2 & = b - 2 \\ 4 & = b \\ \\ \text{Comparing } & \text{the constant term,} \\ -8 & = -2c \\ 4 & = c \\ \\ x^3 + 2x^2 - 4x - 8 & = (x - 2)(x^2 + 4x + 4) \\ & = (x - 2)(x + 2)(x + 2) \\ & = (x - 2)(x + 2)^2 \\ \\ \text{Let } (x - 2)& (x + 2)^2 = 0 , \end{align*} \begin{align*} x - 2 & = 0 && \text{ or } & x + 2 & = 0 \\ x & = 2 &&& x & = -2 \end{align*}

\begin{align} \text{From (a), } x^3 + 2x^2 - 4x - 8 & = 0 \text{ and } x = 2 \text{ or } - 2 \\ \\ \text{Replace } x \text{ by } x^2, & \\ (x^2)^3 + 2(x^2)^2 - 4x^2 - 8 & = 0 \\ x^6 + 2x^4 - 4x^2 - 8 \end{align} \begin{align} \therefore x^2 & = 2 && \text{ or } & x^2 & = -2 \text{ (Reject, since } x^2 \ge 0 ) \\ x & = \pm \sqrt{2} \end{align}

\begin{align} \text{From (a), } x^3 + 2x^2 - 4x - 8 & = 0 \text{ and } x = 2 \text{ or } - 2 \\ \\ \text{Replace } x \text{ by } x + 1, & \\ (x + 1)^3 + 2(x + 1)^2 - 4(x + 1) - 8 & = 0 \\ (x + 1)^3 + 2(x + 1)^2 - 4x - 4 - 8 & = 0 \\ (x + 1)^3 + 2(x + 1)^2 - 4x - 12 & = 0 \end{align} \begin{align} \therefore x + 1 & = 2 && \text{ or } & x + 1 & = -2 \\ x & = 1 &&& x & = -3 \end{align}

\begin{align} \text{From (a), } x^3 + 2x^2 - 4x - 8 & = 0 \text{ and } x = 2 \text{ or } - 2 \\ \\ \text{Replace } x \text{ by } {1 \over x}, & \\ \left(1 \over x\right)^3+ 2\left(1 \over x\right)^2 - 4\left(1 \over x\right) - 8 & = 0 \\ {1 \over x^3} + 2\left(1 \over x^2\right) - {4 \over x} - 8 & = 0 \\ {1 \over x^3} + {2 \over x^2} - {4 \over x} - 8 & = 0 \\ x^3 \left( {1 \over x^3} + {2 \over x^2} - {4 \over x} - 8 \right) & = x^3 (0) \\ 1 + 2x - 4x^2 - 8x^3 & = 0 \\ -8x^3 - 4x^2 + 2x + 1 & = 0 \end{align} \begin{align} \therefore {1 \over x} & = 2 && \text{ or } & {1 \over x} & = -2 \\ 1 & = 2x &&& 1 & = -2x \\ {1 \over 2} & = x &&& - {1 \over 2} & = x \end{align}

\begin{align} x & = -1 && \text{ or } & x & = 3 && \text{ or } & x & = 4 \\ x + 1 & = 0 &&& x - 3 & = 0 &&& x - 4 & = 0 \end{align}
\begin{align} f(x) & = a(x + 1)(x - 3)(x - 4), \text{ where } a \text{ is a constant} \\ \\ f(2) & = a(2 + 1)(2 - 3)(2 - 4) \\ -48 & = a(3)(-1)(-2) \\ -48 & = 6a \\ {-48 \over 6} & = a \\ -8 & = a \\ \\ f(x) & = -8(x + 1)(x - 3)(x - 4) \\ \\ \\ f(-2) & = -8(-2+1)(-2-3)(-2-4) \\ & = 240 \\ \\ \text{Remainder} & = 240 \end{align}

\begin{align} x & = 1 && \text{ or } & x & = -2 \\ x - 1 & = 0 &&& x + 2 & = 0 \end{align}
\begin{align} (x - 1)(x + 2)(x^2 - 5x + 1) & = (x^2 + 2x - x - 2)(x^2 - 5x + 1) \\ & = (x^2 + x - 2)(x^2 - 5x + 1) \\ & = x^4 - 5x^3 + x^2 + x^3 - 5x^2 + x - 2x^2 + 10x - 2 \\ & = x^4 - 4x^3 - 6x^2 + 11x - 2 \\ \\ \\ f(x) & = 2(x^4 - 4x^3 - 6x^2 + 11x - 2) \phantom{000000} [\text{Coefficient of } x^4 = 2] \\ & = 2x^4 - 8x^3 - 12x^2 + 22x - 4 \end{align}